# Heat Engines, Entropy, and the Second Law of Thermodynamics

```chapter
22
Heat Engines, Entropy,
and the Second Law
of Thermodynamics
22.1 Heat Engines and the Second Law
of Thermodynamics
22.2 Heat Pumps and Refrigerators
22.3 Reversible and Irreversible Processes
22.4 The Carnot Engine
22.5 Gasoline and Diesel Engines
22.6 Entropy
22.7 Entropy and the Second Law
22.8 Entropy on a Microscopic Scale
The first law of thermodynamics, which
we studied in Chapter 20, is a statement of
conservation of energy and is a special-case
reduction of Equation 8.2. This law states
that a change in internal energy in a system
can occur as a result of energy transfer by
heat, by work, or by both. Although the first
law of thermodynamics is very important,
it makes no distinction between processes
that occur spontaneously and those that do
not. Only certain types of energy conversion
and energy transfer processes actually take
place in nature, however. The second law
of thermodynamics, the major topic in this
chapter, establishes which processes do and
A Stirling engine from the early nineteenth century. Air is heated in the lower
cylinder using an external source. As this happens, the air expands and pushes
against a piston, causing it to move. The air is then cooled, allowing the cycle to
begin again. This is one example of a heat engine, which we study in this chapter.
do not occur. The following are examples of
625
27819_22_c22_p625-656.indd 625
6/30/09 12:42:49 PM
626
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
J-L Charmet/Science Photo Library/Photo Researchers, Inc.
processes that do not violate the first law of thermodynamics if they proceed in either
direction, but are observed in reality to proceed in only one direction:
• When two objects at different temperatures are placed in thermal contact with each
other, the net transfer of energy by heat is always from the warmer object to the
cooler object, never from the cooler to the warmer.
• A rubber ball dropped to the ground bounces several times and eventually comes to
rest, but a ball lying on the ground never gathers internal energy from the ground and
begins bouncing on its own.
Lord Kelvin
British physicist and mathematician
(1824–1907)
Born William Thomson in Belfast, Kelvin was
the first to propose the use of an absolute
scale of temperature. The Kelvin temperature
scale is named in his honor. Kelvin’s work in
thermodynamics led to the idea that energy
cannot pass spontaneously from a colder
object to a hotter object.
• An oscillating pendulum eventually comes to rest because of collisions with air molecules and friction at the point of suspension. The mechanical energy of the system is
converted to internal energy in the air, the pendulum, and the suspension; the reverse
conversion of energy never occurs.
All these processes are irreversible; that is, they are processes that occur naturally in
one direction only. No irreversible process has ever been observed to run backward. If it
were to do so, it would violate the second law of thermodynamics.1
22.1 Heat Engines and the Second Law
of Thermodynamics
Figure 22.1 A steam-driven locomotive obtains its energy by burning
wood or coal. The generated energy
vaporizes water into steam, which
powers the locomotive. Modern
locomotives use diesel fuel instead of
wood or coal. Whether old-fashioned
or modern, such locomotives can
be modeled as heat engines, which
extract energy from a burning
fuel and convert a fraction of it to
mechanical energy.
27819_22_c22_p625-656.indd 626
A heat engine is a device that takes in energy by heat2 and, operating in a cyclic
process, expels a fraction of that energy by means of work. For instance, in a typical
process by which a power plant produces electricity, a fuel such as coal is burned
and the high-temperature gases produced are used to convert liquid water to
steam. This steam is directed at the blades of a turbine, setting it into rotation. The
mechanical energy associated with this rotation is used to drive an electric generator. Another device that can be modeled as a heat engine is the internal combustion
engine in an automobile. This device uses energy from a burning fuel to perform
work on pistons that results in the motion of the automobile.
A heat engine carries some working substance through a cyclic process during
which (1) the working substance absorbs energy by heat from a high-temperature
energy reservoir, (2) work is done by the engine, and (3) energy is expelled by heat
to a lower-temperature reservoir. As an example, consider the operation of a steam
engine (Fig. 22.1), which uses water as the working substance. The water in a boiler
absorbs energy from burning fuel and evaporates to steam, which then does work
by expanding against a piston. After the steam cools and condenses, the liquid
water produced returns to the boiler and the cycle repeats.
It is useful to represent a heat engine schematically as in Active Figure 22.2. The
engine absorbs a quantity of energy |Q h| from the hot reservoir. For the mathematical discussion of heat engines, we use absolute values to make all energy transfers
1Although a process occurring in the time-reversed sense has never been observed, it is possible for it to occur. As we
shall see later in this chapter, however, the probability of such a process occurring is infinitesimally small. From this
viewpoint, processes occur with a vastly greater probability in one direction than in the opposite direction.
2We
use heat as our model for energy transfer into a heat engine. Other methods of energy transfer are possible in
the model of a heat engine, however. For example, the Earth’s atmosphere can be modeled as a heat engine in which
the input energy transfer is by means of electromagnetic radiation from the Sun. The output of the atmospheric heat
engine causes the wind structure in the atmosphere.
6/30/09 12:42:56 PM
22.1 | Heat Engines and the Second Law of Thermodynamics
627
by heat positive, and the direction of transfer is indicated with an explicit positive or
negative sign. The engine does work Weng (so that negative work W 5 2Weng is done
on the engine) and then gives up a quantity of energy |Q c | to the cold reservoir.
Because the working substance goes through a cycle, its initial and final internal
energies are equal: DE int 5 0. Hence, from the first law of thermodynamics, DE int 5
Q 1 W 5 Q 2 Weng 5 0, and the net work Weng done by a heat engine is equal to
the net energy Q net transferred to it. As you can see from Active Figure 22.2, Q net 5
|Q h| 2 |Q c |; therefore,
Weng 5 |Q h| 2 |Q c |
(22.1)
The thermal efficiency e of a heat engine is defined as the ratio of the net work
done by the engine during one cycle to the energy input at the higher temperature
during the cycle:
e;
Weng
0Qh0
5
0Qh0 2 0Qc0
0Qc0
512
0Qh0
0Qh0
(22.2)
You can think of the efficiency as the ratio of what you gain (work) to what you give
(energy transfer at the higher temperature). In practice, all heat engines expel only
a fraction of the input energy Q h by mechanical work; consequently, their efficiency
is always less than 100%. For example, a good automobile engine has an efficiency
of about 20%, and diesel engines have efficiencies ranging from 35% to 40%.
Equation 22.2 shows that a heat engine has 100% efficiency (e 5 1) only if |Q c | 5
0, that is, if no energy is expelled to the cold reservoir. In other words, a heat engine
with perfect efficiency would have to expel all the input energy by work. Because
efficiencies of real engines are well below 100%, the Kelvin–Planck form of the
second law of thermodynamics states the following:
It is impossible to construct a heat engine that, operating in a cycle, produces
no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work.
This statement of the second law means that during the operation of a heat engine,
Weng can never be equal to |Q h| or, alternatively, that some energy |Q c | must be
rejected to the environment. Every heat engine must have some energy exhaust.
Figure 22.3 is a schematic diagram of the impossible “perfect” heat engine.
W Thermal efficiency of a heat
engine
The engine does
work Weng.
Energy Q h
enters the
engine.
Hot reservoir
att Th
Qh
Weng
Heat
engine
Energy Q c leaves the
engine.
Qc
Cold reservoir
at Tc
ACTIVE FIGURE 22.2
Schematic representation of a heat
engine.
Quick Quiz 22.1 The energy input to an engine is 3.00 times greater than the
work it performs. (i) What is its thermal efficiency? (a) 3.00 (b) 1.00 (c) 0.333
(d) impossible to determine (ii) What fraction of the energy input is expelled
to the cold reservoir? (a) 0.333 (b) 0.667 (c) 1.00 (d) impossible to determine
An impossible heat engine
Hot reservoir
att Th
Qh
Heat
engine
Cold reservoir
at Tc
27819_22_c22_p625-656.indd 627
Weng
Figure 22.3 Schematic diagram of
a heat engine that takes in energy
from a hot reservoir and does an
equivalent amount of work. It is
impossible to construct such a perfect engine.
Pitfall Prevention 22.1
The First and Second Laws
Notice the distinction between the
first and second laws of thermodynamics. If a gas undergoes a one-time
isothermal process, then DE int 5 Q 1
W 5 0 and W 5 2Q. Therefore, the
first law allows all energy input by
heat to be expelled by work. In a
heat engine, however, in which a
substance undergoes a cyclic process,
only a portion of the energy input by
heat can be expelled by work according to the second law.
6/30/09 12:42:58 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
628
The Efficiency of an Engine
Ex a m pl e 22.1
An engine transfers 2.00 3 103 J of energy from a hot reservoir during a cycle and transfers 1.50 3 103 J as exhaust to a
cold reservoir.
(A) Find the efficiency of the engine.
SOLUTION
Conceptualize Review Active Figure 22.2; think about energy going into the engine from the hot reservoir and splitting,
with part coming out by work and part by heat into the cold reservoir.
Categorize This example involves evaluation of quantities from the equations introduced in this section, so we categorize it as a substitution problem.
Find the efficiency of the engine from Equation 22.2:
e512
0Qc0
0Qh0
512
1.50 3 103 J
2.00 3 103 J
5 0.250, or 25.0%
(B) How much work does this engine do in one cycle?
SOLUTION
Find the work done by the engine by taking the difference between the input and output energies:
Weng 5 |Q h | 2 |Q c | 5 2.00 3 103 J 2 1.50 3 103 J
5 5.0 3 102 J
WHAT IF? Suppose you were asked for the power output of this engine. Do you have sufficient information to answer
this question?
Answer No, you do not have enough information. The power of an engine is the rate at which work is done by the engine.
You know how much work is done per cycle, but you have no information about the time interval associated with one
cycle. If you were told that the engine operates at 2 000 rpm (revolutions per minute), however, you could relate this rate
to the period of rotation T of the mechanism of the engine. Assuming there is one thermodynamic cycle per revolution,
the power is
P5
Weng
T
5
5.0 3 102 J 1 min
a
b 5 1.7 3 104 W
1
1 2 000
min 2 60 s
22.2 Heat Pumps and Refrigerators
Work W is done on
the heat pump.
Energy Q h
is expelled
to the hot
reservoir.
Energy Q c is drawn
from the
cold
reservoir.
Hot reservoir
at Th
Qh
Heat
pump
W
Qc
Cold reservoir
at Tc
ACTIVE FIGURE 22.4
Schematic representation of a heat
pump.
27819_22_c22_p625-656.indd 628
In a heat engine, the direction of energy transfer is from the hot reservoir to the
cold reservoir, which is the natural direction. The role of the heat engine is to process the energy from the hot reservoir so as to do useful work. What if we wanted to
transfer energy from the cold reservoir to the hot reservoir? Because that is not the
natural direction of energy transfer, we must put some energy into a device to be
successful. Devices that perform this task are called heat pumps and refrigerators.
For example, homes in summer are cooled using heat pumps called air conditioners.
The air conditioner transfers energy from the cool room in the home to the warm
air outside.
In a refrigerator or a heat pump, the engine takes in energy |Q c | from a cold
reservoir and expels energy |Q h| to a hot reservoir (Active Fig. 22.4), which can be
accomplished only if work is done on the engine. From the first law, we know that
the energy given up to the hot reservoir must equal the sum of the work done and
the energy taken in from the cold reservoir. Therefore, the refrigerator or heat
pump transfers energy from a colder body (for example, the contents of a kitchen
refrigerator or the winter air outside a building) to a hotter body (the air in the
kitchen or a room in the building). In practice, it is desirable to carry out this process with a minimum of work. If the process could be accomplished without doing
any work, the refrigerator or heat pump would be “perfect” (Fig. 22.5). Again, the
6/30/09 12:43:01 PM
22.2 | Heat Pumps and Refrigerators
existence of such a device would be in violation of the second law of thermodynamics, which in the form of the Clausius statement 3 states:
It is impossible to construct a cyclical machine whose sole effect is to transfer
energy continuously by heat from one object to another object at a higher
temperature without the input of energy by work.
In simpler terms, energy does not transfer spontaneously by heat from a cold object
to a hot object. Work input is required to run a refrigerator.
The Clausius and Kelvin–Planck statements of the second law of thermodynamics appear at first sight to be unrelated, but in fact they are equivalent in all respects.
Although we do not prove so here, if either statement is false, so is the other.4
In practice, a heat pump includes a circulating fluid that passes through two sets
of metal coils that can exchange energy with the surroundings. The fluid is cold
and at low pressure when it is in the coils located in a cool environment, where it
absorbs energy by heat. The resulting warm fluid is then compressed and enters
the other coils as a hot, high-pressure fluid. There it releases its stored energy to
the warm surroundings. In an air conditioner, energy is absorbed into the fluid in
coils located in a building’s interior; after the fluid is compressed, energy leaves the
fluid through coils located outdoors. In a refrigerator, the external coils are behind
or underneath the unit (Fig. 22.6). The internal coils are in the walls of the refrigerator and absorb energy from the food.
The effectiveness of a heat pump is described in terms of a number called the
coefficient of performance (COP). The COP is similar to the thermal efficiency
for a heat engine in that it is a ratio of what you gain (energy transferred to or from
a reservoir) to what you give (work input). For a heat pump operating in the cooling mode, “what you gain” is energy removed from the cold reservoir. The most
effective refrigerator or air conditioner is one that removes the greatest amount of
energy from the cold reservoir in exchange for the least amount of work. Therefore, for these devices operating in the cooling mode, we define the COP in terms
of |Q c |:
COP 1 cooling mode 2 5
energy transferred at low temperature
work done on heat pump
5
0Qc0
W
(22.3)
629
An impossible heat pump
Hot reservoir
at Th
Qh Qc
Heat
pump
Qc
Cold reservoir
at Tc
Figure 22.5 Schematic diagram of
an impossible heat pump or refrigerator, that is, one that takes in energy
from a cold reservoir and expels an
equivalent amount of energy to a hot
reservoir without the input of energy
by work.
The coils on the back of
a refrigerator transfer
energy by heat to the air.
A good refrigerator should have a high COP, typically 5 or 6.
In addition to cooling applications, heat pumps are becoming increasingly popular for heating purposes. The energy-absorbing coils for a heat pump are located
outside a building, in contact with the air or buried in the ground. The other set of
coils are in the building’s interior. The circulating fluid flowing through the coils
absorbs energy from the outside and releases it to the interior of the building from
the interior coils.
In the heating mode, the COP of a heat pump is defined as the ratio of the energy
transferred to the hot reservoir to the work required to transfer that energy:
energy transferred at high temperature
work done on heat pump
5
0 Qh 0
W
(22.4)
COP 1 heating mode 2 5
If the outside temperature is 25°F (24°C) or higher, a typical value of the COP
for a heat pump is about 4. That is, the amount of energy transferred to the building
is about four times greater than the work done by the motor in the heat pump. As
the outside temperature decreases, however, it becomes more difficult for the heat
pump to extract sufficient energy from the air and so the COP decreases. Therefore, the use of heat pumps that extract energy from the air, although satisfactory in
Figure 22.6 The back of a house3First
4 See
expressed by Rudolf Clausius (1822–1888).
an advanced textbook on thermodynamics for this proof.
27819_22_c22_p625-656.indd 629
hold refrigerator. The air surrounding the coils is the hot reservoir.
6/30/09 12:43:03 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
630
moderate climates, is not appropriate in areas where winter temperatures are very
low. It is possible to use heat pumps in colder areas by burying the external coils
deep in the ground. In that case, the energy is extracted from the ground, which
tends to be warmer than the air in the winter.
Quick Quiz 22.2 The energy entering an electric heater by electrical transmission can be converted to internal energy with an efficiency of 100%. By
what factor does the cost of heating your home change when you replace
your electric heating system with an electric heat pump that has a COP of
4.00? Assume the motor running the heat pump is 100% efficient. (a) 4.00
(b) 2.00 (c) 0.500 (d) 0.250
Ex a m pl e 22.2
Freezing Water
A certain refrigerator has a COP of 5.00. When the refrigerator is running, its power input is 500 W. A sample of water of
mass 500 g and temperature 20.0°C is placed in the freezer compartment. How long does it take to freeze the water to ice
at 0°C? Assume all other parts of the refrigerator stay at the same temperature and there is no leakage of energy from the
exterior, so the operation of the refrigerator results only in energy being extracted from the water.
SOLUTION
Conceptualize Energy leaves the water, reducing its temperature and then freezing it into ice. The time interval required
for this entire process is related to the rate at which energy is withdrawn from the water, which, in turn, is related to the
power input of the refrigerator.
Categorize We categorize this example as one that combines our understanding of temperature changes and phase
changes from Chapter 20 and our understanding of heat pumps from this chapter.
W
Dt
Analyze Use the power rating of the refrigerator
to find the time interval Dt required for the freezing process to occur:
P5
Use Equation 22.3 to relate the work W done on
the heat pump to the energy |Q c | extracted from
the water:
Dt 5
Use Equations 20.4 and 20.7 to substitute the
amount of energy |Q c | that must be extracted
from the water of mass m:
Dt 5
Recognize that the amount of water that freezes
is Dm 5 2m because all the water freezes:
Dt 5
Subsitute numerical values:
Dt 5
S
Dt 5
W
P
|Q c |
P 1 COP 2
0 mc DT 1 L f Dm 0
P 1 COP 2
0 m 1 c DT 2 L f 2 0
P 1 COP 2
0 1 0.500 kg 2 3 1 4 186 J/kg ? °C 2 1 220.0°C 2 2 3.33 3 105 J/kg 4 0
1 500 W 2 1 5.00 2
5 83.3 s
Finalize In reality, the time interval for the water to freeze in a refrigerator is much longer than 83.3 s, which suggests
that the assumptions of our model are not valid. Only a small part of the energy extracted from the refrigerator interior
in a given time interval comes from the water. Energy must also be extracted from the container in which the water is
placed, and energy that continuously leaks into the interior from the exterior must be extracted.
27819_22_c22_p625-656.indd 630
6/30/09 12:43:06 PM
22.3 | Reversible and Irreversible Processes
631
22.3 Reversible and Irreversible Processes
In the next section, we will discuss a theoretical heat engine that is the most efficient possible. To understand its nature, we must first examine the meaning of
reversible and irreversible processes. In a reversible process, the system undergoing the process can be returned to its initial conditions along the same path on a
PV diagram, and every point along this path is an equilibrium state. A process that
does not satisfy these requirements is irreversible.
All natural processes are known to be irreversible. Let’s examine the adiabatic
free expansion of a gas, which was already discussed in Section 20.6, and show that
it cannot be reversible. Consider a gas in a thermally insulated container as shown
in Figure 22.7. A membrane separates the gas from a vacuum. When the membrane
is punctured, the gas expands freely into the vacuum. As a result of the puncture,
the system has changed because it occupies a greater volume after the expansion.
Because the gas does not exert a force through a displacement, it does no work on
the surroundings as it expands. In addition, no energy is transferred to or from the
gas by heat because the container is insulated from its surroundings. Therefore, in
this adiabatic process, the system has changed but the surroundings have not.
For this process to be reversible, we must return the gas to its original volume
and temperature without changing the surroundings. Imagine trying to reverse the
process by compressing the gas to its original volume. To do so, we fit the container
with a piston and use an engine to force the piston inward. During this process, the
surroundings change because work is being done by an outside agent on the system.
In addition, the system changes because the compression increases the temperature of the gas. The temperature of the gas can be lowered by allowing it to come
into contact with an external energy reservoir. Although this step returns the gas to
its original conditions, the surroundings are again affected because energy is being
added to the surroundings from the gas. If this energy could be used to drive the
engine that compressed the gas, the net energy transfer to the surroundings would
be zero. In this way, the system and its surroundings could be returned to their initial conditions and we could identify the process as reversible. The Kelvin–Planck
statement of the second law, however, specifies that the energy removed from the
gas to return the temperature to its original value cannot be completely converted
to mechanical energy in the form of the work done by the engine in compressing
the gas. Therefore, we must conclude that the process is irreversible.
We could also argue that the adiabatic free expansion is irreversible by relying
on the portion of the definition of a reversible process that refers to equilibrium
states. For example, during the sudden expansion, significant variations in pressure occur throughout the gas. Therefore, there is no well-defined value of the
pressure for the entire system at any time between the initial and final states. In
fact, the process cannot even be represented as a path on a PV diagram. The PV
diagram for an adiabatic free expansion would show the initial and final conditions
as points, but these points would not be connected by a path. Therefore, because
the intermediate conditions between the initial and final states are not equilibrium
states, the process is irreversible.
Although all real processes are irreversible, some are almost reversible. If a real
process occurs very slowly such that the system is always very nearly in an equilibrium state, the process can be approximated as being reversible. Suppose a gas is
compressed isothermally in a piston–cylinder arrangement in which the gas is in
thermal contact with an energy reservoir and we continuously transfer just enough
energy from the gas to the reservoir to keep the temperature constant. For example, imagine that the gas is compressed very slowly by dropping grains of sand onto
a frictionless piston as shown in Figure 22.8. As each grain lands on the piston and
compresses the gas a small amount, the system deviates from an equilibrium state,
but it is so close to one that it achieves a new equilibrium state in a relatively short
time interval. Each grain added represents a change to a new equilibrium state, but
27819_22_c22_p625-656.indd 631
Pitfall Prevention 22.2
All Real Processes Are Irreversible
The reversible process is an idealization; all real processes on the Earth
are irreversible.
Insulating
wall
Vacuum
Membrane
Gas at Ti
Figure 22.7 Adiabatic free expansion of a gas.
The gas is compressed
slowly as individual
grains of sand drop
onto the piston.
Energy reservoir
Figure 22.8 A method for compressing a gas in a reversible isothermal process.
6/30/09 12:43:07 PM
632
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
the differences between states are so small that the entire process can be approximated as occurring through continuous equilibrium states. The process can be
reversed by slowly removing grains from the piston.
A general characteristic of a reversible process is that no dissipative effects (such
as turbulence or friction) that convert mechanical energy to internal energy can
be present. Such effects can be impossible to eliminate completely. Hence, it is not
surprising that real processes in nature are irreversible.
22.4 The Carnot Engine
Pitfall Prevention 22.3
Don’t Shop for a Carnot Engine
The Carnot engine is an idealization; do not expect a Carnot engine
to be developed for commercial use.
We explore the Carnot engine only
for theoretical considerations.
In 1824, a French engineer named Sadi Carnot described a theoretical engine, now
called a Carnot engine, that is of great importance from both practical and theoretical viewpoints. He showed that a heat engine operating in an ideal, reversible
cycle—called a Carnot cycle—between two energy reservoirs is the most efficient
engine possible. Such an ideal engine establishes an upper limit on the efficiencies of all other engines. That is, the net work done by a working substance taken
through the Carnot cycle is the greatest amount of work possible for a given amount
of energy supplied to the substance at the higher temperature. Carnot’s theorem
can be stated as follows:
J.-L. Charmet/Science Photo Library/
Photo Researchers, Inc.
No real heat engine operating between two energy reservoirs can be more
efficient than a Carnot engine operating between the same two reservoirs.
French engineer (1796–1832)
Carnot was the first to show the quantitative
relationship between work and heat. In 1824,
he published his only work, Reflections on
the Motive Power of Heat, which reviewed the
industrial, political, and economic importance
of the steam engine. In it, he defined work as
“weight lifted through a height.”
27819_22_c22_p625-656.indd 632
To prove the validity of this theorem, imagine two heat engines operating
between the same energy reservoirs. One is a Carnot engine with efficiency e C , and
the other is an engine with efficiency e, where we assume e . e C . Because the cycle
in the Carnot engine is reversible, the engine can operate in reverse as a refrigerator. The more efficient engine is used to drive the Carnot engine as a Carnot refrigerator. The output by work of the more efficient engine is matched to the input by
work of the Carnot refrigerator. For the combination of the engine and refrigerator,
no exchange by work with the surroundings occurs. Because we have assumed the
engine is more efficient than the refrigerator, the net result of the combination is
a transfer of energy from the cold to the hot reservoir without work being done on
the combination. According to the Clausius statement of the second law, this process is impossible. Hence, the assumption that e . e C must be false. All real engines
are less efficient than the Carnot engine because they do not operate through a
reversible cycle. The efficiency of a real engine is further reduced by such practical
difficulties as friction and energy losses by conduction.
To describe the Carnot cycle taking place between temperatures Tc and Th , let’s
assume the working substance is an ideal gas contained in a cylinder fitted with a
movable piston at one end. The cylinder’s walls and the piston are thermally nonconducting. Four stages of the Carnot cycle are shown in Active Figure 22.9, and
the PV diagram for the cycle is shown in Active Figure 22.10. The Carnot cycle consists of two adiabatic processes and two isothermal processes, all reversible:
1. Process A S B (Active Fig. 22.9a) is an isothermal expansion at temperature
Th . The gas is placed in thermal contact with an energy reservoir at temperature Th . During the expansion, the gas absorbs energy |Q h| from the
reservoir through the base of the cylinder and does work WAB in raising the
piston.
2. In process B S C (Active Fig. 22.9b), the base of the cylinder is replaced by
a thermally nonconducting wall and the gas expands adiabatically; that is,
no energy enters or leaves the system by heat. During the expansion, the
temperature of the gas decreases from Th to Tc and the gas does work W BC
in raising the piston.
6/30/09 12:43:08 PM
22.4 | The Carnot Engine
633
ACTIVE FIGURE 22.9
ASB
The gas undergoes an
isothermal expansion.
The Carnot cycle. The letters A, B,
C, and D refer to the states of the gas
shown in Active Figure 22.10. The
arrows on the piston indicate the
direction of its motion during each
process.
Qh
Energy reservoir at Th
a
BSC
The gas undergoes
expansion.
DSA
The gas undergoes
compression.
Q0
Q0
Cycle
Thermal insulation
Thermal insulation
d
b
CSD
The gas undergoes
an isothermal
compression.
Qc
P
Energy reservoir at Tc
c
A
3. In process C S D (Active Fig. 22.9c), the gas is placed in thermal contact
with an energy reservoir at temperature Tc and is compressed isothermally
at temperature Tc . During this time, the gas expels energy |Q c | to the reservoir and the work done by the piston on the gas is WCD .
4. In the final process D S A (Active Fig. 22.9d), the base of the cylinder is
replaced by a nonconducting wall and the gas is compressed adiabatically.
The temperature of the gas increases to Th , and the work done by the piston on the gas is W DA .
The thermal efficiency of the engine is given by Equation 22.2:
e512
0Qh0
5
B
Weng
Th
C
D
Qc
Tc
V
ACTIVE FIGURE 22.10
0Qc0
0Qh0
Tc
Th
Qh
PV diagram for the Carnot cycle.
The net work done Weng equals the
net energy transferred into the Carnot engine in one cycle, |Q h | 2 |Q c |.
In Example 22.3, we show that for a Carnot cycle,
0Qc0
The work done
during the cycle
equals the area
enclosed by the path
on the PV diagram.
(22.5)
Hence, the thermal efficiency of a Carnot engine is
eC 5 1 2
27819_22_c22_p625-656.indd 633
Tc
Th
(22.6)
W Efficiency of a Carnot engine
6/30/09 12:43:10 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
634
This result indicates that all Carnot engines operating between the same two temperatures have the same efficiency.5
Equation 22.6 can be applied to any working substance operating in a Carnot
cycle between two energy reservoirs. According to this equation, the efficiency is
zero if Tc 5 Th , as one would expect. The efficiency increases as Tc is lowered and
Th is raised. The efficiency can be unity (100%), however, only if Tc 5 0 K. Such
reservoirs are not available; therefore, the maximum efficiency is always less than
100%. In most practical cases, Tc is near room temperature, which is about 300 K.
Therefore, one usually strives to increase the efficiency by raising Th .
Theoretically, a Carnot-cycle heat engine run in reverse constitutes the most
effective heat pump possible, and it determines the maximum COP for a given combination of hot and cold reservoir temperatures. Using Equations 22.1 and 22.4, we
see that the maximum COP for a heat pump in its heating mode is
COPC 1 heating mode 2 5
5
0Qh0
W
0Qh0
0Qh0 2 0Qc0
5
12
1
5
0Qc0
0Qh0
1
Tc
12
Th
5
Th
Th 2 Tc
The Carnot COP for a heat pump in the cooling mode is
COPC 1 cooling mode 2 5
Tc
Th 2 Tc
As the difference between the temperatures of the two reservoirs approaches zero
in this expression, the theoretical COP approaches infinity. In practice, the low
temperature of the cooling coils and the high temperature at the compressor limit
the COP to values below 10.
Quick Quiz 22.3 Three engines operate between reservoirs separated in
temperature by 300 K. The reservoir temperatures are as follows: Engine A:
Th 5 1 000 K, Tc 5 700 K; Engine B: Th 5 800 K, Tc 5 500 K; Engine C: Th 5
600 K, Tc 5 300 K. Rank the engines in order of theoretically possible efficiency from highest to lowest.
Ex a m pl e 22.3
Efficiency of the Carnot Engine
Show that the ratio of energy transfers by heat in a Carnot engine is equal to the ratio of reservoir temperatures, as given
by Equation 22.5.
SOLUTION
Conceptualize Make use of Active Figures 22.9 and 22.10 to help you visualize the processes in the Carnot cycle.
Categorize Because of our understanding of the Carnot cycle, we can categorize the processes in the cycle as isothermal
5For the processes in the Carnot cycle to be reversible, they must be carried out infinitesimally slowly. Therefore,
although the Carnot engine is the most efficient engine possible, it has zero power output because it takes an infinite
time interval to complete one cycle! For a real engine, the short time interval for each cycle results in the working
substance reaching a high temperature lower than that of the hot reservoir and a low temperature higher than that
of the cold reservoir. An engine undergoing a Carnot cycle between this narrower temperature range was analyzed
by F. L. Curzon and B. Ahlborn (“Efficiency of a Carnot engine at maximum power output,” Am. J. Phys. 43(1), 22,
1975), who found that the efficiency at maximum power output depends only on the reservoir temperatures Tc and
Th and is given by e C-A 5 1 2 (Tc/Th)1/2. The Curzon–Ahlborn efficiency e C-A provides a closer approximation to the
efficiencies of real engines than does the Carnot efficiency.
27819_22_c22_p625-656.indd 634
6/30/09 12:43:10 PM
22.4 | The Carnot Engine
635
22.3 cont.
Analyze For the isothermal expansion (process A S B
in Active Fig. 22.9), find the energy transfer by heat from
the hot reservoir using Equation 20.14 and the first law
of thermodynamics:
0 Q h 0 5 0 DE int 2 WAB 0 5 0 0 2 WAB 0 5 nRTh ln
VB
VA
In a similar manner, find the energy transfer to the cold
reservoir during the isothermal compression C S D:
0 Q c 0 5 0 DE int 2 WCD 0 5 0 0 2 WCD 0 5 nRTc ln
VC
VD
Divide the second expression by the first:
(1)
Apply Equation 21.20 to the adiabatic processes B S C
and D S A:
ThV Bg21 5 TcVCg21
ThVAg21 5 TcV D g21
Divide the first equation by the second:
a
0Qh0
5
Tc ln 1 VC /VD 2
Th ln 1 VB /VA 2
VC g21
VB g21
5a b
b
VA
VD
(2)
0Qc0
Substitute Equation (2) into Equation (1):
0Qc0
0Qh0
VB
VC
5
VA
VD
5
Tc ln 1 VC /VD 2
Tc ln 1 VC /VD 2
Tc
5
5
Th ln 1 VB /VA 2
Th ln 1 VC /VD 2
Th
Finalize This last equation is Equation 22.5, the one we set out to prove.
Ex a m pl e 22.4
The Steam Engine
A steam engine has a boiler that operates at 500 K. The energy from the burning fuel changes water to steam, and this
steam then drives a piston. The cold reservoir’s temperature is that of the outside air, approximately 300 K. What is the
maximum thermal efficiency of this steam engine?
SOLUTION
Conceptualize In a steam engine, the gas pushing on the piston in Active Figure 22.9 is steam. A real steam engine does
not operate in a Carnot cycle, but, to find the maximum possible efficiency, imagine a Carnot steam engine.
Categorize We calculate an efficiency using Equation 22.6, so we categorize this example as a substitution problem.
Substitute the reservoir temperatures into Equation 22.6:
eC 5 1 2
Tc
300 K
512
5 0.400
Th
500 K
or
40.0%
This result is the highest theoretical efficiency of the engine. In practice, the efficiency is considerably lower.
WHAT IF? Suppose we wished to increase the theoretical efficiency of this engine. This increase can be achieved by
raising Th by DT or by decreasing Tc by the same DT. Which would be more effective?
Answer A given DT would have a larger fractional effect on a smaller temperature, so you would expect a larger change
in efficiency if you alter Tc by DT. Let’s test that numerically. Raising Th by 50 K, corresponding to Th 5 550 K, would give
a maximum efficiency of
eC 5 1 2
Tc
300 K
512
5 0.455
Th
550 K
continued
27819_22_c22_p625-656.indd 635
6/30/09 12:43:11 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
636
22.4 cont.
Decreasing Tc by 50 K, corresponding to Tc 5 250 K, would give a maximum efficiency of
eC 5 1 2
Tc
250 K
512
5 0.500
Th
500 K
Although changing Tc is mathematically more effective, often changing Th is practically more feasible.
22.5 Gasoline and Diesel Engines
In a gasoline engine, six processes occur in each cycle; they are illustrated in Active
Figure 22.11. In this discussion, let’s consider the interior of the cylinder above the
piston to be the system that is taken through repeated cycles in the engine’s operation. For a given cycle, the piston moves up and down twice, which represents a
four-stroke cycle consisting of two upstrokes and two downstrokes. The processes
in the cycle can be approximated by the Otto cycle shown in the PV diagram in
Active Figure 22.12. In the following discussion, refer to Active Figure 22.11 for the
pictorial representation of the strokes and Active Figure 22.12 for the significance
on the PV diagram of the letter designations below:
1. During the intake stroke (Active Fig. 22.11a and O S A in Active Figure
22.12), the piston moves downward and a gaseous mixture of air and fuel is
drawn into the cylinder at atmospheric pressure. That is the energy input
part of the cycle: energy enters the system (the interior of the cylinder)
by matter transfer as potential energy stored in the fuel. In this process,
the volume increases from V2 to V1. This apparent backward numbering is
The intake valve
opens, and the air–
fuel mixture enters
as the piston moves
down.
The piston moves
up and compresses
the mixture.
The spark plug
fires and ignites
the mixture.
The hot gas pushes
the piston downward.
The exhaust valve
opens, and the
residual gas escapes.
The piston moves
up and pushes the
remaining gas out.
Spark plug
Air
and
fuel
Exhaust
Piston
Intake
Compression
Spark
Power
Release
Exhaust
a
b
c
d
e
f
ACTIVE FIGURE 22.11
The four-stroke cycle of a conventional gasoline engine. The arrows on the piston indicate the direction of its motion during each process.
27819_22_c22_p625-656.indd 636
6/30/09 12:43:14 PM
22.5 | Gasoline and Diesel Engines
2.
3.
4.
5.
6.
based on the compression stroke (process 2 below), in which the air–fuel
mixture is compressed from V1 to V2.
During the compression stroke (Active Fig. 22.11b and A S B in Active
Fig. 22.12), the piston moves upward, the air–fuel mixture is compressed
adiabatically from volume V1 to volume V2, and the temperature increases
from TA to TB . The work done on the gas is positive, and its value is equal to
the negative of the area under the curve AB in Active Figure 22.12.
Combustion occurs when the spark plug fires (Active Fig. 22.11c and B S C
in Active Fig. 22.12). That is not one of the strokes of the cycle because it
occurs in a very short time interval while the piston is at its highest position.
The combustion represents a rapid energy transformation from potential
energy stored in chemical bonds in the fuel to internal energy associated
with molecular motion, which is related to temperature. During this time
interval, the mixture’s pressure and temperature increase rapidly, with the
temperature rising from TB to TC . The volume, however, remains approximately constant because of the short time interval. As a result, approximately no work is done on or by the gas. We can model this process in the
PV diagram (Active Fig. 22.12) as that process in which the energy |Q h|
enters the system. (In reality, however, this process is a conversion of energy
already in the cylinder from process O S A.)
In the power stroke (Active Fig. 22.11d and C S D in Active Fig. 22.12), the gas
expands adiabatically from V2 to V1. This expansion causes the temperature
to drop from TC to TD . Work is done by the gas in pushing the piston downward, and the value of this work is equal to the area under the curve CD.
Release of the residual gases occurs when an exhaust valve is opened
(Active Fig. 22.11e and D S A in Active Fig. 22.12). The pressure suddenly
drops for a short time interval. During this time interval, the piston is
almost stationary and the volume is approximately constant. Energy is
expelled from the interior of the cylinder and continues to be expelled during the next process.
In the final process, the exhaust stroke (Active Fig. 22.11e and A S O in
Active Fig. 22.12), the piston moves upward while the exhaust valve remains
open. Residual gases are exhausted at atmospheric pressure, and the volume decreases from V1 to V2. The cycle then repeats.
637
P
TA
TC
C
processes
Qh
B
D
O
A
V2
Qc
V1
V
ACTIVE FIGURE 22.12
PV diagram for the Otto cycle, which
approximately represents the processes occurring in an internal combustion engine.
If the air–fuel mixture is assumed to be an ideal gas, the efficiency of the Otto
cycle is
e512
1
1 V1 /V2 2 g21
1 Otto cycle 2
(22.7)
where V1/V2 is the compression ratio and g is the ratio of the molar specific heats
CP/CV for the air–fuel mixture. Equation 22.7, which is derived in Example 22.5,
shows that the efficiency increases as the compression ratio increases. For a typical compression ratio of 8 and with g 5 1.4, Equation 22.7 predicts a theoretical
efficiency of 56% for an engine operating in the idealized Otto cycle. This value
is much greater than that achieved in real engines (15% to 20%) because of such
effects as friction, energy transfer by conduction through the cylinder walls, and
incomplete combustion of the air–fuel mixture.
Diesel engines operate on a cycle similar to the Otto cycle, but they do not employ
a spark plug. The compression ratio for a diesel engine is much greater than that
for a gasoline engine. Air in the cylinder is compressed to a very small volume, and,
as a consequence, the cylinder temperature at the end of the compression stroke is
very high. At this point, fuel is injected into the cylinder. The temperature is high
enough for the air–fuel mixture to ignite without the assistance of a spark plug.
Diesel engines are more efficient than gasoline engines because of their greater
compression ratios and resulting higher combustion temperatures.
27819_22_c22_p625-656.indd 637
6/30/09 12:43:15 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
638
Ex a m pl e 22.5
Efficiency of the Otto Cycle
Show that the thermal efficiency of an engine operating in an idealized Otto cycle (see Active Figs. 22.11 and 22.12) is
given by Equation 22.7. Treat the working substance as an ideal gas.
SOLUTION
Conceptualize Study Active Figures 22.11 and 22.12 to make sure you understand the working of the Otto cycle.
Categorize As seen in Active Figure 22.12, we categorize the processes in the Otto cycle as isovolumetric and adiabatic.
Analyze Model the energy input and output as occurring by heat in processes B S C and D S A. (In reality,
most of the energy enters and leaves by matter transfer
as the air–fuel mixture enters and leaves the cylinder.)
Use Equation 21.8 to find the energy transfers by heat
for these processes, which take place at constant volume:
BSC
|Q h | 5 nCV (TC 2 TB)
DSA
|Q c | 5 nCV (TD 2 TA)
Substitute these expressions into Equation 22.2:
(1) e 5 1 2
Apply Equation 21.20 to the adiabatic processes A S B
and C S D:
ASB
TAVAg21 5 TBV Bg21
CSD
TCVCg21 5 TDV D g21
Solve these equations for the temperatures TA and TD ,
noting that VA 5 V D 5 V1 and V B 5 VC 5 V2:
0Qc0
T D 2 TA
512
0Qh0
TC 2 TB
(2) TA 5 TB a
V2 g21
VB g21
5 TB a b
b
VA
V1
(3) TD 5 TC a
V2 g21
VC g21
5 TC a b
b
VD
V1
V2 g21
TD 2 T A
5a b
TC 2 TB
V1
Subtract Equation (2) from Equation (3) and rearrange:
(4)
Substitute Equation (4) into Equation (1):
e512
1
1 V1/V2 2 g21
Finalize This final expression is Equation 22.7.
Pitfall Prevention 22.4
Entropy Is Abstract
Entropy is one of the most abstract
notions in physics, so follow the
discussion in this and the subsequent sections very carefully. Do not
confuse energy with entropy. Even
though the names sound similar,
they are very different concepts.
27819_22_c22_p625-656.indd 638
22.6 Entropy
The zeroth law of thermodynamics involves the concept of temperature, and the
first law involves the concept of internal energy. Temperature and internal energy
are both state variables; that is, the value of each depends only on the thermodynamic state of a system, not on the process that brought it to that state. Another
state variable—this one related to the second law of thermodynamics—is entropy S.
In this section, we define entropy on a macroscopic scale as it was first expressed by
Clausius in 1865.
Entropy was originally formulated as a useful concept in thermodynamics. Its
importance grew, however, as the field of statistical mechanics developed because
the analytical techniques of statistical mechanics provide an alternative means
of interpreting entropy and a more global significance to the concept. In statistical mechanics, the behavior of a substance is described in terms of the statistical
behavior of its atoms and molecules. An important finding in these studies is that
6/30/09 12:43:15 PM
a
isolated systems tend toward disorder, and entropy is a measure of this disorder.
For example, consider the molecules of a gas in the air in your room. If half the gas
molecules had velocity vectors of equal magnitude directed toward the left and the
other half had velocity vectors of the same magnitude directed toward the right,
the situation would be very ordered. Such a situation is extremely unlikely, however.
If you could view the molecules, you would see that they move haphazardly in all
directions, bumping into one another, changing speed upon collision, some going
fast and others going slowly. This situation is highly disordered.
The cause of the tendency of an isolated system toward disorder is easily
explained. To do so, let’s distinguish between microstates and macrostates of a system.
A microstate is a particular configuration of the individual constituents of the system. For example, the description of the ordered velocity vectors of the air molecules in your room refers to a particular microstate, and the more likely haphazard
motion is another microstate. A macrostate is a description of the system’s conditions from a macroscopic point of view. For a thermodynamic system, macrostates
are described by macroscopic variables such as pressure, density, and temperature.
For any given macrostate of the system, a number of microstates are possible.
Let’s first consider some nonthermodynamic systems for simplicity. For example,
the macrostate of a 4 on a pair of dice can be formed from the possible microstates
1–3, 2–2, and 3–1. The macrostate of 2 has only one microstate, 1–1. It is assumed
all microstates are equally probable. When all possible macrostates are examined,
however, it is found that macrostates associated with disorder have far more possible microstates than those associated with order. Therefore, 4 is a more disordered
macrostate for two dice than 2 because there are three microstates for a 4 and only
one for a 2.
There is only one microstate associated with the macrostate of a royal flush in
a poker hand of five spades, laid out in order from ten to ace (Fig. 22.13a). Figure
22.13b shows another poker hand. The macrostate here is “worthless hand.” The
particular hand (the microstate) in Figure 22.13b is as equally probable as the hand
in Figure 22.13a. There are, however, many other hands similar to that in Figure
22.13b; that is, there are many microstates that also qualify as worthless hands. The
more microstates that belong to a particular macrostate, the higher the probability
that macrostate will occur. The macrostate of a royal flush in spades is ordered, of
low probability, and of high value in poker. The macrostate of a worthless hand is
disordered, of high probability, and of low poker value.
639
22.6 | Entropy
b
Figure 22.13 (a) A royal flush has
low probability of occurring. (b) A
worthless poker hand, one of many.
Quick Quiz 22.4 (a) Suppose you select four cards at random from a standard
deck of playing cards and end up with a macrostate of four deuces. How
many microstates are associated with this macrostate? (b) Suppose you pick
up two cards and end up with a macrostate of two aces. How many microstates are associated with this macrostate?
We can also imagine ordered macrostates and disordered macrostates in physical processes, not just in games of dice and poker. The result of a dice throw or a
poker hand stays fixed once the dice are thrown or the cards are dealt. Physical
systems, on the other hand, are in a constant state of flux, changing from moment
to moment from one microstate to another. Based on the relationship between the
probability of a macrostate and the number of associated microstates, we therefore
see that the probability of a system moving in time from an ordered macrostate to
a disordered macrostate is far greater than the probability of the reverse because
there are more microstates in a disordered macrostate.
The original formulation of entropy in thermodynamics involves the transfer of
energy by heat during a reversible process. Consider any infinitesimal process in
which a system changes from one equilibrium state to another. If dQ r is the amount
of energy transferred by heat when the system follows a reversible path between the
27819_22_c22_p625-656.indd 639
6/30/09 12:43:17 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
640
states, the change in entropy dS is equal to this amount of energy for the reversible
process divided by the absolute temperature of the system:
Change in entropy for an X
infinitesimal process
dS 5
dQ r
(22.8)
T
We have assumed the temperature is constant because the process is infinitesimal.
Because entropy is a state variable, the change in entropy during a process depends
only on the endpoints and therefore is independent of the actual path followed.
Consequently, the entropy change for an irreversible process can be determined by
calculating the entropy change for a reversible process that connects the same initial
and final states.
The subscript r on the quantity dQ r is a reminder that the transferred energy is
to be measured along a reversible path even though the system may actually have
followed some irreversible path. When energy is absorbed by the system, dQ r is positive and the entropy of the system increases. When energy is expelled by the system,
dQ r is negative and the entropy of the system decreases. Notice that Equation 22.8
does not define entropy but rather the change in entropy. Hence, the meaningful
quantity in describing a process is the change in entropy.
To calculate the change in entropy for a finite process, first recognize that T
is generally not constant during the process. Therefore, we must integrate Equation 22.8:
f
Change in entropy X
for a finite process
f
DS 5 3 dS 5 3
i
i
dQ r
T
(22.9)
As with an infinitesimal process, the change in entropy DS of a system going
from one state to another has the same value for all paths connecting the two states.
That is, the finite change in entropy DS of a system depends only on the properties
of the initial and final equilibrium states. Therefore, we are free to choose a particular reversible path over which to evaluate the entropy in place of the actual
path as long as the initial and final states are the same for both paths. This point is
explored further in Section 22.7.
Quick Quiz 22.5 An ideal gas is taken from an initial temperature Ti to a
higher final temperature Tf along two different reversible paths. Path A
is at constant pressure, and path B is at constant volume. What is the relation between the entropy changes of the gas for these paths? (a) DS A . DS B
(b) DS A 5 DS B (c) DS A , DS B
Ex a m pl e 22.6
Change in Entropy: Melting
A solid that has a latent heat of fusion Lf melts at a temperature Tm . Calculate the change in entropy of this substance
when a mass m of the substance melts.
SOLUTION
Conceptualize Imagine placing the substance in a warm environment so that energy enters the substance by heat. The
process can be reversed by placing the substance in a cool environment so that energy leaves the substance by heat.
The mass m of the substance that melts is equal to Dm, the change in mass of the higher-phase (liquid) substance.
Categorize Because the melting takes place at a fixed temperature, we categorize the process as isothermal.
Analyze Use Equation 20.7 in Equation 22.9, noting that
the temperature remains fixed:
DS 5 3
dQ r
T
5
L f Dm
Lfm
Qr
1
5
5
3 dQ r 5
Tm
Tm
Tm
Tm
Finalize Notice that Dm is positive so that DS is positive, representing that energy is added to the ice cube.
27819_22_c22_p625-656.indd 640
6/30/09 12:43:18 PM
22.7 | Entropy and the Second Law
641
22.6 cont.
WHAT IF? Suppose you did not have Equation 22.9 available to calculate an entropy change. How could you argue
from the statistical description of entropy that the changes in entropy should be positive?
Answer When a solid melts, its entropy increases because the molecules are much more disordered in the liquid state
than they are in the solid state. The positive value for DS also means that the substance in its liquid state does not spontaneously transfer energy from itself to the warm surroundings and freeze because to do so would involve a spontaneous
increase in order and a decrease in entropy.
Let’s consider the changes in entropy that occur in a Carnot heat engine that
operates between the temperatures Tc and Th . In one cycle, the engine takes in
energy |Q h| from the hot reservoir and expels energy |Q c | to the cold reservoir. These
energy transfers occur only during the isothermal portions of the Carnot cycle;
therefore, the constant temperature can be brought out in front of the integral
sign in Equation 22.9. The integral then simply has the value of the total amount of
energy transferred by heat. Therefore, the total change in entropy for one cycle is
DS 5
0Qh0
Th
2
0Qc0
Tc
where the minus sign represents that energy is leaving the engine. In Example 22.3,
we showed that for a Carnot engine,
0Qc0
0Qh0
5
Tc
Th
Using this result in the previous expression for DS, we find that the total change in
entropy for a Carnot engine operating in a cycle is zero:
DS 5 0
Now consider a system taken through an arbitrary (non-Carnot) reversible cycle.
Because entropy is a state variable—and hence depends only on the properties of
a given equilibrium state—we conclude that DS 5 0 for any reversible cycle. In general, we can write this condition as
dQ r
C T 50
1 reversible cycle 2
(22.10)
where the symbol r indicates that the integration is over a closed path.
22.7 Entropy and the Second Law
By definition, a calculation of the change in entropy for a system requires information about a reversible path connecting the initial and final equilibrium states.
To calculate changes in entropy for real (irreversible) processes, remember that
entropy (like internal energy) depends only on the state of the system. That is,
entropy is a state variable, and the change in entropy depends only on the initial
and final states.
You can calculate the entropy change in some irreversible process between two
equilibrium states by devising a reversible process (or series of reversible processes)
between the same two states and computing DS 5 edQ r /T for the reversible process. In irreversible processes, it is important to distinguish between Q, the actual
energy transfer in the process, and Q r , the energy that would have been transferred
by heat along a reversible path. Only Q r is the correct value to be used in calculating the entropy change.
27819_22_c22_p625-656.indd 641
6/30/09 12:43:19 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
642
If we consider a system and its surroundings to include the entire Universe, the
Universe is always moving toward a higher-probability macrostate, corresponding
to greater disorder. Because entropy is a measure of disorder, an alternative way of
stating this behavior is as follows:
Entropy statement X
of the second law of
thermodynamics
The entropy of the Universe increases in all real processes.
This statement is yet another wording of the second law of thermodynamics that
can be shown to be equivalent to the Kelvin-Planck and Clausius statements.
When dealing with a system that is not isolated from its surroundings, remember
that the increase in entropy described in the second law is that of the system and
its surroundings. When a system and its surroundings interact in an irreversible
process, the increase in entropy of one is greater than the decrease in entropy of
the other. Hence, the change in entropy of the Universe must be greater than zero
for an irreversible process and equal to zero for a reversible process. Ultimately,
because real processes are irreversible, the entropy of the Universe should increase
steadily and eventually reach a maximum value. At this value, the Universe will be
in a state of uniform temperature and density. All physical, chemical, and biological processes will have ceased at this time because a state of perfect disorder implies
that no energy is available for doing work. This gloomy state of affairs is sometimes
referred to as the heat death of the Universe.
Quick Quiz 22.6 True or False: The entropy change in an adiabatic process
must be zero because Q 5 0.
Entropy Change in Thermal Conduction
Let’s now consider a system consisting of a hot reservoir and a cold reservoir that
are in thermal contact with each other and isolated from the rest of the Universe. A
process occurs during which energy Q is transferred by heat from the hot reservoir
at temperature Th to the cold reservoir at temperature Tc . The process as described
is irreversible (energy would not spontaneously flow from cold to hot), so we must
find an equivalent reversible process. Because the temperature of a reservoir does
not change during the process, we can replace the real process for each reservoir
with a reversible, isothermal process in which the same amount of energy is transferred by heat. Consequently, for a reservoir, the entropy change does not depend
on whether the process is reversible or irreversible.
Because the cold reservoir absorbs energy Q, its entropy increases by Q/Tc . At
the same time, the hot reservoir loses energy Q, so its entropy change is 2Q/Th .
Because Th . Tc , the increase in entropy of the cold reservoir is greater than the
decrease in entropy of the hot reservoir. Therefore, the change in entropy of the
system (and of the Universe) is greater than zero:
DSU 5
Q
Tc
1
2Q
Th
.0
Suppose energy were to transfer spontaneously from a cold object to a hot object,
in violation of the second law. This impossible energy transfer can be described in
terms of disorder. Before the transfer, a certain degree of order is associated with
the different temperatures of the objects. The hot object’s molecules have a higher
average energy than the cold object’s molecules. If energy spontaneously transfers
from the cold object to the hot object, the cold object becomes colder over a time
interval and the hot object becomes hotter. The difference in average molecular
energy becomes even greater, which would represent an increase in order for the
system and a violation of the second law.
In comparison, the process that does occur naturally is the transfer of energy
from the hot object to the cold object. In this process, the difference in average
27819_22_c22_p625-656.indd 642
6/30/09 12:43:20 PM
22.8 | Entropy on a Microscopic Scale
molecular energy decreases, which represents a more random distribution of
energy and an increase in disorder.
Entropy Change in a Free Expansion
Let’s again consider the adiabatic free expansion of a gas occupying an initial volume Vi (Fig. 22.14). In this situation, a membrane separating the gas from an evacuated region is broken and the gas expands to a volume Vf . This process is irreversible; the gas would not spontaneously crowd into half the volume after filling the
entire volume. What are the changes in entropy of the gas and of the Universe
during this process? The process is neither reversible nor quasi-static. As shown in
Section 20.6, the initial and final temperatures of the gas are the same.
To apply Equation 22.9, we cannot take Q 5 0, the value for the irreversible
process, but must instead find Q r ; that is, we must find an equivalent reversible
path that shares the same initial and final states. A simple choice is an isothermal,
reversible expansion in which the gas pushes slowly against a piston while energy
enters the gas by heat from a reservoir to hold the temperature constant. Because T
is constant in this process, Equation 22.9 gives
f
DS 5 3
i
dQ r
T
f
5
1
dQ r
T 3i
643
When the membrane
is ruptured, the gas
will expand freely and
irreversibly into the
full volume.
Insulating
wall
Vacuum
Membrane
Gas at Ti in
volume Vi
Figure 22.14 Adiabatic free expansion of a gas. The container is thermally insulated from its surroundings; therefore, Q 5 0.
f
For an isothermal process, the first law of thermodynamics specifies that ei dQ r is
equal to the negative of the work done on the gas during the expansion from Vi
to Vf , which is given by Equation 20.14. Using this result, we find that the entropy
change for the gas is
Vf
DS 5 nR ln a b
Vi
(22.11)
Because Vf . Vi , we conclude that DS is positive. This positive result indicates that
both the entropy and the disorder of the gas increase as a result of the irreversible,
It is easy to see that the gas is more disordered after the expansion. Instead of
being concentrated in a relatively small space, the molecules are scattered over a
larger region.
Because the free expansion takes place in an insulated container, no energy is
transferred by heat from the surroundings. (Remember that the isothermal, reversible expansion is only a replacement process used to calculate the entropy change for
the gas; it is not the actual process.) Therefore, the free expansion has no effect on
the surroundings, and the entropy change of the surroundings is zero.
22.8 Entropy on a Microscopic Scale
As we have seen, entropy can be approached by relying on macroscopic concepts.
Entropy can also be treated from a microscopic viewpoint through statistical analysis of molecular motions. Let’s use a microscopic model to investigate once again
the free expansion of an ideal gas, which was discussed from a macroscopic point
of view in Section 22.7.
In the kinetic theory of gases, gas molecules are represented as particles moving
randomly. Suppose the gas is initially confined to the volume Vi shown in Figure
22.14. When the membrane is removed, the molecules eventually are distributed
throughout the greater volume Vf of the entire container. For a given uniform distribution of gas in the volume, there are a large number of equivalent microstates,
and the entropy of the gas can be related to the number of microstates corresponding to a given macrostate.
Let’s count the number of microstates by considering the variety of molecular
locations available to the molecules. Let’s assume each molecule occupies some
27819_22_c22_p625-656.indd 643
6/30/09 12:43:21 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
644
microscopic volume Vm . The total number of possible locations of a single molecule
in a macroscopic initial volume Vi is the ratio wi 5 Vi /Vm , which is a huge number.
We use wi here to represent either the number of ways the molecule can be placed
in the initial volume or the number of microstates, which is equivalent to the number of available locations. We assume the probabilities of a molecule occupying any
of these locations are equal.
As more molecules are added to the system, the number of possible ways the
molecules can be positioned in the volume multiplies. For example, if you consider
two molecules, for every possible placement of the first, all possible placements of
the second are available. Therefore, there are wi ways of locating the first molecule,
and for each way, there are wi ways of locating the second molecule. The total number of ways of locating the two molecules is wiwi 5 wi2.
Neglecting the very small probability of having two molecules occupy the same
location, each molecule may go into any of the Vi /Vm locations, and so the number
of ways of locating N molecules in the volume becomes Wi 5 wi N 5 1 Vi /Vm 2 N. (Wi
is not to be confused with work.) Similarly, when the volume is increased to Vf , the
number of ways of locating N molecules increases to Wf 5 wf N 5 1 Vf /Vm 2 N. The
ratio of the number of ways of placing the molecules in the volume for the initial
and final configurations is
Wf
Wi
5
1 Vf /Vm 2 N
Vf N
5a b
1 Vi /Vm 2
Vi
N
Taking the natural logarithm of this equation and multiplying by Boltzmann’s constant gives
k B ln a
Wf
Vf N
Vf
b 5 k B ln a b 5 nNA k B ln a b
Wi
Vi
Vi
where we have used the equality N 5 nNA . We know from Equation 19.11 that NAk B
is the universal gas constant R; therefore, we can write this equation as
Vf
k B lnWf 2 k B lnWi 5 nR ln a b
Vi
(22.12)
From Equation 22.11, we know that when a gas undergoes a free expansion from Vi
to Vf , the change in entropy is
Vf
Sf 2 Si 5 nR ln a b
Vi
(22.13)
Notice that the right sides of Equations 22.12 and 22.13 are identical. Therefore,
from the left sides, we make the following important connection between entropy
and the number of microstates for a given macrostate:
Entropy (microscopic X
definition)
S ; k B ln W
(22.14)
The more microstates there are that correspond to a given macrostate, the greater
the entropy of that macrostate. As discussed previously, there are many more microstates associated with disordered macrostates than with ordered macrostates. Therefore, Equation 22.14 indicates mathematically our earlier statement that entropy is
a measure of disorder. Although our discussion used the specific example of the
free expansion of an ideal gas, a more rigorous development of the statistical interpretation of entropy would lead us to the same conclusion.
We have stated that individual microstates are equally probable. Because there
are far more microstates associated with a disordered macrostate than with an
ordered macrostate, however, a disordered macrostate is much more probable than
an ordered one.
27819_22_c22_p625-656.indd 644
6/30/09 12:43:21 PM
22.8 | Entropy on a Microscopic Scale
645
ACTIVE FIGURE 22.15
a
(a) One molecule in a container has
a 1-in-2 chance of being on the left
side. (b) Two molecules have a 1-in-4
chance of being on the left side at
the same time. (c) Three molecules
have a 1-in-8 chance of being on the
left side at the same time.
b
c
Let’s explore this concept by considering 100 molecules in a container. At any
given moment, the probability of one molecule being in the left part of the container shown in Active Figure 22.15a as a result of random motion is 12. If there are
two molecules as shown in Active Figure 22.15b, the probability of both being in
the left part is 1 12 2 2, or 1 in 4. If there are three molecules (Active Fig. 22.15c), the
probability of them all being in the left portion at the same moment is 1 12 2 3, or 1 in
8. For 100 independently moving molecules, the probability that the 50 fastest ones
will be found in the left part at any moment is 1 12 2 50. Likewise, the probability that
the remaining 50 slower molecules will be found in the right part at any moment
is 1 12 2 50. Therefore, the probability of finding this fast–slow separation as a result
of random motion is the product 1 12 2 50 1 12 2 50 5 1 12 2 100, which corresponds to about 1
in 1030. When this calculation is extrapolated from 100 molecules to the number
in 1 mol of gas (6.02 3 1023), the ordered arrangement is found to be extremely
improbable!
Conceptual Example 22.7
Let’s Play Marbles!
Suppose you have a bag of 100 marbles of which 50 are red and 50 are green. You are allowed to draw four marbles from
the bag according to the following rules. Draw one marble, record its color, and return it to the bag. Shake the bag and
then draw another marble. Continue this process until you have drawn and returned four marbles. What are the possible
macrostates for this set of events? What is the most likely macrostate? What is the least likely macrostate?
SOLUTION
Because each marble is returned to the bag before the
TABLE 22.1
Possible Results of Drawing Four Marbles
next one is drawn and the bag is then shaken, the probfrom a Bag
ability of drawing a red marble is always the same as
Total
the probability of drawing a green one. All the possible
Number of
microstates and macrostates are shown in Table 22.1.
Macrostate
Possible Microstates
Microstates
As this table indicates, there is only one way to draw a
All R
RRRR
1
macrostate of four red marbles, so there is only one
1G, 3R
RRRG, RRGR, RGRR, GRRR
4
microstate for that macrostate. There are, however, four
2G, 2R
RRGG, RGRG, GRRG,
6
RGGR, GRGR, GGRR
possible microstates that correspond to the macrostate
3G, 1R
GGGR, GGRG, GRGG, RGGG
4
of one green marble and three red marbles, six microAll G
GGGG
1
states that correspond to two green marbles and two red
marbles, four microstates that correspond to three green
marbles and one red marble, and one microstate that corresponds to four green marbles. The most likely, and most disordered, macrostate—two red marbles and two green marbles—corresponds to the largest number of microstates. The
least likely, most ordered macrostates—four red marbles or four green marbles—correspond to the smallest number of
microstates.
27819_22_c22_p625-656.indd 645
7/13/09 12:10:13 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
646
Ex a m pl e 22.8
Adiabatic Free Expansion: One Last Time
Let’s verify that the macroscopic and microscopic approaches to the calculation of entropy lead to the same conclusion
for the adiabatic free expansion of an ideal gas. Suppose an ideal gas expands to four times its initial volume. As we have
seen for this process, the initial and final temperatures are the same.
(A) Using a macroscopic approach, calculate the entropy change for the gas.
SOLUTION
Conceptualize Look back at Figure 22.14, which is a diagram of the system before the adiabatic free expansion. Imagine
breaking the membrane so that the gas moves into the evacuated area. The expansion is irreversible.
Categorize We can replace the irreversible process with a reversible isothermal process between the same initial and
final states. This approach is macroscopic, so we use a thermodynamic variable, in particular, the volume V.
Analyze Use Equation 22.11 to evaluate the entropy
change:
Vf
4Vi
DS 5 nR ln a b 5 nR ln a b 5 nR ln 4
Vi
Vi
(B) Using statistical considerations, calculate the change in entropy for the gas and show that it agrees with the answer
you obtained in part (A).
SOLUTION
Categorize This approach is microscopic, so we use variables related to the individual molecules.
Vi N
b
Vm
Analyze The number of microstates available to a single
molecule in the initial volume Vi is wi 5 Vi /Vm . Use this
number to find the number of available microstates for
N molecules:
Wi 5 wi N 5 a
Find the number of available microstates for N molecules in the final volume Vf 5 4Vi :
Wf 5 a
Use Equation 22.14 to find the entropy change:
DS 5 k B ln Wf 2 k B ln Wi 5 k B ln a
Vf
Vm
N
b 5a
5 k B ln a
4Vi N
b
Vm
P
WHAT IF? In part (A), we used Equation 22.11, which was
based on a reversible isothermal process connecting the initial and final states. Would you arrive at the same result if
you chose a different reversible process?
27819_22_c22_p625-656.indd 646
Wi
b
4Vi N
b 5 k B ln 1 4N 2 5 Nk B ln 4 5 nR ln 4
Vi
Finalize The answer is the same as that for part (A), which
dealt with macroscopic parameters.
Answer You must arrive at the same result because entropy
is a state variable. For example, consider the two-step process in Figure 22.16: a reversible adiabatic expansion from
Vi to 4Vi (A S B) during which the temperature drops from
T1 to T2 and a reversible isovolumetric process (B S C) that
takes the gas back to the initial temperature T1. During the
reversible adiabatic process, DS 5 0 because Q r 5 0.
Wf
T1
T2
A
Figure 22.16 (Example
22.8) A gas expands to four
times its initial volume and
back to the initial temperature by means of a two-step
process.
C
B
Vi
4Vi
V
6/30/09 12:43:24 PM
| Summary
647
22.8 cont.
f
For the reversible isovolumetric process (B S C), use
Equation 22.9:
DS 5 3
i
dQ r
T
T1
53
T2
nCV dT
T1
5 nCV ln a b
T
T2
Find the ratio of temperature T1 to T2 from Equation
4Vi g21
T1
5a b
5 1 4 2 g21
T2
Vi
Substitute to find DS:
DS 5 nCV ln 1 4 2 g21 5 nCV 1 g 2 1 2 ln 4
5 nCV a
CP
2 1b ln 4 5 n 1 CP 2 CV 2 ln 4 5 nR ln 4
CV
and you do indeed obtain the exact same result for the entropy change.
Summary
Definitions
The thermal efficiency e of a heat engine is
e;
Weng
0Qh0
5
0Qh0 2 0Qc0
0Qc0
512
0Qh0
0Qh0
From a microscopic viewpoint, the entropy of a given macrostate
is defined as
(22.2)
S ; k B ln W
(22.14)
where k B is Boltzmann’s constant and W is the number of microstates of the system corresponding to the macrostate.
In a reversible process, the system can be returned to its initial conditions along the same path on a PV diagram, and
every point along this path is an equilibrium state. A process that does not satisfy these requirements is irreversible.
Concepts and Principles
A heat engine is a device that takes in energy by
heat and, operating in a cyclic process, expels a
fraction of that energy by means of work. The net
work done by a heat engine in carrying a working
substance through a cyclic process (DE int 5 0) is
Weng 5 |Q h | 2 |Q c |
(22.1)
where |Q h | is the energy taken in from a hot reservoir and |Q c | is the energy expelled to a cold
reservoir.
Two ways the second law of thermodynamics can be stated
are as follows:
• It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of
energy by heat from a reservoir and the performance of an
equal amount of work (the Kelvin–Planck statement).
• It is impossible to construct a cyclical machine whose sole
effect is to transfer energy continuously by heat from one
object to another object at a higher temperature without
the input of energy by work (the Clausius statement).
continued
27819_22_c22_p625-656.indd 647
6/30/09 12:43:26 PM
648
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
Carnot’s theorem states that no real heat engine
operating (irreversibly) between the temperatures Tc
and Th can be more efficient than an engine operating reversibly in a Carnot cycle between the same two
temperatures.
The thermal efficiency of a heat engine operating in
the Carnot cycle is
eC 5 1 2
Tc
Th
(22.6)
The second law of thermodynamics states that when real (irreversible) processes occur, the degree of disorder in the
system plus the surroundings increases. When a process occurs in an isolated system, the state of the system becomes
more disordered. The measure of disorder in a system is called entropy S. Therefore, yet another way the second law
can be stated is as follows:
• The entropy of the Universe increases in all real processes.
The change in entropy dS of a system during a process
between two infinitesimally separated equilibrium
states is
dS 5
dQ r
T
(22.8)
where dQ r is the energy transfer by heat for the system
for a reversible process that connects the initial and
final states.
Objective Questions
1. A steam turbine operates at a boiler temperature of
450 K and an exhaust temperature of 300 K. What is the
maximum theoretical efficiency of this system? (a) 0.240
(b) 0.500 (c) 0.333 (d) 0.667 (e) 0.150
2. An engine does 15.0 kJ of work while exhausting 37.0 kJ
to a cold reservoir. What is the efficiency of the engine?
(a) 0.150 (b) 0.288 (c) 0.333 (d) 0.450 (e) 1.20
3. A refrigerator has 18.0 kJ of work done on it while 115 kJ
of energy is transferred from inside its interior. What is
its coefficient of performance? (a) 3.40 (b) 2.80 (c) 8.90
(d) 6.40 (e) 5.20
4. Of the following, which is not a statement of the second
law of thermodynamics? (a) No heat engine operating
in a cycle can absorb energy from a reservoir and use it
entirely to do work. (b) No real engine operating between
two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs.
(c) When a system undergoes a change in state, the change
in the internal energy of the system is the sum of the
energy transferred to the system by heat and the work done
on the system. (d) The entropy of the Universe increases
in all natural processes. (e) Energy will not spontaneously
transfer by heat from a cold object to a hot object.
27819_22_c22_p625-656.indd 648
The change in entropy of a system during an arbitrary
process between an initial state and a final state is
f
DS 5 3
i
dQ r
T
(22.9)
The value of DS for the system is the same for all paths
connecting the initial and final states. The change in
entropy for a system undergoing any reversible, cyclic
process is zero, and when such a process occurs, the
entropy of the Universe remains constant.
Solutions Manual/Study Guide
5. Consider cyclic processes completely characterized by each
of the following net energy inputs and outputs. In each
case, the energy transfers listed are the only ones occurring. Classify each process as (a) possible, (b) impossible
according to the first law of thermodynamics, (c) impossible according to the second law of thermodynamics, or
(d) impossible according to both the first and second laws.
(i) Input is 5 J of work, and output is 4 J of work. (ii) Input is
5 J of work, and output is 5 J of energy transferred by heat.
(iii) Input is 5 J of energy transferred by electrical transmission, and output is 6 J of work. (iv) Input is 5 J of energy
transferred by heat, and output is 5 J of energy transferred
by heat. (v) Input is 5 J of energy transferred by heat, and
output is 5 J of work. (vi) Input is 5 J of energy transferred
by heat, and output is 3 J of work plus 2 J of energy transferred by heat.
6. A compact air-conditioning unit is placed on a table inside
a well-insulated apartment and is plugged in and turned
on. What happens to the average temperature of the apartment? (a) It increases. (b) It decreases. (c) It remains constant. (d) It increases until the unit warms up and then
decreases. (e) The answer depends on the initial temperature of the apartment.
6/30/09 12:43:29 PM
| Conceptual Questions
7. The second law of thermodynamics implies that the coefficient of performance of a refrigerator must be what?
(a) less than 1 (b) less than or equal to 1 (c) greater than or
equal to 1 (d) finite (e) greater than 0
8. A thermodynamic process occurs in which the entropy of a
system changes by 28 J/K. According to the second law of
thermodynamics, what can you conclude about the entropy
change of the environment? (a) It must be 18 J/K or less.
(b) It must be between 18 J/K and 0. (c) It must be equal
to 18 J/K. (d) It must be 18 J/K or more. (e) It must be
zero.
9. A sample of a monatomic ideal gas is contained in a cylinder with a piston. Its state is represented by the dot in the
PV diagram shown in Figure OQ22.9. Arrows A through E
represent isobaric, isothermal, adiabatic, and isovolumetric processes that the sample can undergo. In each process
C
P
649
except D, the volume changes by a factor of 2. All five processes are reversible. Rank the processes according to the
change in entropy of the gas from the largest positive value
to the largest-magnitude negative value. In your rankings,
display any cases of equality.
10. Assume a sample of an ideal gas is at room temperature.
What action will necessarily make the entropy of the sample
increase? (a) Transfer energy into it by heat. (b) Transfer energy into it irreversibly by heat. (c) Do work on it.
(d) Increase either its temperature or its volume, without letting the other variable decrease. (e) None of those
choices is correct.
11. The arrow OA in the PV diagram shown in Figure OQ22.11
represents a reversible adiabatic expansion of an ideal gas.
The same sample of gas, starting from the same state O,
now undergoes an adiabatic free expansion to the same
final volume. What point on the diagram could represent
the final state of the gas? (a) the same point A as for the
reversible expansion (b) point B (c) point C (d) any of
those choices (e) none of those choices
D
B
P
O
A
E
V
Figure OQ22.9
Conceptual Questions
1. What are some factors that affect the efficiency of automobile engines?
2. A steam-driven turbine is one major component of an electric power plant. Why is it advantageous to have the temperature of the steam as high as possible?
3. Does the second law of thermodynamics contradict or correct the first law? Argue for your answer.
4. “The first law of thermodynamics says you can’t really win,
and the second law says you can’t even break even.” Explain
how this statement applies to a particular device or process; alternatively, argue against the statement.
B
A
C
Figure OQ22.11
Solutions Manual/Study Guide
electric potential energy is produced. When one leg is at
a higher temperature than the other as shown in the photograph on the right, however, electric potential energy is
produced as the device extracts energy from the hot reservoir and drives a small electric motor. (a) Why is the difference in temperature necessary to produce electric potential energy in this demonstration? (b) In what sense does
this intriguing experiment demonstrate the second law of
thermodynamics?
Courtesy of PASCO Scientific Company
5. Is it possible to construct a heat engine that creates no
thermal pollution? Explain.
6. (a) Give an example of an irreversible process that occurs
in nature. (b) Give an example of a process in nature that
is nearly reversible.
7. The device shown in Figure CQ22.7, called a thermoelectric converter, uses a series of semiconductor cells to transform internal energy to electric potential energy, which
we will study in Chapter 25. In the photograph on the left,
both legs of the device are at the same temperature and no
27819_22_c22_p625-656.indd 649
V
Figure CQ22.7
6/30/09 12:43:31 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
650
8. Discuss three different common examples of natural
processes that involve an increase in entropy. Be sure to
account for all parts of each system under consideration.
9. Discuss the change in entropy of a gas that expands (a) at
room after a big party. Because she is creating more order,
does this process represent a violation of the second law of
thermodynamics?
11. “Energy is the mistress of the Universe, and entropy is her
for this statement with at least two examples. Alternatively,
argue for the view that entropy is like an executive who
instantly determines what will happen, whereas energy
is like a bookkeeper telling us how little we can afford.
(Arnold Sommerfeld suggested the idea for this question.)
12. (a) If you shake a jar full of jelly beans of different sizes, the
larger beans tend to appear near the top and the smaller
ones tend to fall to the bottom. Why? (b) Does this process
violate the second law of thermodynamics?
13. The energy exhaust from a certain coal-fired electric
generating station is carried by “cooling water” into Lake
Ontario. The water is warm from the viewpoint of living
things in the lake. Some of them congregate around the
outlet port and can impede the water flow. (a) Use the theory of heat engines to explain why this action can reduce
the electric power output of the station. (b) An engineer
says that the electric output is reduced because of “higher
back pressure on the turbine blades.” Comment on the
accuracy of this statement.
Problems
The problems found in this chapter may be assigned
online in Enhanced WebAssign
1. denotes straightforward problem; 2. denotes intermediate problem;
3. denotes challenging problem
1. full solution available in the Student Solutions Manual/Study Guide
1. denotes problems most often assigned in Enhanced WebAssign;
these provide students with targeted feedback and either a Master It
tutorial or a Watch It solution video.
Section 22.1 Heat Engines and
the Second Law of Thermodynamics
1. An engine absorbs 1.70 kJ from a hot reservoir at 277°C
and expels 1.20 kJ to a cold reservoir at 27°C in each cycle.
(a) What is the engine’s efficiency? (b) How much work is
done by the engine in each cycle? (c) What is the power
output of the engine if each cycle lasts 0.300 s?
2. The work done by an engine equals one-fourth the energy it
absorbs from a reservoir. (a) What is its thermal efficiency?
(b) What fraction of the energy absorbed is expelled to the
cold reservoir?
3. A heat engine takes in 360 J of energy from a hot reservoir
and performs 25.0 J of work in each cycle. Find (a) the efficiency of the engine and (b) the energy expelled to the
cold reservoir in each cycle.
4. A gun is a heat engine. In particular, it is an internal combustion piston engine that does not operate in a cycle, but
comes apart during its adiabatic expansion process. A certain gun consists of 1.80 kg of iron. It fires one 2.40-g bullet at 320 m/s with an energy efficiency of 1.10%. Assume
the body of the gun absorbs all the energy exhaust—the
27819_22_c22_p625-656.indd 650
denotes asking for quantitative and conceptual reasoning
denotes symbolic reasoning problem
denotes Master It tutorial available in Enhanced WebAssign
denotes guided problem
shaded denotes “paired problems” that develop reasoning with
symbols and numerical values
other 98.9%—and increases uniformly in temperature for
a short time interval before it loses any energy by heat into
the environment. Find its temperature increase.
5. A particular heat engine has a mechanical power output
of 5.00 kW and an efficiency of 25.0%. The engine expels
8.00 3 103 J of exhaust energy in each cycle. Find (a) the
energy taken in during each cycle and (b) the time interval
for each cycle.
6. A multicylinder gasoline engine in an airplane, operating at 2.50 3 103 rev/min, takes in energy 7.89 3 103 J and
exhausts 4.58 3 103 J for each revolution of the crankshaft.
(a) How many liters of fuel does it consume in 1.00 h of
operation if the heat of combustion of the fuel is equal to
4.03 3 107 J/L? (b) What is the mechanical power output
of the engine? Ignore friction and express the answer in
horsepower. (c) What is the torque exerted by the crankshaft on the load? (d) What power must the exhaust and
cooling system transfer out of the engine?
7. Suppose a heat engine is connected to two energy reservoirs, one a pool of molten aluminum (660°C) and the
other a block of solid mercury (238.9°C). The engine runs
6/30/09 12:43:34 PM
| Problems
by freezing 1.00 g of aluminum and melting 15.0 g of mercury during each cycle. The heat of fusion of aluminum
is 3.97 3 105 J/kg; the heat of fusion of mercury is 1.18 3
104 J/kg. What is the efficiency of this engine?
Section 22.2 Heat Pumps and Refrigerators
8. A refrigerator has a coefficient of performance equal to
5.00. The refrigerator takes in 120 J of energy from a cold
reservoir in each cycle. Find (a) the work required in each
cycle and (b) the energy expelled to the hot reservoir.
9. During each cycle, a refrigerator ejects 625 kJ of energy to
a high-temperature reservoir and takes in 550 kJ of energy
from a low-temperature reservoir. Determine (a) the work
done on the refrigerant in each cycle and (b) the coefficient of performance of the refrigerator.
10. A heat pump has a coefficient of performance of 3.80
and operates with a power consumption of 7.03 3 103 W.
(a) How much energy does it deliver into a home during
8.00 h of continuous operation? (b) How much energy
does it extract from the outside air?
17. What is the coefficient of performance of a refrigerator
that operates with Carnot efficiency between temperatures
23.00°C and 127.0°C?
18. Why is the following situation impossible? An inventor comes to
a patent office with the claim that her heat engine, which
employs water as a working substance, has a thermodynamic efficiency of 0.110. Although this efficiency is low
compared with typical automobile engines, she explains
that her engine operates between an energy reservoir at
room temperature and a water–ice mixture at atmospheric
pressure and therefore requires no fuel other than that to
make the ice. The patent is approved, and working prototypes of the engine prove the inventor’s efficiency claim.
19.
A heat engine is being designed to have a Carnot
efficiency of 65.0% when operating between two energy
reservoirs. (a) If the temperature of the cold reservoir is
20.0°C, what must be the temperature of the hot reservoir?
(b) Can the actual efficiency of the engine be equal to
65.0%? Explain.
20.
An ideal refrigerator or ideal heat pump is equivalent
to a Carnot engine running in reverse. That is, energy |Q c |
is taken in from a cold reservoir and energy |Q h | is rejected
to a hot reservoir. (a) Show that the work that must be supplied to run the refrigerator or heat pump is
11. A freezer has a coefficient of performance of 6.30. It
is advertised as using electricity at a rate of 457 kWh/yr.
(a) On average, how much energy does it use in a single
day? (b) On average, how much energy does it remove
from the refrigerator in a single day? (c) What maximum
mass of water at 20.0°C could the freezer freeze in a single
day? Note: One kilowatt-hour (kWh) is an amount of energy
equal to running a 1-kW appliance for one hour.
12. A heat pump has a coefficient of performance equal to 4.20
and requires a power of 1.75 kW to operate. (a) How much
energy does the heat pump add to a home in one hour?
(b) If the heat pump is reversed so that it acts as an air
conditioner in the summer, what would be its coefficient of
performance?
Section 22.3 Reversible and Irreversible Processes
W5
14. A heat engine operates between a reservoir at 25.0°C and
one at 375°C. What is the maximum efficiency possible for
this engine?
15. A Carnot engine has a power output of 150 kW. The engine
operates between two reservoirs at 20.0°C and 500°C.
(a) How much energy enters the engine by heat per hour?
(b) How much energy is exhausted by heat per hour?
16.
A Carnot engine has a power output P. The engine
operates between two reservoirs at temperature Tc and Th .
(a) How much energy enters the engine by heat in a time
interval Dt ? (b) How much energy is exhausted by heat in
the time interval Dt ?
27819_22_c22_p625-656.indd 651
Th 2 Tc
0Qc0
Tc
(b) Show that the coefficient of performance (COP) of the
ideal refrigerator is
COP 5
Tc
Th 2 Tc
21. What is the maximum possible coefficient of performance of a heat pump that brings energy from outdoors at
23.00°C into a 22.0°C house? Note: The work done to run
the heat pump is also available to warm the house.
22.
Section 22.4 The Carnot Engine
13. One of the most efficient heat engines ever built is a coalfired steam turbine in the Ohio River valley, operating
between 1 870°C and 430°C. (a) What is its maximum theoretical efficiency? (b) The actual efficiency of the engine
is 42.0%. How much mechanical power does the engine
deliver if it absorbs 1.40 3 105 J of energy each second from
its hot reservoir?
651
How much work does an ideal Carnot refrigerator
require to remove 1.00 J of energy from liquid helium at
4.00 K and expel this energy to a room-temperature (293-K)
environment?
23. If a 35.0%-efficient Carnot heat engine (Active Fig. 22.2)
is run in reverse so as to form a refrigerator (Active Fig.
22.4), what would be this refrigerator’s coefficient of
performance?
24.
A Carnot heat engine operates between temperatures Th and Tc . (a) If Th 5 500 K and Tc 5 350 K, what is
the efficiency of the engine? (b) What is the change in its
efficiency for each degree of increase in Th above 500 K?
(c) What is the change in its efficiency for each degree of
change in Tc ? (d) Does the answer to part (c) depend on Tc ?
Explain.
25. An ideal gas is taken through a Carnot cycle. The isothermal expansion occurs at 250°C, and the isothermal compression takes place at 50.0°C. The gas takes in 1.20 3 103 J
of energy from the hot reservoir during the isothermal
expansion. Find (a) the energy expelled to the cold reservoir in each cycle and (b) the net work done by the gas in
each cycle.
6/30/09 12:43:34 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
652
26.
27. Argon enters a turbine at a rate of 80.0 kg/min, a temperature of 800°C, and a pressure of 1.50 MPa. It expands
at pressure 300 kPa. (a) Calculate its temperature at exit.
(b) Calculate the (maximum) power output of the turning turbine. (c) The turbine is one component of a model
closed-cycle gas turbine engine. Calculate the maximum
efficiency of the engine.
28.
pressures, volumes, and temperatures as you fill in the following table:
An electric power plant that would make use of the
temperature gradient in the ocean has been proposed.
The system is to operate between 20.0°C (surface-water
temperature) and 5.00°C (water temperature at a depth of
about 1 km). (a) What is the maximum efficiency of such
a system? (b) If the electric power output of the plant is
75.0 MW, how much energy is taken in from the warm
reservoir per hour? (c) In view of your answer to part (a),
explain whether you think such a system is worthwhile.
Note that the “fuel” is free.
Suppose you build a two-engine device with the
exhaust energy output from one heat engine supplying the
input energy for a second heat engine. We say that the two
engines are running in series. Let e 1 and e 2 represent the
efficiencies of the two engines. (a) The overall efficiency
of the two-engine device is defined as the total work output divided by the energy put into the first engine by heat.
Show that the overall efficiency e is given by
A
B
C
D
29.
An electric generating station is designed to have
an electric output power of 1.40 MW using a turbine with
two-thirds the efficiency of a Carnot engine. The exhaust
energy is transferred by heat into a cooling tower at 110°C.
(a) Find the rate at which the station exhausts energy by
heat as a function of the fuel combustion temperature Th .
(b) If the firebox is modified to run hotter by using more
advanced combustion technology, how does the amount
of energy exhaust change? (c) Find the exhaust power for
Th 5 800°C. (d) Find the value of Th for which the exhaust
power would be only half as large as in part (c). (e) Find
the value of Th for which the exhaust power would be onefourth as large as in part (c).
30. At point A in a Carnot cycle, 2.34 mol of a monatomic ideal
gas has a pressure of 1 400 kPa, a volume of 10.0 L, and
a temperature of 720 K. The gas expands isothermally to
point B and then expands adiabatically to point C, where
its volume is 24.0 L. An isothermal compression brings it
to point D, where its volume is 15.0 L. An adiabatic process
returns the gas to point A. (a) Determine all the unknown
27819_22_c22_p625-656.indd 652
V
T
1 400 kPa
10.0 L
720 K
24.0 L
15.0 L
(b) Find the energy added by heat, the work done by the
engine, and the change in internal energy for each of
the steps A S B, B S C, C S D, and D S A. (c) Calculate the efficiency Wnet /|Q h |. (d) Show that the efficiency is
equal to 1 2 TC /TA , the Carnot efficiency.
31. A heat pump used for heating shown in Figure P22.31 is
essentially an air conditioner installed backward. It extracts
energy from colder air outside and deposits it in a warmer
room. Suppose the ratio of the actual energy entering the
room to the work done by the device’s motor is 10.0% of
the theoretical maximum ratio. Determine the energy
entering the room per joule of work done by the motor
given that the inside temperature is 20.0°C and the outside
temperature is 25.00°C.
Heat
pump
Qc
e 5 e 1 1 e 2 2 e 1e 2
What If? For parts (b) through (e) that follow, assume
the two engines are Carnot engines. Engine 1 operates
between temperatures Th and Ti . The gas in engine 2 varies
in temperature between Ti and Tc . In terms of the temperatures, (b) what is the efficiency of the combination engine?
(c) Does an improvement in net efficiency result from the
use of two engines instead of one? (d) What value of the
intermediate temperature Ti results in equal work being
done by each of the two engines in series? (e) What value
of Ti results in each of the two engines in series having the
same efficiency?
P
Qh
Outside
Tc
Inside
Th
Figure P22.31
32. An ideal (Carnot) freezer in a kitchen has a constant temperature of 260 K, whereas the air in the kitchen has a constant temperature of 300 K. Suppose the insulation for the
freezer is not perfect but rather conducts energy into the
freezer at a rate of 0.150 W. Determine the average power
required for the freezer’s motor to maintain the constant
temperature in the freezer.
Section 22.5 Gasoline and Diesel Engines
Note: For problems in this section, assume the gas in the
engine is diatomic with g 5 1.40.
33.
In a cylinder of an automobile engine, immediately after combustion the gas is confined to a volume of
50.0 cm3 and has an initial pressure of 3.00 3 106 Pa. The
piston moves outward to a final volume of 300 cm3, and the
gas expands without energy transfer by heat. (a) What is
the final pressure of the gas? (b) How much work is done
by the gas in expanding?
34. A gasoline engine has a compression ratio of 6.00. (a) What
is the efficiency of the engine if it operates in an idealized
Otto cycle? (b) What If? If the actual efficiency is 15.0%,
what fraction of the fuel is wasted as a result of friction and
energy transfers by heat that could be avoided in a revers-
7/1/09 2:16:08 PM
| Problems
ible engine? Assume complete combustion of the air–fuel
mixture.
35.
An idealized diesel engine operates in a cycle known
as the air-standard diesel cycle shown in Figure P22.35. Fuel
is sprayed into the cylinder at the point of maximum compression, B. Combustion occurs during the expansion
B S C, which is modeled as an isobaric process. Show that
the efficiency of an engine operating in this idealized diesel cycle is
653
41. A 2.00-L container has a center partition that divides it into
two equal parts as shown in Figure P22.41. The left side
contains 0.044 0 mol of H2 gas, and the right side contains
0.044 0 mol of O2 gas. Both gases are at room temperature
and at atmospheric pressure. The partition is removed, and
the gases are allowed to mix. What is the entropy increase
of the system?
0.044 0 mol
H2
1 TD 2 TA
e512 a
b
g TC 2 TB
0.044 0 mol
O2
Figure P22.41
P
Qh
C
B
processes
42. How fast are you personally making the entropy of the Universe increase right now? Compute an order-of-magnitude
estimate, stating what quantities you take as data and the
values you measure or estimate for them.
D
Qc
A
V2 VB
VC
V1 VA
V
Figure P22.35
43. When an aluminum bar is connected between a hot reservoir at 725 K and a cold reservoir at 310 K, 2.50 kJ of energy
is transferred by heat from the hot reservoir to the cold
reservoir. In this irreversible process, calculate the change
in entropy of (a) the hot reservoir, (b) the cold reservoir,
and (c) the Universe, neglecting any change in entropy of
the aluminum rod.
44.
Section 22.6 Entropy
Section 22.7 Entropy and the Second Law
36. An ice tray contains 500 g of liquid water at 0°C. Calculate
the change in entropy of the water as it freezes slowly and
completely at 0°C.
37. A Styrofoam cup holding 125 g of hot water at 100°C
cools to room temperature, 20.0°C. What is the change in
entropy of the room? Neglect the specific heat of the cup
and any change in temperature of the room.
38. Two 2.00 3 103 -kg cars both traveling at 20.0 m/s undergo
a head-on collision and stick together. Find the change in
entropy of the surrounding air resulting from the collision
if the air temperature is 23.0°C. Ignore the energy carried
away from the collision by sound.
39. A 70.0-kg log falls from a height of 25.0 m into a lake. If the
log, the lake, and the air are all at 300 K, find the change
in entropy of the air during this process.
40. A 1.00-mol sample of H2 gas is contained in the left side
of the container shown in Figure P22.40, which has equal
volumes on the left and right. The right side is evacuated.
When the valve is opened, the gas streams into the right
side. (a) What is the entropy change of the gas? (b) Does
the temperature of the gas change? Assume the container
is so large that the hydrogen behaves as an ideal gas.
Valve
H2
When a metal bar is connected between a hot reservoir
at Th and a cold reservoir at Tc , the energy transferred by
heat from the hot reservoir to the cold reservoir is Q. In
this irreversible process, find expressions for the change in
entropy of (a) the hot reservoir, (b) the cold reservoir, and
(c) the Universe, neglecting any change in entropy of the
metal rod.
45. The temperature at the surface of the Sun is approximately
5 800 K, and the temperature at the surface of the Earth
is approximately 290 K. What entropy change of the Universe occurs when 1.00 3 103 J of energy is transferred by
radiation from the Sun to the Earth?
Section 22.8 Entropy on a Microscopic Scale
46. If you roll two dice, what is the total number of ways in
which you can obtain (a) a 12 and (b) a 7?
47. Prepare a table like Table 22.1 by using the same procedure (a) for the case in which you draw three marbles from
your bag rather than four and (b) for the case in which you
draw five marbles rather than four.
48. (a) Prepare a table like Table 22.1 for the following occurrence. You toss four coins into the air simultaneously and
then record the results of your tosses in terms of the numbers of heads (H) and tails (T) that result. For example,
HHTH and HTHH are two possible ways in which three
heads and one tail can be achieved. (b) On the basis of
your table, what is the most probable result recorded for a
toss? In terms of entropy, (c) what is the most ordered macrostate, and (d) what is the most disordered?
Vacuum
Figure P22.40
27819_22_c22_p625-656.indd 653
49. The energy absorbed by an engine is three times greater
than the work it performs. (a) What is its thermal efficiency?
6/30/09 12:43:36 PM
654
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
(b) What fraction of the energy absorbed is expelled to the
cold reservoir?
ety of applications ever since, including the solar power
application discussed on the cover of this textbook. Fuel
is burned externally to warm one of the engine’s two cylinders. A fixed quantity of inert gas moves cyclically between
the cylinders, expanding in the hot one and contracting
in the cold one. Figure P22.57 represents a model for its
thermodynamic cycle. Consider n moles of an ideal monatomic gas being taken once through the cycle, consisting
of two isothermal processes at temperatures 3Ti and Ti and
two constant-volume processes. Let us find the efficiency
of this engine. (a) Find the energy transferred by heat into
the gas during the isovolumetric process AB. (b) Find the
energy transferred by heat into the gas during the isothermal process BC. (c) Find the energy transferred by heat
into the gas during the isovolumetric process CD. (d) Find
the energy transferred by heat into the gas during the isothermal process DA. (e) Identify which of the results from
parts (a) through (d) are positive and evaluate the energy
input to the engine by heat. (f) From the first law of thermodynamics, find the work done by the engine. (g) From
the results of parts (e) and (f), evaluate the efficiency of
the engine. A Stirling engine is easier to manufacture than
an internal combustion engine or a turbine. It can run on
burning garbage. It can run on the energy transferred by
sunlight and produce no material exhaust. Stirling engines
are not currently used in automobiles due to long startup
times and poor acceleration response.
50. A steam engine is operated in a cold climate where the
exhaust temperature is 0°C. (a) Calculate the theoretical
maximum efficiency of the engine using an intake steam
temperature of 100°C. (b) If, instead, superheated steam at
200°C is used, find the maximum possible efficiency.
51. Find the maximum (Carnot) efficiency of an engine that
absorbs energy from a hot reservoir at 545°C and exhausts
energy to a cold reservoir at 185°C.
52. Every second at Niagara Falls, some 5.00 3 103 m3 of water
falls a distance of 50.0 m. What is the increase in entropy of
the Universe per second due to the falling water? Assume
the mass of the surroundings is so great that its temperature and that of the water stay nearly constant at 20.0°C.
Also assume a negligible amount of water evaporates.
53.
54.
Energy transfers by heat through the exterior walls
and roof of a house at a rate of 5.00 3 103 J/s 5 5.00 kW
when the interior temperature is 22.0°C and the outside
temperature is 25.00°C. (a) Calculate the electric power
required to maintain the interior temperature at 22.0°C if
the power is used in electric resistance heaters that convert
all the energy transferred in by electrical transmission into
internal energy. (b) What If? Calculate the electric power
required to maintain the interior temperature at 22.0°C if
the power is used to drive an electric motor that operates
the compressor of a heat pump that has a coefficient of
performance equal to 60.0% of the Carnot-cycle value.
In 1993, the U.S. government instituted a requirement that all room air conditioners sold in the United States
must have an energy efficiency ratio (EER) of 10 or higher.
The EER is defined as the ratio of the cooling capacity of
the air conditioner, measured in British thermal units per
hour, or Btu/h, to its electrical power requirement in watts.
(a) Convert the EER of 10.0 to dimensionless form, using
the conversion 1 Btu 5 1 055 J. (b) What is the appropriate name for this dimensionless quantity? (c) In the 1970s,
it was common to find room air conditioners with EERs
of 5 or lower. State how the operating costs compare for
10 000-Btu/h air conditioners with EERs of 5.00 and 10.0.
Assume each air conditioner operates for 1 500 h during
the summer in a city where electricity costs 17.0¢ per kWh.
55. An airtight freezer holds n moles of air at 25.0°C and
1.00 atm. The air is then cooled to 218.0°C. (a) What is the
change in entropy of the air if the volume is held constant?
(b) What would the entropy change be if the pressure were
maintained at 1.00 atm during the cooling?
56.
Suppose an ideal (Carnot) heat pump could be constructed for use as an air conditioner. (a) Obtain an expression for the coefficient of performance (COP) for such an
air conditioner in terms of Th and Tc . (b) Would such an
air conditioner operate on a smaller energy input if the
difference in the operating temperatures were greater or
smaller? (c) Compute the COP for such an air conditioner
if the indoor temperature is 20.0°C and the outdoor temperature is 40.0°C.
57.
In 1816, Robert Stirling, a Scottish clergyman,
patented the Stirling engine, which has found a wide vari-
27819_22_c22_p625-656.indd 654
P
Isothermal
processes
B
C
3Ti
A
D Ti
Vi
2Vi
V
Figure P22.57
58.
A firebox is at 750 K, and the ambient temperature is
300 K. The efficiency of a Carnot engine doing 150 J of work
as it transports energy between these constant-temperature
baths is 60.0%. The Carnot engine must take in energy
150 J/0.600 5 250 J from the hot reservoir and must put
out 100 J of energy by heat into the environment. To follow Carnot’s reasoning, suppose some other heat engine
S could have an efficiency of 70.0%. (a) Find the energy
input and exhaust energy output of engine S as it does 150 J
of work. (b) Let engine S operate as in part (a) and run the
Carnot engine in reverse between the same reservoirs. The
output work of engine S is the input work for the Carnot
refrigerator. Find the total energy transferred to or from
the firebox and the total energy transferred to or from the
environment as both engines operate together. (c) Explain
how the results of parts (a) and (b) show that the Clausius
statement of the second law of thermodynamics is violated.
(d) Find the energy input and work output of engine S as it
puts out exhaust energy of 100 J. Let engine S operate as in
part (c) and contribute 150 J of its work output to running
6/30/09 12:43:37 PM
| Problems
the Carnot engine in reverse. Find (e) the total energy the
firebox puts out as both engines operate together, (f) the
total work output, and (g) the total energy transferred to
the environment. (h) Explain how the results show that
the Kelvin–Planck statement of the second law is violated.
Therefore, our assumption about the efficiency of engine S
must be false. (i) Let the engines operate together through
one cycle as in part (d). Find the change in entropy of the
Universe. (j) Explain how the result of part (i) shows that
the entropy statement of the second law is violated.
59. Review. This problem complements Problem 84 in Chapter
10. In the operation of a single-cylinder internal combustion piston engine, one charge of fuel explodes to drive the
piston outward in the power stroke. Part of its energy output
is stored in a turning flywheel. This energy is then used
to push the piston inward to compress the next charge
of fuel and air. In this compression process, assume an
original volume of 0.120 L of a diatomic ideal gas at atmospheric pressure is compressed adiabatically to one-eighth
of its original volume. (a) Find the work input required to
compress the gas. (b) Assume the flywheel is a solid disk
of mass 5.10 kg and radius 8.50 cm, turning freely without friction between the power stroke and the compression
stroke. How fast must the flywheel turn immediately after
the power stroke? This situation represents the minimum
angular speed at which the engine can operate without
stalling. (c) When the engine’s operation is well above the
point of stalling, assume the flywheel puts 5.00% of its
maximum energy into compressing the next charge of fuel
and air. Find its maximum angular speed in this case.
60. A biology laboratory is maintained at a constant temperature of 7.00°C by an air conditioner, which is vented to
the air outside. On a typical hot summer day, the outside
temperature is 27.0°C and the air-conditioning unit emits
energy to the outside at a rate of 10.0 kW. Model the unit as
having a coefficient of performance (COP) equal to 40.0%
of the COP of an ideal Carnot device. (a) At what rate
does the air conditioner remove energy from the laboratory? (b) Calculate the power required for the work input.
(c) Find the change in entropy of the Universe produced by
the air conditioner in 1.00 h. (d) What If? The outside temperature increases to 32.0°C. Find the fractional change in
the COP of the air conditioner.
61. A heat engine operates between two reservoirs at T2 5
600 K and T1 5 350 K. It takes in 1.00 3 103 J of energy
from the higher-temperature reservoir and performs 250 J
of work. Find (a) the entropy change of the Universe DSU
for this process and (b) the work W that could have been
done by an ideal Carnot engine operating between these
two reservoirs. (c) Show that the difference between the
amounts of work done in parts (a) and (b) is T1 DSU .
62.
A 1.00-mol sample of a monatomic ideal gas is
taken through the cycle shown in Figure P22.62. At point
A, the pressure, volume, and temperature are Pi , Vi , and Ti ,
respectively. In terms of R and Ti , find (a) the total energy
entering the system by heat per cycle, (b) the total energy
leaving the system by heat per cycle, and (c) the efficiency
of an engine operating in this cycle. (d) Explain how the
efficiency compares with that of an engine operating in a
Carnot cycle between the same temperature extremes.
27819_22_c22_p625-656.indd 655
655
Q2
P
B
3Pi
C
Q3
Q1
2Pi
Pi
D
A
Q4
2Vi
Vi
V
Figure P22.62
63. A power plant, having a Carnot efficiency, produces
1.00 GW of electrical power from turbines that take in
steam at 500 K and reject water at 300 K into a flowing
river. The water downstream is 6.00 K warmer due to the
output of the power plant. Determine the flow rate of the
river.
64.
A power plant, having a Carnot efficiency, produces
electric power P from turbines that take in energy from
steam at temperature Th and discharge energy at temperature Tc through a heat exchanger into a flowing river. The
water downstream is warmer by DT due to the output of the
power plant. Determine the flow rate of the river.
65.
A sample consisting of n moles of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi . Find the change in entropy of the gas by calculatf
ing ei dQ /T, where dQ 5 nCP dT.
66.
An athlete whose mass is 70.0 kg drinks 16.0 ounces
(454 g) of refrigerated water. The water is at a temperature of 35.0°F. (a) Ignoring the temperature change of the
body that results from the water intake (so that the body is
regarded as a reservoir always at 98.6°F), find the entropy
increase of the entire system. (b) What If? Assume the
entire body is cooled by the drink and the average specific
heat of a person is equal to the specific heat of liquid water.
Ignoring any other energy transfers by heat and any metabolic energy release, find the athlete’s temperature after
she drinks the cold water given an initial body temperature
of 98.6°F. (c) Under these assumptions, what is the entropy
increase of the entire system? (d) State how this result compares with the one you obtained in part (a).
67.
A 1.00-mol sample of an ideal monatomic gas is taken
through the cycle shown in Figure P22.67. The process
A S B is a reversible isothermal expansion. Calculate
(a) the net work done by the gas, (b) the energy added to
P (atm)
A
5
1
Isothermal
process
C
10
B
50
V (liters)
Figure P22.67
6/30/09 12:43:37 PM
CHAPTER 22 | Heat Engines, Entropy, and the Second Law of Thermodynamics
656
the gas by heat, (c) the energy exhausted from the gas by
heat, and (d) the efficiency of the cycle. (e) Explain how
the efficiency compares with that of a Carnot engine operating between the same temperature extremes.
68.
69.
A system consisting of n moles of an ideal gas
with molar specific heat at constant pressure CP undergoes
two reversible processes. It starts with pressure Pi and volume Vi , expands isothermally, and then contracts adiabatically to reach a final state with pressure Pi and volume 3Vi .
(a) Find its change in entropy in the isothermal process.
(The entropy does not change in the adiabatic process.)
(b) What If? Explain why the answer to part (a) must be
the same as the answer to Problem 65. (You do not need to
solve Problem 65 to answer this question.)
A sample of an ideal gas expands isothermally,
doubling in volume. (a) Show that the work done on the
gas in expanding is W 5 2nRT ln 2. (b) Because the internal energy E int of an ideal gas depends solely on its temperature, the change in internal energy is zero during
the expansion. It follows from the first law that the energy
input to the gas by heat during the expansion is equal to
the energy output by work. Does this process have 100%
efficiency in converting energy input by heat into work
output? (c) Does this conversion violate the second law?
Explain.
70. Why is the following situation impossible? Two samples of
water are mixed at constant pressure inside an insulated
container: 1.00 kg of water at 10.0°C and 1.00 kg of water
at 30.0°C. Because the container is insulated, there is no
exchange of energy by heat between the water and the environment. Furthermore, the amount of energy that leaves
the warm water by heat is equal to the amount that enters
the cool water by heat. Therefore, the entropy change
of the Universe is zero for this process.
27819_22_c22_p625-656.indd 656
Challenge Problems
71. A 1.00-mol sample of an ideal gas (g 5 1.40) is carried
through the Carnot cycle described in Active Figure 22.10.
At point A, the pressure is 25.0 atm and the temperature is
600 K. At point C, the pressure is 1.00 atm and the temperature is 400 K. (a) Determine the pressures and volumes at
points A, B, C, and D. (b) Calculate the net work done per
cycle.
72. The compression ratio of an Otto cycle as shown in Active
Figure 22.12 is VA/V B 5 8.00. At the beginning A of the
compression process, 500 cm3 of gas is at 100 kPa and
20.0°C. At the beginning of the adiabatic expansion, the
temperature is TC 5 750°C. Model the working fluid as an
ideal gas with g 5 1.40. (a) Fill in this table to follow the
states of the gas:
T (K)
A
B
C
D
293
P (kPa)
V (cm3)
100
500
1 023
(b) Fill in this table to follow the processes:
Q
W
DE int
ASB
BSC
CSD
DSA
ABCDA
(c) Identify the energy input |Q h |, (d) the energy exhaust
|Q c |, and (e) the net output work Weng. (f) Calculate the
thermal efficiency. (g) Find the number of crankshaft revolutions per minute required for a one-cylinder engine to
have an output power of 1.00 kW 5 1.34 hp. Note: The thermodynamic cycle involves four piston strokes.
6/30/09 12:43:38 PM
```