Lab-27-(Heating Water with Microwaves)

Lab-27-(Heating Water with Microwaves).doc
Rev: 05/2003
H W
eating
ater with
µ
icrowaves
Name _________________________________________________________________ Box _________________
Due Date ____________________
Theory
How does a microwave oven warm (add energy to, increase the temperature of)
water? It begins with the fact that water is a polar molecule, which means that it has
positive and negative ends. The positive end is on the side with the hydrogen atoms.
If a water molecule is placed in an electric field, it will rotate until the positive
end is pointing in the direction of the field. In other words, an electric field
will exert a torque on the molecule. The microwave is a form of electromagnetic
radiation. Like all electromagnetic radiation it has oscillating magnetic and electric
fields at right angles to each other. The changing electric field of the microwave
radiation causes the water molecules to rotate. As the electric field in the wave
alternates very rapidly, the water molecules vibrate back and forth trying to follow
the changing electric field. Microwaves are produced by a magnetron.
Basic Equations:
Energy of a microwave (or any other) photon
E=hƒ
(where h is Planck’s Constant and f is the frequency)
The velocity of any wave, including the waves in electromagnetic radiation is given by
v = ƒ λ (where f is the frequency and λ is the wavelength)
Quantity of heat needed to change the absolute temperature (in K) of any other material with a known specific
heat, c, is given by
Q = m c ∆T (where m is the mass and ∆T is the temperature change)
You will use a direct measurement of the energy absorbed by two water samples in the oven to estimate the
efficiency with which the microwave energy (in the form of electromagnetic waves or photons) is absorbed by
water in these particular samples in this particular oven.
06/06/2003
Instructions: (See the Technical Specifications on page 8 for the base data on the oven.)
Record the frequency of the microwaves produced:
f = _______________________________ Hz (or 1/s or s-1)
Record the power output of the oven in watts:
POven = _______________________________ W
Record the mass and initial temperature of the water in each trial. Do two trials.
Trial I
Trial II
M1 = Mass = __________ g = __________ kg
M2 = Mass = __________ g = __________ kg
Initial Temperature = ___________________ ºC
Initial Temperature = __________________ ºC
Run the microwave oven with one water sample at a time.
Trial I
Trial II
Time1 = Heating time = _______________ sec
Time2 = Heating time = _______________ sec
Record the final temperature of the water in each sample and calculate the temperature changes.
Trial I
Trial II
Final Temperature = ______________________ ºC Final Temperature = _______________________ ºC
∆T1 = Temperature change = ________________ K ∆T2 = Temperature change = _________________ K
Analysis - Part I:
A. Does the Microwave Oven Deliver what It Promises? Calculate the total energy delivered to the water as
heat in each trial. (The specific heat of water is cWater = 4186 Joules/(kilogram•Kelvin), J/kg•K.)
Trial I
Trial II
Q1 = ___________________________________ J
Q2 = __________________________________ J
= M1·cWater·∆T1
= M2·cWater·∆T2
Now determine the power put into the water. This will be the total heat energy put into the water divided by the
time (watts = joules / seconds).
P1 = Q1 / Time1 = _____________________ watts
P2 = Q2 / Time2 = _____________________ watts
Compare the power output listed in the Technical Specifications with your measurement of the power actually
absorbed by the water. Calculate the ratio of measured to listed power and record it as a percentage efficiency (you
could consider this efficiency a measure of the oven’s ability to deliver power to the water in your samples.).
%Efficiency = _________________________ %
%Efficiency = _________________________ %
This might be called the power efficiency. It is only the simplest and most obvious form of analysis and by itself
does not provide much insight into how the microwave radiation and the molecules of water interact with each
other. The analyses in the following sections are a bit more theoretical and, it is hoped, a bit more insightful. You
should pay careful attention to the story behind these calculations because you will be expected to explain it in a
written narrative, as though trying to teach someone who was not there and did not understand the physics, how all
this was accomplished and what it showed about how matter and electromagnetic radiation interact.
Page 2
B. How “Big” (an energy question, not a size question) is a Microwave Photon?
The energy of a single microwave photon is calculated using E = hƒ. From the listed frequency find the energy
of one photon (where h is Planck's Constant = 6.626E−34 joule-sec and ƒ = frequency of the radiation in hertz).
From the listed power output in the oven specification, P1, find the number of photons produced each second.
Energy per photon = hf = ________________________ J
PP = # of Photons / second = ___________________
= POven / hf
Produced by the oven
C. Determine the number of microwave photons absorbed by the water during each trial. Use the power out
put of the oven, the duration of the heating, the total heat absorbed by each sample and the known energy of a single
photon to calculate the following:
Trial I
Trial II
N1P = # of photons produced = ________________
N2P = # of photons produced = ________________
N1A = # of photons absorbed = ________________
= (Q1 / hf )
During the heating time
N2A = # of photons absorbed = ________________
= (Q2 / hf)
During the heating time
%Efficiency = ____________________ %
%Efficiency = ____________________ %
= PP·Time1
During the heating time
= PP·Time2
During the heating time
This photon efficiency should be equal to the efficiency reported at the end of Part A.
D. How many photons were produced per water molecule and how many absorbed per water molecules?
Given the mass of water in each cup, use the following expression to calculate the number of water molecules in
each sample.
mass of water in grams x
6.02E23 molecules
1 mole
= molecules of water
x
mole
18 grams
Trial I
Trial II
N1W = # of water molecules = __________________ N2W = # of water molecules = __________________
From the total number of photons produced and the known number of molecules in each sample, calculate the
number photons produced for each molecule. This is the maximum number of photons that each molecule might
have absorbed during the time the oven was running.
Trial I
Trial II
Photons/molecule = ________________________
Photons/molecule = _________________________
N1P / N1W = Produced during the heating time
N2P / N2W = Produced during the heating time
Number of photons absorbed per molecule = Q / #molecules / (energy per photon)
Trial I
Trial II
Photons/molecule = ________________________
N1A / N1W =
Photons/molecule = __________________________
Absorbed during the heating time
N2A / N2W =
Absorbed during the heating time
%Efficiency = ___________________________ % %Efficiency = _____________________________ %
This total photon absorption efficiency should be equal to the efficiency reported at the end of Part A.
Page 3
E. How many photons were produced for each molecule of water each second and how many photons were
absorbed by each water molecule each second? The number of photons produced for each molecule each
second = (# photons produced / # molecules) / (duration of the heating). The number of photons absorbed per
water molecule per second = (# of photons absorbed / #molecules) / (duration of the heating).
Trial I
Trial II
Photons/molecule/sec = _______________________ Photons/molecule/sec = ________________________
N1P / N1W / Time1 = Photons produced/molec each second
N2P / N2W / Time2 = Photons produced/molec each
second
Photons/molecule/sec = _______________________ Photons/molecule/sec = ________________________
N1A / N1W / Time1 = Photons absorbed/molec each second
N2A / N2W / Time2 = Photons absorbed/molec each
second
From the total number of photons produced for each molecule each second and the number of photons actually
absorbed by each molecule each second, calculate the photon absorption %Efficiency.
Trial I
Trial II
%Efficiency = ____________________ %
%Efficiency = ____________________ %
This absorption efficiency calculated here should be equal to the efficiency reported at the end of Part A.
Analysis - Part II - Calculations in Part II are all on a per molecule basis.
The basic unit of time will be
either one period or one quarter of the period of oscillation of the electromagnetic radiation. We are trying to
understand how one molecule interacts with the electric field in an electromagnetic wave.
We’ll try some fundamental calculations describing how a water molecule responds to the oscillating electric field
of a microwave.
F. First, we need to calculate the average value of the oscillating electric field.
The amplitude of the electric field is given by
E MAX =
2 Pµ o c
A
where P = POven is the power output of the oven, c is the speed of light, A is the total surface area enclosing the
cavity, and µo is the permeability of free space (a constant found in the back of your textbook). Calculate the
electric field amplitude using the lab data.
(Calculate the area of the interior of the oven as “Area = 2H•W+2W•D+2D•H”)
Area = A = ___________________________ m2
Amplitude = EMAX = _______________________ N/C
G. Next we need to find the torque this field exerts on each water molecule.
This electric field produces a torque on the water molecule according to the equation
The number in this equation is called the electric dipole moment of water. This number has been measured
experimentally. The optimum torque occurs when E = EMAX and θ equals 90°, i.e. when sin θ equals 1.00.
Thus,
Maximum |Torque| = |τMAX | = __________________________ N•m
The average torque during a 180º rotation (one half-rotation) of the molecule, is
Average |Torque| = |τAverage | = π½ • |τMAX | / 2 = __________________________ N•m
Page 4
H. The Moment of Inertia for a single water molecule.
Next, we would like to determine if the acceleration produced by this torque is able to rotate the water fast enough
for it to keep up with the changing electric field of the microwaves. We begin by estimating the moment of inertia
of the water molecule. We need the moment of inertia about an axis that interchanges the positive and negative
ends of the molecule. Rotations will accomplish that interchange about two principle axes. There are only three
principle axes of rotation in a three dimensional molecule. All of the principle axes of rotation pass through the
center of mass in the following diagram. In these three sections we check the possibilities.
Rotations about both the x-axis and the y-axis will interchange the positive and negative ends of the molecular
dipole. The microwaves cannot cause the molecule to rotate around the z-axis. Make sure you understand why
rotations about x and y interact with the microwaves, while rotation about z cannot interact with the microwaves.
To keep things simple, we will use the average of the two moments of inertia; Ixx and Iyy . The three principle
moments of inertia of water are
Ixx = 1.9274 x 10-47 kg•m2
Iyy = 2.9528 x 10-47 kg•m2
Izz = 1.0254 x 10-47 kg•m2
Assuming that rotations about both the x and y axes are equally likely, we will use an average of these two moments
of inertia as our average moment of inertia for the molecule.
Moment of Inertia = (Ixx + Iyy )/2= I = ____________________________ kg•m2
Page 5
I. Now, use Newton’s 2nd Law (recast in angular form) to calculate the average angular acceleration.
Use the rotational form of Newton's 2nd Law (|τAverage| = I αAverage ) to find the average angular acceleration of the
water molecule from the known average torque on the water molecule.
Average α = αAverage = (|τAverage| / I = _________________________ rad/s2
We assume this average acceleration applies to each half-rotation of a water molecule in a microwave field.
J. Find the angular distance a water molecule could rotate during each half-cycle.
Frequency of the microwave = f = ____________________________ s-1 (from the instructions on page 2)
Wavelength of the microwave = λ= ___________________________ m = ____________________________ mm
Period of the microwave = 1/f = T = ____________________________ s
Acceleration time = Period / 2 = T/2 = ____________________________ s
Now we are ready to calculate the angular distance the hydrogen nuclei could rotate during each half-cycle. For the
water to keep up with the changing electric field of the microwave, it must be able to rotate at least one-half of a
rotation during each T/2 seconds. As long as it can rotate at least that far that fast we can safely assume that is
actually will rotate exactly one half-rotation per half-period.
Angular distance = θ = ½ αAve (T/2) = ____________ radians; Rotations = θ / (2π) = ____________ rotations
2
Be sure to understand the significance of this result. What it shows is that the molecule can rotate fast enough, in
the absence of collisions, to keep up with the changing electric field. Furthermore, it shows that the molecule will
pick up more than enough energy to keep up and therefore has excess rotational kinetic energy that it can transfer to
the other molecules. What it does not answer is how much energy it will transfer. That is a separate question
addressed in the next two sections.
K. Energy transfer from the microwave field to a water molecule.
There is Quantum Mechanical consideration that must be taken into account at this point. A molecule must absorb
energy from the microwaves in increments of hf and must pass that energy along in increments of hf. The WorkEnergy Theorem tells us that the change in internal energy of a system equals its change in kinetic energy. In other
words;
∆U½ = ∆K½ = ½Iωf2 – ½Iωi2
therefore
ωf2 – ωi2 = 2∆K½ / I
Assuming constant angular acceleration, we can use the rotational form of our equations of motion to calculate how
far the molecule rotates as it absorbs one photon of energy. Thus,
ωf2 = ωi2 + 2 αMAX ∆θ
or
ωf2 − ωi2 = 2 αMAX ∆θ
or
2∆K½ / I = 2∆θ |τMAX | / I
therefore, for one half-rotation (= π radians)
∆U½ = ∆K½ = |τMAX | ∆θ = |τMAX | π =___________________ J
So, in absorbing one photon of energy, the molecule must complete
∆θrotations = hf / (2 • ∆K½ ) = ___________________ rotations
Page 6
L. Finding the work done by the microwaves on a water molecule.
The classic textbook solution to this problem in the presence of a constant, external electric field is to calculate the
change in potential energy of the water molecule that occurs when an external agent rotates the molecule against the
direction of the electric field. In this solution one integrates force times distance, or as in this case, torque times
angular distance. Assuming the molecule undergoes one half-rotation, or π radians, in the presence of an electric
field of magnitude EAverage = EMAX / √2, thereby averaging over all possible phase shifts, we find
∆U½ = 2 • (EAverage N/C) • (6.3 x 10-30 C·m) = _____________________ J = W½
In absorbing one photon, however, the molecule makes more than one half-rotation. We know from the last section
that is makes
∆θrotations = ___________________ rotations
Therefore, the work done on the molecule by the electric field in one photon is the number of rotations times the
work per rotation,
WE-photon = (∆θrotations) • (2•W½ ) = _______________________ J
Have we overlooked anything? Well, in fact, we have.
Recall that an electromagnetic wave contains both electric and magnetic fields. The magnetic field carries exactly
the same amount of energy as the electric field. Not only that, but it is an oscillating magnetic field which, as we
know, can cause charges to move. In the case of a dipolar molecule, like water, the charged ends will try to move in
opposite directions with the result that the magnetic field will tend to make the molecule rotate. To take full account
of the energy in the magnetic field, we should double the last result. Thus
Wphoton = WE-photon + WB-photon = 2 WE-photon = _______________________ J
Comparing this to the energy in one photon we find that the following fraction of the energy of the photon
accomplishes work on the water molecule:
%Efficiency = 100% • Wphoton / hf = _____________________%
The efficiency should be close to 100%. Deviation from 100% is probably due to our use of simplifying
assumptions. These allow us to complete the calculations without the use of calculus but are also less accurate than
a more rigorous mathematic treatment.
Still, given the constraints this seems to me to be in excellent agreement with expectations; namely that 100% of the
energy in the microwaves is transmissible to the water. In our experiment, the efficiency was less than 100%. That
is probably attributable to absorption by the walls of the oven when the amount of water is too small to capture all
the photons. Our expectation is that with a larger sample of water, we will get those experimental efficiencies up
near 100%.
Page 7
M. Further experimentation
To verify our expectation that the %Efficiency should indeed be 100%, we might try one additional experiment.
Heat a larger volume of water in an attempt to capture as many of the photons as possible. If we can get the
%Efficiency calculated in section A up close to 100%, that should be a sufficient verification that the %Efficiency
in section L should also be 100%. A kilogram of water and 2 minutes heating time should provide a sufficient test.
The idea is to capture all the photons in order to verify that all the energy in the photons (in both the electric and
magnetic fields) is capable of being converted into heat.
Trial III
Mass = ___________________________ gram
Initial Temperature = ___________________ ºC
Heating time = _______________ sec
Record the final temperature of the water and calculate the temperature change.
Final Temperature = ______________________ ºC
Temperature change = _____________________ K
Calculate the total energy delivered to the water as heat in each trial. (The specific heat of water is 4186
Joules/(kilogram•Kelvin), J/kg•K.)
Q = ___________________________________ J
Now determine the power put into the water. This will be the total heat energy put into the water divided by the
time (watts = joules / seconds).
Q / t = _____________________________ watts
%Efficiency = _________________________ %
This little test should give us the maximum absorption of microwaves by the water. If this is 100% then our
expectation that the answer in Section L should be 100% will be vindicated. If it is less than 100%, then it is also
possible that leaks from the cavity or a phase shift of something other than 90º are responsible.
Page 8
Questions:
1. If you heat 500 grams of water with this microwave, how long will you need to run the microwave to
heat the water from 25°C to 100°C?
2. If you continue heating the water, how much longer will it take to boil all the water away?
3. How big (a size question this time) are the microwave photons generated by this oven:
a) What is the wavelength of the microwaves generated by the oven?
b) Does one of these waves fit inside the oven?
c) Could the internal cavity of the oven by considered a resonant cavity? Why or why not?
4. What is the overall efficiency of the oven? (Decide how this should be calculated and justify your
method. Then calculate the overall efficiency.)
5. According to the information in the Technical Specifications what do you calculate is the operating
voltage of this oven?
6. According to the information in the Technical Specifications what do you calculate is the overall
equivalent electrical resistance of this oven?
7. Why is the power consumption of this oven, or any oven, greater than its power output?
Essay Assignment - Extra Credit
Write a lucid and carefully considered description of the microwave oven's ability to deliver power to enclosed
foods. You should provide an introduction to the essay describing the scope of your article, discuss in outline form
the type of measurements and calculations you performed, and explain how your conclusions follow from those
measurements and calculations. Provide a quantitative basis for your conclusions. In a few clearly written
paragraphs explain to someone who was not present at the time what you did, how you did it and what you learned
from doing it. Illegible, incoherent, or disorganized essays receive no credit.
Page 9
FAQ - Microwave Ovens
Q. Do Microwave ovens cook from the inside out?
No. Food is partially transparent to the electromagnetic waves in the microwave region of the spectrum, so the
energy is able to shine through it. At the same time some of the energy is absorbed by the food. Usually most of
the absorption occurs, and therefore most of the heat is produced, in an outer layer about an inch thick. So, for
example, large pieces of meat will quickly cook to a depth of about an inch, while the inside portions are cooked
more slowly by heat conduction; just like in a conventional over. The effect can be dramatically different for
different foods. The differences depend on the amount of water each food contains. If a food is mostly water, only
the outside inch cooks at all. If a food contains both air and water (like bread, cake, etc.) then the energy penetrates
all the way through and the food gets heated everywhere, even deep inside.
Q. If I put a fork in the Microwave oven, will it destroy the oven?
No. This is a myth, but it has some roots in reality. In order to safely use metals inside a microwave oven, the cook
has to learn numerous complex and mysterious rules in order to avoid fires and undercooked food. For example,
thin metal will heat up fast in the oven and may cause a fire. The famous staple in the paper popcorn bag comes to
mind here. The staple heats up and sets fire to the bag. Another rule is that if a metal object in the oven is touched
even lightly to another one, or touched to the metal wall of the oven, an electric arc might ignite at the point of
contact. This can set fire to the other items in the oven or even to the oven itself. Another rule is to avoid putting
metal objects with sharp points in the microwave oven. Sharp points on metal objects can initiate a coronal
discharge, also sometimes known as Saint Elmo’s Fire, which behaves the same as a flame and can set fire to the
oven if allowed to continue.
Given the complexity of the rules for safe use, and this list is not complete, it is much easier and safer to totally ban
the use of metals in microwave ovens. The alternative would be to send everyone to a safety school to learn these
complex rules!
Q. Aren’t these ovens tuned to a special frequency so they heat only the water?
No. The usual operating frequency of a microwave oven is nowhere near the resonant frequency of water, and the
energy in the radiation will heat other substances. For example, drops of grease on a plastic microwave dish can be
heated far hotter than 100°C, and this causes the mysterious scarring which frequently occurs on plastic utensils.
Any molecule that is “polar” (has positive and negative ends) will be rotated to align with the electric field of the
microwaves in the oven. The vibrating electric field rotates (vibrates) the water molecules and any other polar
molecules within the food. Microwave ovens have difficulty melting ice, presumably because the molecules are
bound together and cannot be rotated by the electric field of the microwaves.
If the oven was tuned specifically to the water resonance frequency the water in the thin outer surface of the food
would absorb all the energy and only the outside surface would be heated. The thin outer surface of meat, for
example, would become a blast of steam while the inside remained ice cold. Because the microwaves are not at the
resonant frequency for water, the waves are able to penetrate an inch or so into the food before it is completely
absorbed.
School Microwave Oven
Home Microwave Oven
Technical Specifications ----------------------------|-----------------------------------Model Numbers
MW8779W
MT 0130SJB-0
-------------------------------|--------------------------------------------------|------------------------------------------------Power Consumption:
11.0 Amps; 1250 W
13.0 Amps; 1,560 W
Output: *
800 W
1000 W
mm(H)x
mm(W)x
mm(D)
Outside Dimensions:
248mm(H)x456mm(W)x305mm(D)
-------------------------------|--------------------------------------------------|------------------------------------------------mm(H)x
mm(W)x
mm(D)
Oven Cavity Dimensions: 175mm(H)x292mm(W)x285mm(D)
Operating Frequency:
2,450 MHz
2,450 MHz
Uncrated Weight: Approx. 25 lbs.
Approx. 40 lbs.
-------------------------------|--------------------------------------------------|------------------------------------------------* IEC 705-88 Test procedure
Specifications subject to change without notice
Page 10
Instructions: (Keep a copy of your numerical results on these 3 pages and data look-up will be much easier.)
Record the frequency of the microwaves produced:
f = _______________________________ Hz (or 1/s or s-1)
Record the power output of the oven in watts:
POven = _______________________________ W
Trial I
Trial II
M1 = Mass = __________ g = __________ kg
M2 = Mass = __________ g = __________ kg
Initial Temperature = ___________________ ºC
Initial Temperature = __________________ ºC
Trial I
Trial II
Time1 = Heating time = _______________ sec
Time2 = Heating time = _______________ sec
Trial I
Trial II
Final Temperature = ______________________ ºC Final Temperature = _______________________ ºC
∆T1 = Temperature change = ________________ K ∆T2 = Temperature change = _________________ K
Analysis - Part I:
A. Does the Microwave Oven Deliver what It Promises?
Trial I
Trial II
Q1 = ___________________________________ J
Q2 = __________________________________ J
= M1·cWater·∆T1
= M2·cWater·∆T2
P1 = Q1 / Time1 = _____________________ watts
P2 = Q2 / Time2 = _____________________ watts
%Efficiency = _________________________ %
%Efficiency = _________________________ %
B. How “Big” (an energy question, not a size question) is a Microwave Photon?
Energy per photon = hf = ________________________ J
PP = # of Photons / second = ___________________
= POven / hf
Produced by the oven
C. Determine the number of microwave photons absorbed by the water during each trial.
Trial I
Trial II
N1P = # of photons produced = ________________
N2P = # of photons produced = ________________
N1A = # of photons absorbed = ________________
= (Q1 / hf )
During the heating time
N1A = # of photons absorbed = ________________
= (Q2 / hf)
During the heating time
%Efficiency = ____________________ %
%Efficiency = ____________________ %
= PP·Time1
During the heating time
= PP·Time2
Page 11
During the heating time
D. How many photons were produced per water molecule and how many absorbed per water molecules?
mass of water in grams x
1 mole
6.02E23 molecules
x
= molecules of water
18 grams
mole
Trial I
Trial II
N1W = # of water molecules = __________________ N2W = # of water molecules = __________________
Trial I
Trial II
Photons/molecule = ________________________
Photons/molecule = _________________________
N1P / N1W = Produced during the heating time
N2P / N2W = Produced during the heating time
Number of photons absorbed per molecule = Q / #molecules / (energy per photon)
Trial I
Trial II
Photons/molecule = ________________________
N1A / N1W =
Photons/molecule = __________________________
Absorbed during the heating time
N2A / N2W =
Absorbed during the heating time
%Efficiency = ___________________________ % %Efficiency = _____________________________ %
E. How many photons were produced for each molecule of water each second and how many photons were
absorbed by each water molecule each second?
Trial I
Trial II
Photons/molecule/sec = _______________________ Photons/molecule/sec = ________________________
N1P / N1W / Time1 = Photons produced/molec each second
N2P / N2W / Time2 = Photons produced/molec each
second
Photons/molecule/sec = _______________________ Photons/molecule/sec = ________________________
N1A / N1W / Time1 = Photons absorbed/molec each second
N2A / N2W / Time2 = Photons absorbed/molec each
second
Trial I
Trial II
%Efficiency = ____________________ %
%Efficiency = ____________________ %
Analysis - Part II F. First, we need to calculate the average value of the oscillating electric field.
(Calculate the area of the interior of the oven as “Area = 2H•W+2W•D+2D•H”)
Area = A = ___________________________ m2
Amplitude = EMAX = _______________________ N/C
G. Next we need to find the torque this field exerts on each water molecule.
Maximum |Torque| = |τMAX | = __________________________ N•m
Average |Torque| = |τAverage | = π½ • |τMAX | / 2 = __________________________ N•m
Page 12
H. The Moment of Inertia for a single water molecule.
Ixx = 1.9274 x 10-47 kg•m2
Iyy = 2.9528 x 10-47 kg•m2
Izz = 1.0254 x 10-47 kg•m2
Moment of Inertia = (Ixx + Iyy )/2= I = ____________________________ kg•m2
I. Now, use Newton’s 2nd Law (recast in angular form) to calculate the average angular acceleration.
Average α = αAverage = (|τAverage| / I = _________________________ rad/s2
J. Find the angular distance a water molecule could rotate during each half-cycle.
Frequency of the microwave = f = ____________________________ s-1 (from the instructions on page 2)
Wavelength of the microwave = λ= ___________________________ m = ____________________________ mm
Period of the microwave = 1/f = T = ____________________________ s
Acceleration time = Period / 2 = T/2 = ____________________________ s
Angular distance = θ = ½ αAve (T/2) = ____________ radians; Rotations = θ / (2π) = ____________ rotations
2
K. Energy transfer from the microwave field to a water molecule.
∆U½ = ∆K½ = |τMAX | ∆θ = |τMAX | π =___________________ J
∆θrotations = hf / (2 • ∆K½ ) = ___________________ rotations
L. Finding the work done by the microwaves on a water molecule.
EAverage = EMAX / v2
∆U½ = 2 • (EAverage N/C) • (6.3 x 10-30 C·m) = _____________________ J = W½
∆θrotations = ___________________ rotations
WE-photon = (∆θrotations) • (2•W½ ) = _______________________ J
Wphoton = WE-photon + WB-photon = 2 WE-photon = _______________________ J
%Efficiency = 100% • Wphoton / hf = _____________________%
Page 13
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