Microwave
Oven
Observations
ART BY CHARLES BEYL
U s i n g a m i c ro w a v e o v e n t o t e a c h
about transitions of matter
38
T H E
S C I E N C E
T E A C H E R
S
TUDENTS WHO ARE GOOD AT SOLVING
word problems may be able to plug the right
numbers into the right equations but may
not have a conceptual understanding of the
terms these equations include. Multistep word
problems about transitions of matter from the
solid through the liquid to the gaseous states may
include such terms as specific heat of ice, heat of
fusion, and specific heat of water, which many
students find confusing. Students may also have
trouble seeing the relevance of this information to
real-life applications, such as determining the cost
of energy when heating or cooling solids, liquids,
or gases.
BY WILLIAM J. SUMRALL,
DENISE RICHARDSON, AND YUAN YAN
F E B R U A R Y
C O N C E P T U A L U N D E R S TA N D I N G
Microwave ovens are effective tools for helping students
understand word problems that relate to states of matter.
Microwave ovens allow a quick and relatively safe method
of collecting data; provide an accurate means of comparing electrical energy released by a microwave to heat
energy absorbed in water; and allow students to calculate
and compare electrical costs to heat energy costs of
heating items in a microwave.
Prior to using a microwave in the classroom, teachers should inform students about safety concerns and
correct ways to use a microwave oven. Because water
can be heated to boiling in a microwave, students should
use heat-resistant kitchen pads or gloves when handling
heated objects. Only microwaveable bowls should be
used, and no metal objects should be placed in the
microwave oven.
In addition to addressing safety concerns, teachers
should do the experiments ahead of time to determine
the approximate length of time it takes for a certain mass
of water to boil in the microwave at the maximum power
rating prior to beginning an activity. This will help in
assigning different types of problems relating to the
various states of matter.
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39
Some microwave ovens do not have numerical
power ratings but instead set power values at high,
medium, and low. In this case, teachers may want students to calculate a “quantitative” power setting from the
given “qualitative” rating. The calculations made below
are for a microwave that has power settings that range
from 1 to 10.
Q U A N T I TAT I V E U N D E R S TA N D I N G
The first problem involving states of matter that I assign
to students is to measure the effect of heating various
masses of water for 1 minute. I assign students to groups
of four and stagger their schedules so everyone has
access to the microwave. Using 100-, 200-, 300-, and 400gram beakers of water, students determine the amount of
heat energy (kJ) absorbed by the water after it is heated
in a microwave for 1 minute at a power setting of 10. It
is not necessary to determine exact masses of water for
this activity, but it should be pointed out that even
though different masses of water are heated, the amount
of energy absorbed by the water remains fairly constant;
students will also see that the change in temperature is
close to linear when plotted against change in mass.
Some microwaves have power settings that regulate the release of energy. The power-setting capability of
microwaves relates to the speed at which they cook food.
This feature provides an excellent way to compare energy costs. To show students this, I give them the following data obtained using a microwave in the laboratory:
Mass of beaker = 61.1 g;
Temperature of water heated for 1 minute at a power
level of 10 = 34°C, where room temperature is 19°C; and
■ Mass of water and beaker = 384.1 g.
■
■
I plug these numbers into the following equation:
Energy = (mass) (change in temperature) (specific heat
of water) = (384.1 g) (15°C) (4.184 J/g°C) = 24 106 joules
= 24.1 kilojoules.
Q U E S T I O N I N G T H E D ATA
One area of difficulty for students is getting a feel for the
accuracy of measurements. My students use the above
equation for different masses of water and get answers
such as 24.1 kJ, 25.8 kJ, 24.7 kJ, and 24.0 kJ. So, I ask them
why the amount of heat energy absorbed by water
seemed to differ as the mass of the water changed. Before
they can answer, I explain the significance of experimental errors that occur during an activity and how important
this component of science is. Students realize that the
difference between the numbers 24 016 and 25 779 is
only 7 percent, which is small when recording measurements in the thousands. Requiring students to do more
than one trial and using statistics are two strategies that
can improve accuracy and show if an experiment provides significant information.
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T H E
S C I E N C E
Next, I ask students what the difference is between
electrical energy released by the microwave and heat
energy absorbed by different masses of the water. The
output in watts is listed on the back of most microwave
ovens. The microwave I use has a 500-watt output at a
power rating of 10. Once the different masses of water
heated for 1 minute, the following calculation is made to
determine electrical energy released by the microwave:
500 watts × 60 seconds = 30 000 watt-seconds = 30 000
joules = 30kJ. The difference between heat and electrical
energy range from 6 to 4.2 kJ, with different masses
tested. I point out to students that there is a difference
between heat and electrical energy and that heat is lost
through the thermometer, the beaker, and the water’s
surface. However, the heat energy measured is between
80 and 86 percent of the electrical energy put out by the
microwave oven; this is due primarily to the heat lost by
the microwave “generator” in the oven.
Finally, I ask students to find the average cost of the
electrical energy released during the heating of the
water. To calculate the average cost, students use the
equation: Electrical energy released by the microwave =
(500 watts)(60 seconds) = 30 000 watt-seconds.
(30 000 watt-seconds)(1 kW/1000 watt)(1 minute/60
seconds)(1 hour/60 minutes) = .0083 kWh
Price of 1 kWh = 5 cents. (Because the price of 1 kWh
varies, students should call the electric company.)
(5 cents/kWh)(.0083 kWh) = .0415 cents
Once the cost of the energy released has been determined, students can then calculate the cost of the electricity used in heating four masses of water. This cost is
(4)(.0083 kWh)(5 cents) = 0.166 cents.
E F F E C T S O F VA R I O U S P O W E R R AT I N G S
Once students are comfortable manipulating the equations above, I ask them to determine the increase in
temperature and the increase in heat energy absorbed by
approximately 200 grams of water heated in a microwave
for 1 minute at power ratings of 3, 5, 7, and 10. The
following calculations must be performed (the temperature of the room was 20°C):
Power rating 3:
Mass of water = 202.5 grams
Temperature change = 30°C – 20°C = 10°C
Energy (J) = (mass)(change in temperature)(specific heat
of water)
= (202.5g)(10°C)(4.184 J/g°C)
= 8,473 joules
= 8.47 kJ
Power rating 5:
Mass of water = 203.0 grams
Temperature change = 34°C – 20°C=14°C
Energy (J) = 11.9 kJ
T E A C H E R
FIGURE 1.
■
■
Power rating plotted with change in temperature.
■
■
Power 10 = 500 watts
Power 7 = 425 watts
Power 5 = 375 watts
Power 3 = 325 watts
To calculate the cost of heat energy absorbed by the water, students used the following numbers:
8.47kJ + ll.9 kJ + 16.2kJ + 23.0kJ = 59.6kJ
1 joule = 1 watt-second
59 600 joules = 59 600 watt-seconds
(59 600 watt-seconds)(1 kW/1000 watt)(1 min/
60 seconds)(1 hr/60 min) = .017 kWh
(.017 kWh)(5 cents/kWh) = .085 cents
To calculate the cost of the electrical
energy released by the microwave oven, students used the following numbers:
Power rating 7:
Mass of water = 203.3 grams
Temperature change = 39°C – 20°C = 19°C
Energy (J) = 16.2 kJ
Power rating 10:
Mass of water = 196.4 grams
Temperature change = 48°C – 20°C = 28°C
Energy (J) = 23.0 kJ
Students use these results to determine whether or
not there is a linear relationship between temperature
change and power rating if the mass of the water is held
fairly constant. The graph in Figure 1 shows that temperature increases are directly proportional to energy increases (power rating increases) if everything else is held
constant.
Next, students determine if it is cheaper to use the
microwave at a lower power rating. They should be able
to figure out that the cost of operating the microwave is
dependent on both power rating and time setting. Because the relationship between power rating and changes
in temperature is fairly linear for my microwave, the cost
is the same if the same end-temperature is obtained
regardless of power rating. That is because the time
setting will increase proportionally to the decrease in
power rating for this particular microwave. Not all microwaves have proportional ratings, so the graph may not be
linear and different conclusions must by drawn.
Finally, the total cost of the heat energy absorbed
by the water should be calculated. Students can determine how much this cost differs from the total actual cost
of the electrical energy released by the microwave.
There was a 25-watt difference in power output with
each change in power rating on the microwave used in
my class.
F E B R U A R Y
(500 watt)(60 seconds)(lkW/1000 watt)(lhr/3600 seconds) = .0083 kWh
(425 watt)(60 seconds)(lkW/1000 watt)(lhr/3600 seconds) = .0071 kWh
(375 watt)(60 seconds)(lkW/1000 watt)(lhr/3600 seconds) = .0063 kWh
(325 watt)(60 seconds)(lkW/1000 watt)(lhr/3600 seconds = .0054 kWh
Total = .0271 kWh
(.0271 kWh)(5 cents/kWh)= .14 cents
So, the percentage of energy lost to the system is [(.14 .08)/(.14)](100) = 43%
SOLID PHASE PROBLEMS
Students’ next task is to determine how much heat
energy is absorbed if ice is heated for 3 minutes at a
power rating of 10. Students begin by determining the
mass of five or six ice cubes, making sure to subtract the
mass of the beaker from the mass of the ice. Teachers
should test the microwave to determine time and power
rating for a one-phase change problem (ice to water)
before giving this assignment. My students collected the
following data while doing this exercise:
Mass of ice before heating = 61.5 g
Mass of water after heating = 61 .5 g
Change in temperature = 32°C-0°C = 32°C
These data were then plugged into the following equation:
Energy = (heat of fusion)(mass of ice)(molecular weight
of water) + (change in temperature)(total mass of
water)(specific heat of water)
= (6.02 kJ/mol)(61.5 g)(1 mol/18g) + (32°C)(61.5 g)(4.184
J/g°C)(1 kJ/1000 J)
= 28.8 kJ
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41
U s i n g a m i c ro w a v e o v e n
t o s o l v e p ro b l e m s t h a t
parallel those found in
physical science
t e x t b o o k s i s a c o n c re t e
way to teach students a
variety of physical
change concepts.
H E AT I N G I C E AT
VA RY I N G T I M E I N T E RVA L S
The next assignment I give students is to find out how
much heat energy is absorbed if ice is heated at a power
rating of 10 for various time intervals. The following data
are typical of what students will find:
Mass of ice before heating = 223.0 g
Mass of water after heating = 223.0 g
Change in temperature = 28°C – 0°C = 28°C
Time interval = 4 minutes
Heat energy = (6.02 kJ/mol)(223.0 g)(1 mol/18 g) +
(28°C) (223 g)(4.184J/g°C)(1 kJ/1000 J) = 100.7 kJ
Mass of ice before heating = 224.0 g
Mass of water after heating = 224.0 g
Change in temperature = 32°C – 0°C = 32°C
Time interval = 6 minutes
Heat energy = 104.9 kJ
Mass of ice before heating = 229.2 g
Mass of water after heating = 224.0 g
Change in temperature = 82°C – 0°C = 82°C
Time interval = 8 minutes
The loss of mass indicates evaporation has occurred, so
the heat of vaporization (Hvap = 40.7 kJ) must be added to
the equation to take into account the water that evaporated.
Therefore, heat energy = (6.02 kJ/mol)(229.2 g)(1 mol/
18g) + (82°C)(229.2 g)(4.184 J/g°C)(1 kJ/1000 J) + (5.2
g)(1 mol/18 g) (40.7 kJ/1 mol) = 167.1 kJ
Mass of ice before heating = 233.3 g
Mass of water after heating = 217.3 g
Change in temperature = 88°C – 0°C = 88°C
Time interval = 10 minutes
Heat energy = 200.1 kJ
T H E
At this point, I discuss phase changes and expected
temperatures during a phase change. The fact that a
significant loss of mass occurred prior to a phase change
provides a good opportunity to discuss convection currents and how evaporation can be accelerated by wind,
heat, and increased surface area. Also, because the phase
change for water to steam (when water begins to boil)
did not occur at the expected 100°C, I explain to the class
that atmospheric pressure and experimental error must
be taken into account.
LASTING LESSONS
As the heating time increases, students will find
that some of the mass “disappears.” To take this phenomena into account, we use the following data:
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Mass of ice before heating = 227.7 g
Mass of water after heating = 152.5 g
Change in temperature = 98°C – 0°C = 98°C
Time interval = 15 minutes
Heat energy = 339.5 kJ
S C I E N C E
Although I do not find that using a microwave is a
practical method for measuring superheated steam, I like
to use this opportunity to ask students to think of ways to
design a device that uses the microwave to measure
superheated steam.
Using a microwave oven to solve problems that
parallel those found in physical science textbooks is a
concrete way to teach students a variety of physical
change concepts. Concepts such as heat of fusion, heat
of vaporization, and specific heat become clearer to
students after they do hands-on exercises. Students will
be able to go beyond the “number crunching/plug and
chug” requirements of the traditional word problem and
gain a better understanding of the actual processes that
take place during physical changes of matter. ✧
William J. Sumrall is a professor in science education
(e-mail: [email protected]), Denise Richardson
is a graduate assistant in science education(e-mail:
[email protected]), and Yuan Yan is a graduate
assistant in physics (e-mail: [email protected]), all
at Mississippi State University, Box 9705, Mississippi
State, MS 39762.
T E A C H E R