Interactive Lecture Demonstrations
Demonstration 1: 100 grams of
water is initially at room temperature in a perfectly insulated cup (no heat
can leak in or out). Heat energy is
transferred to the water at a steady rate for 80 seconds, and then no more heat
is transferred. Sketch to the right
your prediction for the graph of the temperature of the water as a function
of time. Only after you have made your predictions, click here to download and view a video of the experiment.
(Note that
each on the graph represents an equal quantity of
heat transferred by a heating coil immersed in the water.) Then, click here to view the starting and final
temperatures. Compare the resulting graph to your prediction and
explain its shape. How is the change in temperature related to the quantity
of heat transferred to the water? |
Record Temperatures Below: Starting Temperature: _______ Ending Temperature: _______ Change in Temperature: _______ |
Calculation
1: Using the actual starting and ending temperatures
of the water, determine the amount of heat energy that was transferred to the
water in this process. The specific
heat of water is 4186 J/kg Cº. Heat
Energy Transferred = Q1
= Only after you have made your calculation, click here to compare it. |
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Demonstration 2: 50 grams of
aluminum is added to 100 grams of
water. The aluminum and the water are
both initially at room temperature in a perfectly insulated cup. Heat energy is transferred to the system at
the same steady rate as in the previous demonstration for 80 seconds, and
then no more heat energy is transferred.
(Note that the specific heat of water is 4186 J/kg Cº and the specific
heat of aluminum is 900 J/kg Cº.) Do you think that the change in temperature
of the water plus aluminum will be greater than, less than or equal to the
change in temperature in the previous demonstration? Only after you have made your predictions, click here to download and view a video of the
experiment. Then, click here to view the
starting and final temperatures. Compare the results to your predictions and
explain any differences. Why is the temperature change smaller than in
Demonstration 1? |
Record Temperatures Below: Starting Temperature: ______ Ending Temperature: ______ Change in Temperature: ______ |
Calculation 2: Using the actual starting and
ending temperatures of the system, determine the amount of heat that was
transferred to the water and aluminum in this process. The specific heat of water is 4186 J/kg Cº
and the specific heat of aluminum is 900 J/kg Cº. Heat Transferred = Q2 = Only after you have made your calculation, click here to compare it. |
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Demonstration 3: Now 50 grams
of aluminum are used to replace 50
grams of water, so that we start with 50 grams of water and 50 grams of
aluminum, both initially at room temperature in a perfectly insulated
cup. Heat energy is transferred to the
water and aluminum at the same steady rate as in the previous demonstrations
for 80 seconds, and then no more heat energy is transferred. (Again, the specific heat of water is 4186
J/kg Cº and the specific heat of aluminum is 900 J/kg Cº.) Do
you think that the change in temperature of the water plus aluminum will be
greater than, less than or equal to the change in temperature in
Demonstration 2 (where there was 100 grams water with 50 grams aluminum)? Do
you think that the change in temperature of the water plus aluminum will be
greater than, less than or equal to the change in temperature in
Demonstration 1 (where there was only 100 grams of water)? Only after you have made your predictions, click here to download and view a video of the
experiment. Then, click here to view the
starting and final temperatures. Compare the results to your predictions and
explain any differences. Why is the temperature
change larger than in Demonstration 1? Demonstration 2? |
Record Temperatures Below: Starting Temperature: ______ Ending Temperature: ______ Change in Temperature: ______ |
Calculation 3: Now let’s assume that the
same amount of heat energy will be transferred to this system as was
transferred in the previous two demonstrations. Since you probably found that these amounts
were slightly different, first take the average of the two amounts of heat: Heat
Transferred = Q3 = 1/2 (Q1 + Q2) Since
this heat will be transferred to the water and the aluminum, we can write
down an equation that will allow us to predict the theoretical change in
temperature of our system. The amount
of heat energy gained by each part of the system will be of the form mc /\ T. In the space below, write down an equation
which sets Q3 equal to the heat energy gained by the water and
aluminum, and then solve it for /\
T: Q3 = /\T = How
does the theoretical numerical answer you just calculated compare with the
actual change in temperature of the system?
Can you think of a possible explanation for any discrepancy? Only after you have made your calculation, click here to compare it. |