Vocabulary: calorie, calorimeter, joule, specific heat capacity
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
1. The Latin word calor means “heat,” and meter comes from the Greek word meaning “to
measure.” What do you think a calorimeter does? _
measuring the heat of chemical reactions or physical changes as well as heat
capacity.
2. Where have you heard the word calorie before? What do you think a calorie is?
Calorie, a unit of energy or heat variously defined. the calorie was originally defined as
the amount of heat required at a pressure of 1 standard atmosphere to raise the
temperature of 1 gram of water 1 degree Celsius.
Gizmo Warm-up
A calorimeter is an insulated container filled with a liquid,
usually water. When a hot object is placed in the calorimeter,
heat energy is transferred from the object to the water and the
water heats up.
Calorimeters can be used to find a substance’s specific heat
capacity. You will use the Calorimetry Lab Gizmo to determine
the specific heat capacities of various substances.
1. On the SIMULATION pane, select Copper. Use the slider to set its Mass to 200 g. Set the
Water mass to 200 g. Check that the Water temp is set to 30.0 °C and the copper’s Temp
is 90 °C. Select the GRAPH tab, and click Play (
).
A.
What was the Final temperature of the copper and the water?
35.06 degrees celsius
B.
How much did the temperature of the copper change?
54.94 degrees celsius
C.
How much did the temperature of the water change?
5.06 degrees celsius
2. Specific heat capacity can be described as a substance’s resistance to temperature
changes. Which substance has a greater specific heat capacity, copper or water? Explain.
Water has a higher specific heat capacity because it has less of a tendency to change in
temperature. copper with the temperature of 60 degrees Celsius. and the same mass as
that of the water in which is was placed, cooled off to 23.4 degrees Celsius.
Activity A:
Get the Gizmo ready:
Heat transfer
 Click Reset (
).
Question: What factors determine how heat energy transfers between objects?
1. Predict: In the Gizmo warm-up, you saw how 200 g of 90 °C copper transfers heat to 200 g
of 30.0 °C water.
A. How do you think increasing the water’s mass would affect the final temperature?
The larger the amount of water, the higher the heat capacity.
B. How do you think decreasing the copper’s mass would affect the final temperature?
Decreasing the temperature of the copper made the final temperature
decrease.
C. How do you think increasing or decreasing the copper’s initial temperature would
affect the final temperature?
I believe increasing or decreasing the copper’s initial temperature would do
the same to the final temperature.
2. Collect data: Use the Gizmo to determine the final temperature for each set-up listed below.
Record your results in the tables. In the first table, you experiment with changing the water’s
mass. In the second table, you change the copper’s mass. In the third table, you change the
initial temperature of the copper. The first row of each table has been completed for you.
Copper
Water
Initial Temp. (°C)
Mass (g)
Initial Temp. (°C)
Mass (g)
Final Temp.
(°C)
90 °C
200 g
30.0 °C
200 g
34.96 °C
90 °C
200 g
30.0 °C
2,000 g
30.54 °C
90 °C
200 g
30.0 °C
200 g
34.96 °C
90 °C
20 g
30.0 °C
200 g
30.54 °C
90 °C
200 g
30.0 °C
200 g
34.96 °C
100 °C
200 g
30.0 °C
200 g
35.79 °C
50 °C
200 g
30.0 °C
200 g
31.65 °C
3.
4.
5.
(Activity A continued on next page)
Activity A (continued from previous page)
3. Analyze: For each factor listed in the chart below, explain how the final temperature was
changed and why you think that change occurred.
A. What was the effect of increasing the water’s mass?
Final temp decreases.
B. What was the effect of decreasing the copper’s mass?
Final temp decreases.
C. What was the effect of changing the initial temperature of the copper?
Increasing the temperature of the copper made the final temperature increase and
decreasing the temperature of the copper made the final temperature decrease .
4. Draw conclusions: The amount that the water’s temperature increases depends on the
mass of the water and the amount of heat energy in the copper.
A. How does changing the initial mass of the copper affect how much heat energy it
has?
The more copper, the more heat energy.
B. How does changing the initial temperature of the copper affect how much heat
energy it has?
The higher temperature, the more heat energy.
5. Apply: Many gyms and health clubs have steam saunas, which are small steam-filled rooms.
Traditionally, steam saunas have a container of heated rocks. A small ladle of water is
poured on the rocks in order to make the steam.
A. Use what you have learned so far about heat transfer to explain how hot rocks can
be used to make steam?
When the water comes in contact with the hot rocks the heat energy of the
rocks is transferred to the water causing it to become steam.
B. Why do you think only a small ladle-full of water is poured on the rocks at one time?
The heat vaporization of water is very high with small amounts of water you can vaporize
the water as it hits the rocks into steam.
Activity B:
Specific heat
Get the Gizmo ready:
​ Click Reset.
​ Deselect Copper, and select Granite.
Question: How can you compare the specific heat capacities of various substances?
1. Explain: How do you think you can use the calorimeter to compare the specific heat
capacities of the substances listed on the Gizmo?
You can compare how much different substances will heat up the water in the
calorimeter.
2. Predict: Which substance do you think will have the highest specific heat capacity? Why?
39.35ºC 40.59ºC Granite had the greatest change in water temperature showing the
specific heat is higher than copper and lead. Granite, copper, lead give you the high and
low specific heat energy.
3. Experiment: Use the Gizmo to determine the final temperature for each set-up listed below.
Record your results in the table. The first row has been completed for you.
Substance
Substance initial
temp. (°C)
Substance
mass
Water initial
temp. (°C)
Water
mass
Final
temp. (°C)
Copper
90 °C
200 g
30.0 °C
200 g
34.96 °C
Granite
90 °C
200 g
30.0 °C
200 g
39.59 °C
Lead
90 °C
200 g
30.0 °C
200 g
31.75 °C
4. Analyze: Of the three substances, which caused the largest temperature change in the
water? What does this indicate about its relative specific heat capacity?
Granite. It indicates that Granite has a higher specific heat than copper and lead.
5. Interpret: Remember that specific heat capacity is a measure of a substance’s resistance to
temperature change. The more resistant a substance is to temperature change, the higher is
its specific heat capacity.
Rank the three substances in order of their specific heat capacities, from highest to lowest.
Granite, copper, lead.
(Activity B continued on next page)
Activity B (continued from previous page)
6. Predict: How do you think the specific heat capacity of ice will compare to that of copper,
granite, and lead?
I think the specific heat capacity of ice will be higher
7.
Experiment: Deselect Lead, and select Ice. Use the default values for Temp (-30 °C) and
Mass (50 g). Set the Water temp to 60 °C and the Water mass to 200 g. Click Play.
A. What was the final temperature?
32.26 °C
B. What do you think is happening when the ice line on the graph is at 0 °C for a
long period of time? Why do you think the line disappears after that? The ice
melts. Therefore the line disappears.
C. How much of a temperature change did the water experience? _27.74 degrees
celsius. _
D. How does this change in the water’s temperature compare to the change caused by
the other substances you tested?
The temperature change is much greater
than the others.
8. Extend your thinking: A lot of energy is needed to heat a substance with a high specific
heat capacity. However, even more energy is needed to cause a phase change (such as
the melting of ice). Click Reset. Set the ice’s Temp to -100 °C and its Mass to 50 g. Set the
Water temp to 50 °C and Water mass to 200 g. Click Play.
A. What was the final temperature?
15.60 degrees Celsius
B. Do you think all the ice melted? Explain.
_Yes because the line disappeared.
C. Look at the GRAPH. The graph shows two separate stages: the heating of the ice
and then the melting of the ice. How much did the water’s temperature change while
the ice was heating? How much did it change while the ice was melting?
The water temperature changed by 14 °C when the ice was heating and
the water temperature changed by 22 °C when the ice was melting.
D. How did this experiment demonstrate ice’s high specific heat capacity?
During the process of just heating the ice, it caused a greater
change in temperature than any of the other substances.
Activity C:
Calculating
specific heat
And Extra Credit
Opportunity
Get the Gizmo ready:
 Click Reset.
Introduction: The specific heat capacity of a substance is the amount of energy needed to
change the temperature of that substance by 1 °C. Specific heat capacity can be calculated
using the following equation:
q = mc∆T
In the equation q represents the amount of heat energy gained or lost (in joules), m is the mass
of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and ∆T
is the temperature change of the substance (in °C).
Goal: Calculate the specific heat capacities of copper, granite, lead, and ice.
1. Solve: When you mix two substances, the heat gained by one substance is equal to the heat
lost by the other substance. Suppose you place 125 g of aluminum in a calorimeter with
1,000 g of water. The water changes temperature by 2 °C and the aluminum changes
temperature by –74.95 °C.
A.
Water has a known specific heat capacity of 4.184 J/g °C. Use the
specific heat equation to find out how much heat energy the water gained (q).
∆T Q= 1000 x 4.18 x 2Q = 8360 J8360 J; 8.4 kJ
B.
Assume that the heat energy gained by the water is equal to the heat
energy lost by the aluminum. Use the specific heat equation to solve for the specific
heat of aluminum. (Hint: Because heat energy is lost, the value of q is negative.)
_0.893 J/g oC_
Aluminum’s accepted specific heat value is 0.900 J/g °C. Use this value to check your work.
2. Calculate: Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C.
A. What is the final temperature?
21.45
B. Calculate the temperature change of each substance by subtracting the initial
temperature from the final temperature.
∆Twater:
C.
1.45
∆Tcopper:
-78.55
How much heat energy (q) did the water gain?
5941.28J
D.
Now solve for the specific heat (c) of copper 0.378 J/g oC
(Activity C continued on next page) Activity C
(continued from previous page)
3. Calculate: Use the Gizmo to mix 200 g of granite at 100 °C with 1,000 g of water at 20 °C.
A. What is the final temperature? 22.91
B. Calculate the temperature change of each substance by subtracting the initial
temperature from the final temperature.
Twater: 2.91
Tgranite: -77.09
C. How much heat energy (q) did the water gain? 12,259.12J
D.
Now solve for the specific heat (c) of granite: 0.795 J/g oC
E.
Repeat steps A through D to find the specific heat (c) of lead: 0.130 J/g oC
4. Challenge: Use the specific heat capacity that you calculated for granite to determine how
many grams of granite at the initial temperature of 80 °C must mix with 3,000 g of water at
the initial temperature of 20 °C to result in a final system temperature of 20.45 °C. (Hint:
Start by calculating how much heat energy is needed to change the water’s temperature by
0.45 °C). Show your work. Use the Gizmo to check your answer.
Mass of granite = 119.31 g
5. Extend your thinking: In addition to calculating specific heat capacities, some calorimeters
can be used to determine how much energy is in food. The energy in food is usually
expressed in calories or kilocalories (Calories). A calorie is the amount of energy needed
to change the temperature of 1 g of water by 1 C. There are 1,000 calories in a Calorie.
A.
How many joules are in 1 calorie? (The specific heat of water is 4.184 J/g °C.)
4.184 J
B. Suppose a snack bar is burned in a calorimeter and heats 2,000 g water by 20 °C.
How much heat energy was released? (Hint: Use the specific heat equation.) Give
your answer in both joules and calories.
167,369 J (40,000 calories) The specific heat of water is 1 calorie per
gram-ºC. The heat required to raise the temperature of 2000 g water by 20ºC
is 1.00 cal/g-C x 2000 g x 20ºC = 40000 calories or 167,369 joules.
C. How many kilocalories (Calories) does the snack bar contain? It has 1000
cal/kcal. #kcal = 1 kcal/1000 cal x 40000 cal = 40kca