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Student Exploration Calorimetry Lab
Accounting Information system 10th edition hall (9781337619202)
Bank Street College of Education
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Name: Date:
Student Exploration: Calorimetry Lab
Directions: Follow the instructions to go through the simulation. Respond to the questions and prompts in the orange boxes.
Vocabulary: calorie, calorimeter, joule, specific heat capacity
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
- The Latin word calor means “heat,” and meter comes from the Greek word meaning “to measure.” What do
you think a calorimeter does?
To measure Heat
- Where have you heard the word calorie before? What do you think a calorie is?
When looking at the nutritional facts on the back of a food container
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.
- 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 °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? 34 °C
B. How much did the temperature of the copper change? 55 °C
C. How much did the temperature of the water change? 4 °C
- 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, because its temperature only changed 4 °C compared to 55 °C
Activity A:
Heat transfer
Get the Gizmo ready:
● Click Reset ( ).
Question: What factors determine how heat energy transfers between objects?
- Predict: In the Gizmo warm-up, you saw how 200 g of90 °C copper transfers heat to 200 g of 30 °C water.
A. How do you think increasing the water’s mass would affect the final temperature?
I think that will make the temperature drop
B. How do you think decreasing the copper’s mass would affect the final temperature?
I think temperature will also drop
C. How do you think increasing or decreasing the copper’s initial temperature would affect the final temperature?
It wouldn't affect it that much since the specific heat capacity is larger on water
- 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 Final Temp. Initial Temp. (°C) Mass (g) Initial Temp. (°C) Mass (g) (°C) 3. 90 °C 200 g 30 °C 200 g 34 °C 90 °C 200 g 30 °C 2,000 g 30. 4. 90 °C 200 g 30 °C 200 g 34 °C 90 °C 20 g 30 °C 200 g 30. 5. 90 °C 200 g 30 °C 200 g 34 °C 100 °C 200 g 30 °C 200 g 35 °C 50 °C 200 g 30 °C 200 g 31 °C
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?
- 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 compart how much different substances will heat up the water in the calorimeter
- Predict: Which substance do you think will have the highest specific heat capacity? Why?
39ºC40ºC 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.
- 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 °C 200 g 34 °C Granite 90 °C 200 g 30 °C 200 g 39. Lead 90 °C 200 g 30 °C 200 g 31.
- 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
- 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
- Predict: How do you think the specific heat capacity of ice will compare to that of copper, granite, and lead?
Less
- 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?
28.
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?
Evaporates
C. How much of a temperature change did the water experience?
It went down then stayed the same
D. How does this change in the water’s temperature compare to the change caused by the other substances you tested?
The two lines don't meet
- 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? 13.
B. Do youthinkall the ice melted? Explain.
Yes because it is way above 0
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?
Activity C:
Calculating specific heat
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.
- 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 °C.
A. Water has a known specific heat capacity of 4 J/g °C. Use the specific heat equation to find out how much heat energy the water gained ( q ).
∆T Q= 1000 x 4 x 2Q = 8360 J8360 J; 8 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.
Aluminum’s accepted specific heat value is 0 J/g °C. Use this value to check your work.
- 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.
B. Calculate the temperature change of each substance by subtracting the initial temperature from the final temperature.
∆Twater : 1.. ∆Tcopper : 20
C. How much heat energy ( q ) did the water gain? 18..
D. Now solve for the specific heat ( c ) of copper:
it is 20ºC
- 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?
21.
B. Calculate the temperature change of each substance by subtracting the initial temperature from the final temperature.
∆Twater : 1 ∆Tgranite : 30
C. How much heat energy ( q ) did the water gain?
D. Now solve for the specific heat ( c ) of granite: Same as above
E. Repeat steps A through D to find the specific heat ( c ) of lead: Done
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 °C. (Hint: Start by calculating how much heat energy is needed to change the water’s temperature by 0 °C). Show your work. Use Gizmo to check your answer.
specific heat of granite = .79 J/g/k
let the mass of granite = m
heat lost by granite = heat gained by water
heat lost = mass x specific heat x drop in temp
= m x .79 x (80-20)
heat gained by water
= 3000 x 4 x (20-20)
heat lost by granite = heat gained by water
m x .79 x 59 = 3000 x v x.
m = 120 g
Mass of granite = 120 g
- Extend your thinking: In addition to calculating specificheat 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 J/g °C.)
One calorie is equal to 4 joules
Student Exploration Calorimetry Lab
Course: Accounting Information system 10th edition hall (9781337619202)
University: Bank Street College of Education
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