Gizmo Answer Key: Calorimetry Lab

Name: Catherine Bennett Date: 05-20-2022 Student Exploration: Calorimetry Lab Directions: Follow the instructions to go through the simulation. Respond to the questions andprompts in the orange boxes. 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? Heat measurement 2. Where have you heard the word calorie before? What do you think a calorie is? When examining the nutritional information on a food container's rear Gizmo Warm-up A calorimeter is an insulated container filled with a liquid, usually water. Whena hot object is placed in the calorimeter, heat energy is transferred from theobject 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? 34.96 °C B. How much did the temperature of the copper change? 55.04 °C C. How much did the temperature of the water change? 4.96 °C 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, as it only saw a 4.96 °C shift as opposed to a 55.04 °C change. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

Activity A: Heat transfer Get the Gizmo ready: ● 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? That, I believe, will cause the temperature to drop. B. How do you think decreasing the copper’s mass would affect the final temperature? I anticipate a decrease in temperature. C. How do you think increasing or decreasing the copper’s initial temperature would affect the final temperature? Given that water has a higher specific heat capacity, it wouldn't have a significantimpact. 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 firstrow of each table has been completed for you. Copper Water Final Temp. (°C) Initial Temp. (°C) Mass (g) Initial Temp. (°C) Mass (g) 3. 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 4. 90 °C 200 g 30.0 °C 200 g 34.96 °C 90 °C 20 g 30.0 °C 200 g 30.54 °C 5. 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 Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

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? Decreases B. What was the effect of decreasing the copper’s mass? Decreases by an amount equal to the mass of water. C. What was the effect of changing the initial temperature of the copper? Decreases 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? Heat energy increases with copper content. B. How does changing the initial temperature of the copper affect how much heat energy it has? More heat energy is produced at higher temperatures. 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 tomake the steam. A. Use what you have learned so far about heat transfer to explain how hot rocks can be used to make steam? Heat energy from the rocks is transferred to the water as it comes into contact withthem, turning the water into steam. B. Why do you think only a small ladle-full of water is poured on the rocks at one time? Small amounts of water can vaporize into steam as they hit the rocks due to the highheat that causes water to vaporize. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

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? The amount that various compounds will heat the water in the calorimeter can becompared. 2. Predict: Which substance do you think will have the highest specific heat capacity? Why? 39.35ºC 40.59ºC The biggest change in water temperature was observed in granite,indicating that the temperature is higher than that of copper and lead. Lead, copper, andgranite all provide both 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 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. Copper, lead, granite 6. Predict: How do you think the specific heat capacity of ice will compare to that of copper, granite, and lead? Less Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

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? 28.83 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 decreased and then remained unchanged. D. How does this change in the water’s temperature compare to the change caused by the other substances you tested? Two lines do not intersect here. 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? 13.39 B. Do you think all the ice melted? Explain. Yes, since it is significantly higher than 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? Howmuch did it change while the ice was melting? The amount between the two is about 15 D. How did this experiment demonstrate ice’s high specific heat capacity? The heating of the ice and then the melting of the ice. how much temperaturechange while the ice Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

Activity C: Calculatingspecific 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. 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 waterchanges 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 ). 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 heatenergy is lost, the value of q is negative.) 0.893 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? 21.42 B. Calculate the temperature change of each substance by subtracting the initial temperature from the final temperature. ∆T water : 1.49 ∆T copper : 20 C. How much heat energy ( q ) did the water gain? 18 D. Now solve for the specific heat ( c ) of copper: it is 20°C Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

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? 21.45 B. Calculate the temperature change of each substance by subtracting the initial temperature from the final temperature. ∆T water : 1.49 ∆T granite : 30 C. How much heat energy ( q ) did the water gain? 31.9 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 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 °Cto result in a final system temperature of 20.45 °C. (Hint: Start by calculating how much heat energy isneeded to change the water’s temperature by 0.45 °C). Show your work. Use the Gizmo to check youranswer. 5. specific heat of granite = .79 J/g/k6. let the mass of granite = m7. heat lost by granite = heat gained by water8. heat lost = mass x specific heat x drop in temp9. = m x .79 x (80-20.45)10. heat gained by water11. = 3000 x 4.186 x (20.45-20)12. heat lost by granite = heat gained by water13. m x .79 x 59.55 = 3000 x v.186 x .4514. m = 120.12 g Mass of granite = 120.12 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 joules are needed to make one calorie. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

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 andcalories. The specific heat of water is 1 calorie per gram-ºC. The heat required to raise thetemperature of 2000 g water by 20ºC is 1.00 cal/g-C x 2000 g x 20ºC = 40000 caloriesor 167360 joules C. How many kilocalories (Calories) does the snack bar contain? There are 1000 cal/kcal. #kcal = 1 kcal/1000 cal x 40000 cal = 40kcal Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved