Student Exploration: Calorimetry Lab

Name: Angelina Dicintio Date: July 10 2025 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.) 1.​ The Latin word calor means “heat,” and meter comes from the Greek word meaning “to measure.” What do you think a calorimeter does? I think a calorimeter measures the changes in heat in order to have a clear indication of the calories. 2.​ Where have you heard the word calorie before? What do you think a calorie is? I have heard of calories from food and know from chemistry that it is the amount of energy needed to raise the temperature. 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 °C B. How much did the temperature of the copper change? down by 54.94 °C C. How much did the temperature of the water change? up by 5.06 °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. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved Since the water's temperature only changed by 5.06 °C as opposed to copper's 54.94 °C, water has a higher specific heat capacity. Compared to the copper, the water was more resistant to temperature changes. 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? I believe that the final temperature will drop as the mass of the water increases. B.​ How do you think decreasing the copper’s mass would affect the final temperature? I believe that the final temperature will drop as the total amount of copper decreases. C.​ How do you think increasing or decreasing the copper’s initial temperature would affect the final temperature? The ultimate temperature would rise with an increase in the original temperature of the copper and drop with a reduction in the initial 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 35.06 °C 90 °C 200 g 30.0 °C 2,000 g 30.55 °C 90 °C 200 g 30.0 °C 200 g 35.06 °C 90 °C 20 g 30.0 °C 200 g 30.55 °C 90 °C 200 g 30.0 °C 200 g 35.06 °C 100 °C 200 g 30.0 °C 200 g 35.90 °C 50 °C 200 g 30.0 °C 200 g 31.69 °C 3.​ 4.​ 5.​ 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? ​ Because the water's volume was greater, the end temperature was lower. B.​ What was the effect of decreasing the copper’s mass? Because the reduced mass had less energy, there was originally less heat, which resulted in a lower final temperature. C.​ What was the effect of changing the initial temperature of the copper? Because of the increase in heat energy, the final temperature was greater when the initial temperature was raised, and vice versa when the initial temperature was dropped. 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? It will have more heat energy the more mass it contains. B.​ How does changing the initial temperature of the copper affect how much heat energy it has? The amount of heat energy will increase with the initial temperature. 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? The water vaporizes and produces steam as a result of the heat from the rocks. B.​ Why do you think only a small ladle-full of water is poured on the rocks at one time? Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved​ ​ If there is too much water, the rocks' temperature will be too greatly impacted and they will cool down rather than producing steam from their heat. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved​ ​ Get the Gizmo ready: Activity B: ●​ Click Reset. ●​ Deselect Copper, and select Granite. Specific heat 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? With the same mass and temperature, you can compare the outcomes of the various substances for each one. 2.​ Predict: Which substance do you think will have the highest specific heat capacity? Why? The material that reaches the highest final temperature is the one with the highest specific heat capacity. Since ice is water in a different phase, I believe it will have the maximum specific heat capacity. 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 35.06 °C Granite 90 °C 200 g 30.0 °C 200 g 39.53 °C Lead 90 °C 200 g 30.0 °C 200 g 31.78 °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? Since lead had the lowest resistance to temperature change, it experienced the highest temperature change, suggesting that it had the lowest specific heat of all the substances. 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 6.​ Predict: How do you think the specific heat capacity of ice will compare to that of copper, granite, and lead? I believe that ice will have a greater specific head capacity than lead, granite, and copper. 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? 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? Since the ice has a melting point of 0°C, I believe the line vanishes because the substance underwent a phase transition into water throughout the melting process, leaving the substance unchanged. C.​ How much of a temperature change did the water experience? a 27.74 °C temperature change D.​ How does this change in the water’s temperature compare to the change caused by the other substances you tested? The substance with the highest specific heat (and greater resistance to change) is ice. Despite having a much smaller mass, ice altered the water's temperature much more than the other substances. 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? B. Do you think all the ice melted? Explain. 15.60 °C Since the line on the graph that shows ice still halted or vanished at the ice's melting threshold of 0°C, I do believe that all of the ice melted. 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? Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved​ ​ It was during the melting of the ice that the water's temperature dropped from 50°C to 25°C. As the ice melted, the water's temperature increased by 9.40 °C, from 25°C to 15.60 °C D. How did this experiment demonstrate ice’s high specific heat capacity? This experiment confirmed ice's high specific heat capacity because it showed ice's high resistance to temperature change Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved​ ​ Activity C: Get the Gizmo ready: Calculating specific heat ●​ 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 2.00 °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). q=? m = 1000 g c = 4.184 J/g°C ∆T = 2°C q = mc∆T q = (1000g) (4.184 J/g °C) (2 °C) the heat gained is 8368 J (8.368 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.) q = mc∆T ----------- (divided by) ∆T ∆T c = q / m∆T = - 8368 J / (125 g) (-74.95 °C) = 0.893 J/gºC 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. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved​ ​ A. What is the final temperature? B. Calculate the temperature change of each substance by subtracting the initial temperature from the final temperature. ∆Twater: 21.42 °C 21.42 °C - 20.0 °C = 1.42 °C ∆Tcopper: q = mc∆T = (1000g) (4.184 J/g °C) (1.42 °C) = 5941 J C. How much heat energy (q) did the water gain? q = mc∆T = (1000g) (4.184 J/g °C) (1.42 °C) = 5941 J D. Now solve for the specific heat (c) of copper: c = q / m∆T = - 5941 J / (200g) (-78.58 °C ) = 0.378 J/g °C 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? 22.93 °C B.​ Calculate the temperature change of each substance by subtracting the initial temperature from the final temperature. ∆Twater: 22.93 °C - 20 = 2.93 °C ∆Tgranite: 22.93 °C - 100 °C = - 77.07 °C C. How much heat energy (q) did the water gain? q = mc∆T = (1000g) (4.184 J/g °C) (2.93 °C) = 12259 J D. Now solve for the specific heat (c) of granite: c = q / m∆T = -12259 J / (200g) (-77.07) = 0.795 J/g °C E. Repeat steps A through D to find the specific heat (c) of lead: A: 20.49°C B: ∆T water: 20.49°C - 20 = 0.49°C ∆T lead: 20.49°C - 100°C = -79.51 Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved​ ​ C: q = mc∆T = (1000g) (4.184 J/g °C) (0.49 °C) = 2050 J D: c = q / m∆T = -2050 J / (200g) (-79.51) = 0.129 J/g °C specific heat (c) of lead: 0.129 J/g °C 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. heat gained by water = heat lost by granite q = -q (m water) (c water) (∆T) = (m granite) (c granite) (∆T) m granite = (m water) (c water) (∆T) / (c granite) (∆T) m granite = (3000g) (4.184 J/g °C) (0.45°C) / (0.795 J/g °C) (-59.55°C) m granite = 5648.4 g / 47.34225 m granite = 119.3 g Mass of granite = 119 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.) 1 cal / g°C = 4.184 J/g °C 1 cal = 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. q = mc∆T = (2000g) (4.184 J/g °C) (20°C ) = 167360 J = 40,000 cal Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved​ ​ C.​ How many kilocalories (Calories) does the snack bar contain? 40 kilocalories 167360 J = 40,000 calories 1000 calories = 1 kilocalories 40,000 calories = 40 kilocalories Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved