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Gizmos Calorimetry Lab Activity C

Student Exploration: Calorimetry Lab Activity C 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 represe ... nts 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). 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) B. Assume that the heat energy gained by the water is equal to the heat ener7e 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.) 2. Calculate: Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C 3. Calculate: Use the Gizmo to mix 200 g of granite at 100 °C with 1,000 g of water at 20 °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 5. .3of 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. [Show More]