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Calculate the specific heat of a metal (J/ °C g) if the temperature change of a 9.250 g of copper is 172.5 °C; and the temperature change of 72.5 g of water (AT water is 5.75 °C when the hot copper piece is mixed with cold water in the calorimeter. (Water has specific heat of 4.184 J °C g)

Calculate the specific heat of a metal (J/ °C g) if the temperature change of a 9.250 g of copper is 172.5 °C; and the temperature change of 72.5 g of water (AT water is 5.75 °C when the hot copper piece is mixed with cold water in the calorimeter. (Water has specific heat of 4.184 J °C g)

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Problem Statement: Calculation of Specific Heat of a Metal** Calculate the specific heat of a metal (J/°C·g) if the temperature change of a 9.250 g of copper is 172.5 °C, and the temperature change of 72.5 g of water (ΔT_water) is 5.75 °C when the hot copper piece is mixed with cold water in the calorimeter. (Water has a specific heat of 4.184 J/°C·g). --- ### Explanation and Calculation: Given: - Mass of copper (m_copper) = 9.250 g - Temperature change of copper (ΔT_copper) = 172.5 °C - Mass of water (m_water) = 72.5 g - Temperature change of water (ΔT_water) = 5.75 °C - Specific heat of water (C_water) = 4.184 J/°C·g We need to calculate the specific heat of copper (C_copper). ### Formula: The heat gained or lost by a substance can be calculated using the formula: \[ q = m \cdot C \cdot \Delta T \] For copper: \[ q_{copper} = m_{copper} \cdot C_{copper} \cdot \Delta T_{copper} \] For water: \[ q_{water} = m_{water} \cdot C_{water} \cdot \Delta T_{water} \] In a calorimeter, the heat lost by the metal (copper) is equal to the heat gained by the water: \[ q_{copper} = q_{water} \] \[ m_{copper} \cdot C_{copper} \cdot \Delta T_{copper} = m_{water} \cdot C_{water} \cdot \Delta T_{water} \] ### Calculation: \[ 9.250 \, \text{g} \cdot C_{copper} \cdot 172.5 \, \text{°C} = 72.5 \, \text{g} \cdot 4.184 \, \text{J/°C·g} \cdot 5.75 \, \text{°C} \] \[ 1594.125 \, \text{g·
Transcribed Image Text:**Problem Statement: Calculation of Specific Heat of a Metal** Calculate the specific heat of a metal (J/°C·g) if the temperature change of a 9.250 g of copper is 172.5 °C, and the temperature change of 72.5 g of water (ΔT_water) is 5.75 °C when the hot copper piece is mixed with cold water in the calorimeter. (Water has a specific heat of 4.184 J/°C·g). --- ### Explanation and Calculation: Given: - Mass of copper (m_copper) = 9.250 g - Temperature change of copper (ΔT_copper) = 172.5 °C - Mass of water (m_water) = 72.5 g - Temperature change of water (ΔT_water) = 5.75 °C - Specific heat of water (C_water) = 4.184 J/°C·g We need to calculate the specific heat of copper (C_copper). ### Formula: The heat gained or lost by a substance can be calculated using the formula: \[ q = m \cdot C \cdot \Delta T \] For copper: \[ q_{copper} = m_{copper} \cdot C_{copper} \cdot \Delta T_{copper} \] For water: \[ q_{water} = m_{water} \cdot C_{water} \cdot \Delta T_{water} \] In a calorimeter, the heat lost by the metal (copper) is equal to the heat gained by the water: \[ q_{copper} = q_{water} \] \[ m_{copper} \cdot C_{copper} \cdot \Delta T_{copper} = m_{water} \cdot C_{water} \cdot \Delta T_{water} \] ### Calculation: \[ 9.250 \, \text{g} \cdot C_{copper} \cdot 172.5 \, \text{°C} = 72.5 \, \text{g} \cdot 4.184 \, \text{J/°C·g} \cdot 5.75 \, \text{°C} \] \[ 1594.125 \, \text{g·
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