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
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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·](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b0fedb7-adbc-492c-aa14-9f5d12342848%2Fc143046d-c8c3-48a0-9d10-b5b3519b01ab%2Fhgkt0a_processed.jpeg&w=3840&q=75)
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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