7. 200g of copper is placed in an oven. The hot metal is removed from the oven and quickly added to 300g of water. The starting temperature of the water was 12°C. After adding the metal, the final temperature was 37°C. Assuming that the heat lost by the metal = heat gained by the water, what was the temperature of the oven? The specific heat of copper is 0.385 J/g/°C, and the specific heat of water is 4.18 J/g/°C.
7. 200g of copper is placed in an oven. The hot metal is removed from the oven and quickly added to 300g of water. The starting temperature of the water was 12°C. After adding the metal, the final temperature was 37°C. Assuming that the heat lost by the metal = heat gained by the water, what was the temperature of the oven? The specific heat of copper is 0.385 J/g/°C, and the specific heat of water is 4.18 J/g/°C.
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:**
7. 200g of copper is placed in an oven. The hot metal is removed from the oven and quickly added to 300g of water. The starting temperature of the water was 12°C. After adding the metal, the final temperature was 37°C. Assuming that the heat lost by the metal = heat gained by the water, what was the temperature of the oven? The specific heat of copper is 0.385 J/g/°C, and the specific heat of water is 4.18 J/g/°C.
**Solution Explanation:**
To solve this problem, we can use the principle of conservation of energy. The heat lost by the metal (copper) as it cools down will be equal to the heat gained by the water as it heats up.
1. **Define Variables:**
- \( m_{copper} = 200 \, \text{g} \) (mass of copper)
- \( m_{water} = 300 \, \text{g} \) (mass of water)
- \( T_{initial, water} = 12 \, \text{°C} \) (initial temperature of the water)
- \( T_{final} = 37 \, \text{°C} \) (final temperature of the system)
- \( c_{copper} = 0.385 \, \text{J/g/°C} \) (specific heat of copper)
- \( c_{water} = 4.18 \, \text{J/g/°C} \) (specific heat of water)
- \( T_{initial, copper} = T_{oven} \) (initial temperature of the copper, which is what we want to find)
2. **Heat Lost by Copper:**
\[
q_{copper} = m_{copper} \cdot c_{copper} \cdot (T_{initial, copper} - T_{final})
\]
3. **Heat Gained by Water:**
\[
q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water})
\]
4. **Equating Heat Lost to Heat Gained:**
\[
m_{copper} \cdot c_{copper}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09cc29b2-b8f1-46a4-8709-ee61d656c9a7%2F915938e7-82dd-48b6-81c1-a516737bbc7d%2Fnu7utb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
7. 200g of copper is placed in an oven. The hot metal is removed from the oven and quickly added to 300g of water. The starting temperature of the water was 12°C. After adding the metal, the final temperature was 37°C. Assuming that the heat lost by the metal = heat gained by the water, what was the temperature of the oven? The specific heat of copper is 0.385 J/g/°C, and the specific heat of water is 4.18 J/g/°C.
**Solution Explanation:**
To solve this problem, we can use the principle of conservation of energy. The heat lost by the metal (copper) as it cools down will be equal to the heat gained by the water as it heats up.
1. **Define Variables:**
- \( m_{copper} = 200 \, \text{g} \) (mass of copper)
- \( m_{water} = 300 \, \text{g} \) (mass of water)
- \( T_{initial, water} = 12 \, \text{°C} \) (initial temperature of the water)
- \( T_{final} = 37 \, \text{°C} \) (final temperature of the system)
- \( c_{copper} = 0.385 \, \text{J/g/°C} \) (specific heat of copper)
- \( c_{water} = 4.18 \, \text{J/g/°C} \) (specific heat of water)
- \( T_{initial, copper} = T_{oven} \) (initial temperature of the copper, which is what we want to find)
2. **Heat Lost by Copper:**
\[
q_{copper} = m_{copper} \cdot c_{copper} \cdot (T_{initial, copper} - T_{final})
\]
3. **Heat Gained by Water:**
\[
q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water})
\]
4. **Equating Heat Lost to Heat Gained:**
\[
m_{copper} \cdot c_{copper}
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