8. A 0.26 kg sample of an alloy is heated to 450°C. It is then placed in 0.52 kg of water that is at 20.0° C which is contained in a 0.18 kg aluminum calorimeter cup. If the final temperature of the system (alloy, cup and water) is 28.5°C then what is the specific heat of the alloy?
8. A 0.26 kg sample of an alloy is heated to 450°C. It is then placed in 0.52 kg of water that is at 20.0° C which is contained in a 0.18 kg aluminum calorimeter cup. If the final temperature of the system (alloy, cup and water) is 28.5°C then what is the specific heat of the alloy?
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![**Problem 8: Calorimetry and Specific Heat Calculation**
A 0.26 kg sample of an alloy is heated to 450°C. It is then placed in 0.52 kg of water that is at 20.0°C, which is contained in a 0.18 kg aluminum calorimeter cup. If the final temperature of the system (alloy, cup, and water) is 28.5°C, what is the specific heat of the alloy?
**Approach to Solution:**
To find the specific heat of the alloy, apply the principle of conservation of energy. The heat lost by the alloy will be equal to the heat gained by the water and the calorimeter cup.
Use the formula:
\[ Q = mc\Delta T \]
where \( Q \) is the heat transfer, \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature.
For the alloy:
- Initial temperature \( T_i = 450°C \)
- Final temperature \( T_f = 28.5°C \)
- Change in temperature \( \Delta T = T_f - T_i \)
For the water and the aluminum cup, do similar calculations to find the heat gained.
Finally, solve for the specific heat of the alloy using:
\[ c_{\text{alloy}} = \frac{Q_{\text{alloy}}}{m_{\text{alloy}} \times \Delta T_{\text{alloy}}} \]
No graphs or diagrams are associated with this problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe591ff8b-7fbf-41eb-ab1a-65fc8e854653%2F8bf49615-9724-4250-9990-e4e0283295d8%2F1yx9mbl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 8: Calorimetry and Specific Heat Calculation**
A 0.26 kg sample of an alloy is heated to 450°C. It is then placed in 0.52 kg of water that is at 20.0°C, which is contained in a 0.18 kg aluminum calorimeter cup. If the final temperature of the system (alloy, cup, and water) is 28.5°C, what is the specific heat of the alloy?
**Approach to Solution:**
To find the specific heat of the alloy, apply the principle of conservation of energy. The heat lost by the alloy will be equal to the heat gained by the water and the calorimeter cup.
Use the formula:
\[ Q = mc\Delta T \]
where \( Q \) is the heat transfer, \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature.
For the alloy:
- Initial temperature \( T_i = 450°C \)
- Final temperature \( T_f = 28.5°C \)
- Change in temperature \( \Delta T = T_f - T_i \)
For the water and the aluminum cup, do similar calculations to find the heat gained.
Finally, solve for the specific heat of the alloy using:
\[ c_{\text{alloy}} = \frac{Q_{\text{alloy}}}{m_{\text{alloy}} \times \Delta T_{\text{alloy}}} \]
No graphs or diagrams are associated with this problem.
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