Question:
Temperature of a cold reservoir of a Carnot engine is 127 °C. If the efficiency of the Carnot engine is 20%, find the temperature of the hot reservoir.
Temperature of a cold reservoir of a Carnot engine is 127 °C. If the efficiency of the Carnot engine is 20%, find the temperature of the hot reservoir.
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Carnot efficiency: \(\eta = 1 - T_C/T_H\). Always convert temperatures to Kelvin before calculations.
Updated On: Jul 18, 2026
- 500 °C
- 227 °C
- 273 °C
- 400 °C
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The Correct Option is B
Solution and Explanation
Step 1: Recall Carnot efficiency formula.
\[ \eta = 1 - \frac{T_C}{T_H} \]
where \(\eta\) is efficiency, \(T_C\) cold reservoir, \(T_H\) hot reservoir in Kelvin.
Step 2: Convert cold reservoir to Kelvin.
\[ T_C = 127 + 273 = 400 \, \text{K} \]
Step 3: Solve for \(T_H\).
\[ 0.20 = 1 - \frac{400}{T_H} \implies \frac{400}{T_H} = 0.8 \implies T_H = \frac{400}{0.8} = 500 \, \text{K} \]
Step 4: Convert back to Celsius.
\[ T_H = 500 - 273 = 227 \, °\text{C} \]
Step 5: Verify reasoning.
\(\eta = 1 - T_C/T_H = 1 - 400/500 = 0.2 = 20\%\), consistent with problem statement.
Step 6: Final conclusion.
Hence, the temperature of the hot reservoir is:
\[ \boxed{227 \, °\text{C}} \]
\[ \eta = 1 - \frac{T_C}{T_H} \]
where \(\eta\) is efficiency, \(T_C\) cold reservoir, \(T_H\) hot reservoir in Kelvin.
Step 2: Convert cold reservoir to Kelvin.
\[ T_C = 127 + 273 = 400 \, \text{K} \]
Step 3: Solve for \(T_H\).
\[ 0.20 = 1 - \frac{400}{T_H} \implies \frac{400}{T_H} = 0.8 \implies T_H = \frac{400}{0.8} = 500 \, \text{K} \]
Step 4: Convert back to Celsius.
\[ T_H = 500 - 273 = 227 \, °\text{C} \]
Step 5: Verify reasoning.
\(\eta = 1 - T_C/T_H = 1 - 400/500 = 0.2 = 20\%\), consistent with problem statement.
Step 6: Final conclusion.
Hence, the temperature of the hot reservoir is:
\[ \boxed{227 \, °\text{C}} \]
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