Question:
A Carnot engine operates between heat reservoirs differing in temperature by 80 °C. The efficiency of the Carnot engine is \(20\%\). The temperature of the cold reservoir is:
A Carnot engine operates between heat reservoirs differing in temperature by 80 °C. The efficiency of the Carnot engine is \(20\%\). The temperature of the cold reservoir is:
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For a Carnot engine, use \(\eta = 1 - \frac{T_c}{T_h}\) with temperatures in Kelvin. If given temperature difference, express \(T_h = T_c + \Delta T\) and solve for \(T_c\).
Updated On: Jun 19, 2026
- 440 K
- 400 K
- 250 K
- 320 K
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The Correct Option is D
Solution and Explanation
Step 1: Recall Carnot efficiency formula.
\[ \eta = 1 - \frac{T_c}{T_h} \] where \(\eta\) is efficiency, \(T_c\) and \(T_h\) are absolute temperatures of cold and hot reservoirs.
Step 2: Convert temperature difference.
The temperature difference is \(\Delta T = T_h - T_c = 80~^\circ\text{C} = 80~\text{K}\)
Step 3: Express \(T_h\) in terms of \(T_c\).
\[ T_h = T_c + 80 \]
Step 4: Solve for \(T_c\) using efficiency.
\[ \eta = 0.20 = 1 - \frac{T_c}{T_h} = 1 - \frac{T_c}{T_c + 80} \] \[ 0.20 = \frac{80}{T_c + 80} \Rightarrow T_c + 80 = \frac{80}{0.20} = 400 \] \[ T_c = 400 - 80 = 320~\text{K} \]
Step 5: Conclusion.
The temperature of the cold reservoir is 320 K.
\[ \eta = 1 - \frac{T_c}{T_h} \] where \(\eta\) is efficiency, \(T_c\) and \(T_h\) are absolute temperatures of cold and hot reservoirs.
Step 2: Convert temperature difference.
The temperature difference is \(\Delta T = T_h - T_c = 80~^\circ\text{C} = 80~\text{K}\)
Step 3: Express \(T_h\) in terms of \(T_c\).
\[ T_h = T_c + 80 \]
Step 4: Solve for \(T_c\) using efficiency.
\[ \eta = 0.20 = 1 - \frac{T_c}{T_h} = 1 - \frac{T_c}{T_c + 80} \] \[ 0.20 = \frac{80}{T_c + 80} \Rightarrow T_c + 80 = \frac{80}{0.20} = 400 \] \[ T_c = 400 - 80 = 320~\text{K} \]
Step 5: Conclusion.
The temperature of the cold reservoir is 320 K.
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