Question:
The efficiency of a Carnot heat engine is 25%. If the absolute temperature of the sink is increased by 10%, then the efficiency of the engine becomes:
The efficiency of a Carnot heat engine is 25%. If the absolute temperature of the sink is increased by 10%, then the efficiency of the engine becomes:
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Carnot efficiency depends only on reservoir temperatures — increasing sink temperature reduces efficiency.
Be careful: “increase by 10%” means multiply by 1.10 (relative), not add 10 K.
Always work with absolute temperatures (Kelvin).
Small relative changes in temperatures can noticeably change efficiency.
Be careful: “increase by 10%” means multiply by 1.10 (relative), not add 10 K.
Always work with absolute temperatures (Kelvin).
Small relative changes in temperatures can noticeably change efficiency.
Updated On: Mar 6, 2026
- $27.5%$
- $37.5%$
- $17.5%$
- $21.5%$
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The Correct Option is C
Solution and Explanation
• Carnot efficiency $\eta = 1 - \dfrac{T_c}{T_h}$, where $T_c$ is sink temperature, $T_h$ source temperature (absolute scales).
• Given $\eta = 0.25 \Rightarrow \dfrac{T_c}{T_h} = 0.75$.
• Sink temperature increased by 10%: $T_c' = 1.10\,T_c = 1.10 \times 0.75\,T_h = 0.825\,T_h$.
• New efficiency $\eta' = 1 - \dfrac{T_c'}{T_h} = 1 - 0.825 = 0.175 = 17.5%$.
• Hence the new efficiency is 17.5%.
• Given $\eta = 0.25 \Rightarrow \dfrac{T_c}{T_h} = 0.75$.
• Sink temperature increased by 10%: $T_c' = 1.10\,T_c = 1.10 \times 0.75\,T_h = 0.825\,T_h$.
• New efficiency $\eta' = 1 - \dfrac{T_c'}{T_h} = 1 - 0.825 = 0.175 = 17.5%$.
• Hence the new efficiency is 17.5%.
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