Carnot Engine Temperature Change

Question

The efficiency of a Carnot engine is 40 percent when the sink temperature is 300 K, so to what temperature must the source be raised to increase the efficiency to 50 percent?

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Using the framework of NCERT Class 11, Chapter 12 (Thermodynamics), the efficiency of a Carnot engine is governed solely by the absolute temperatures through $\eta = 1 - \dfrac{T_C}{T_H}$, and in this problem the sink temperature $T_C$ is held fixed at 300 K while the source temperature is varied. First find the original source temperature from the 40 percent condition: $0.40 = 1 - \dfrac{300}{T_H}$ gives $\dfrac{300}{T_H} = 0.60$, so $T_H = \dfrac{300}{0.60} = 500$ K. Now impose the new requirement of 50 percent efficiency: $0.50 = 1 - \dfrac{300}{T_H'}$ gives $\dfrac{300}{T_H'} = 0.50$, so $T_H' = \dfrac{300}{0.50} = 600$ K. Thus the source must be raised from 500 K to 600 K. The option 500 K is wrong because that is the original source temperature, which only yields 40 percent efficiency. The option 450 K is wrong because lowering the source would reduce efficiency below 40 percent, the wrong direction entirely. The option 550 K is wrong because $1 - 300/550 = 0.455$, which is not 50 percent. A plausibility check confirms the logic: with a fixed sink, raising the source temperature must raise efficiency, and indeed 600 K > 500 K corresponds to 50 percent > 40 percent, consistent in both direction and magnitude.

Carnot Engine Temperature Change

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