Discussion
To increase the efficiency of a Carnot engine working between a source at 500 K and a sink at 300 K by 10 % , with the same sink temperature the source temperature should be increased to
Answer: B
The source temperature should be raised to 600 K.
The present efficiency is \(1 - \frac{300}{500} = 0.4\), that is 40%. Increasing it by 10% here means by 10 percentage points, to 50%.
\[ \begin{aligned} 1 - \frac{300}{T_1} &= 0.5 \\ T_1 &= 600\ \text{K} \end{aligned} \]
Reading '10%' as a relative increase (to 44%) gives about 536 K, which is not offered, so the percentage-point reading is intended. 550 K is a 10% rise in the source temperature itself.
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