Discussion
A Carnot engine is working between 127 °C and 27 °C. Keeping the sink temperature unaltered, the temperature at which the source has to be kept so as to double its efficiency is
Answer: C
The source must be kept at 327 °C.
With temperatures in kelvin, the efficiency is \(\eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{400} = 0.25\). Doubling it to 0.5 with the sink still at 300 K:
\[ \begin{aligned} 1 - \frac{300}{T_{1}'} &= 0.5 \\ &\Rightarrow T_{1}' \\ &= 600\ \text{K} \\ &= 327^{\circ}\text{C} \end{aligned} \]
600 °C is the trap of forgetting to convert back from kelvin, and the other temperatures do not give an efficiency of 0.5.
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