A carnot engine whose sink is at 300K has an efficiency of 40%. By how much should the temperature of the source be increased, so as to increase the efficiency by 50% of original efficiency?

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325 K

250 K

275 K

380 K

answer is C.

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Detailed Solution

$\begin{array}{l} 1 - \frac{300}{T_{1}} = 0.4 \Rightarrow T_{1} = 500 K \\ Now 1 - \frac{300}{T_{1}^{1}} = 1 .5 \times 0 .4 = 0 .6 \Rightarrow T_{1}^{1} = 750 K \\ ∴ T_{1}^{1} - T_{1} = 250 K \end{array}$

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