A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its 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

$η = 1 - \frac{T_{2}}{T_{1}} \Rightarrow 1 - \frac{300}{T_{1}} = 0 .4 \Rightarrow T_{1} = 500 K$
now, $η' = 0 .4 + 0 .4 \times \frac{50}{100} = 0 .6$
Therefore, $0 .6 = 1 - \frac{300}{500 + ΔT}$
$\Rightarrow 500 + ΔT = 750 \Rightarrow ΔT = 250 K$

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