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A Carnot engine whose sink is at a temperature of 300 K has an efficiency of 40%, By how much should the temperature of the source be increased so as to increase the efficiency to 60%

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380 K

325 K

250 K

275 K

answer is C.

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Detailed Solution

$\frac{40}{100} = 1 â \frac{300}{T_{1}} or T_{1} = 500 K$
For 60% efficiency
$\frac{60}{100} = 1 â \frac{300}{T_{1}} or T_{1} = 750 K$
So the temperature should be increased by
750 - 500 = 250 K

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