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NEET 2006 (AIPMT) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from NEET 2006 (AIPMT)

Q81NEET 2006 · AIPMTMedium

Heat Engines, Refrigerators & Second Law

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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Answer: (d)

$T_1 = \frac{T_2}{1 - \eta} = \frac{300}{0.6} = 500\text{ K}$
New efficiency $= 0.4 \times 1.5 = 0.6$ , so $T_1' = \frac{300}{0.4} = 750\text{ K}$ .
Increase $= 750 - 500 = 250\text{ K}$
So the answer is (d) 250 K.

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