Q.

The temperature of the sink of an engine operating at an efficiency of 0.25 is reduced by 100° C. As a result, the efficiency of heat engine is doubled. Then, the temperature of the source is:

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a

150°C

b

222K

c

242°C

d

400 K

answer is D.

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Detailed Solution

Hence, η1=1−T2T2or 0.25=1−T2T2⇒14=1−T2T2T2T2=1−14=34     ….(i)According to question,η2=2η1, and T2'=T2−100∘C∴2×14=1−T2−100T1⇒1−12=T2−100T112=T2T1−100T1⇒34−12=100T1T1=400K
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