Q.

A heat engine has an efficiency n.Temperature of source and sink are each decreased by 100K. The efficiency of the engine

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a

Decreases

b

Remains constant

c

Increases

d

Becomes one

answer is C.

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Detailed Solution

efficiency of heat engine=η=1−T2T1⇒η=T1−T2T1⇒η=T1−T2T1temperature of source reduced by 100=T1−100temperature of sink reduced by 100=T2−100new efficiency is η1=T1−100−T2−100T1−100=T1−T2T1−100>ηη1=T1−T2-200T1−100When T1 and T2 both are decreased by 100 K eachT1−T2 remains constant T1 decreases η1 increasesefficiency increases
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