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Heat engine

Question

An ideal heat engine working between temperatures T1 and T2(T1>T2) has efficiency  η . If both temperatures are lowered by 100K each, The new efficiency of the engine will be

Moderate
Solution

η=T1T2T1η'=1T2100T1100=T1100(T2100)T1100=T1T2T1100>η



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