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A reversible engine converts one –sixth of the heat input into work. When the temperature of the sink is reduced by 620C,  that efficiency of the engine is doubled. Now the temperature of the source and sink are

a
800C,  370C
b
950C,   280C
c
900C,   370C
d
990C,   370C

detailed solution

Correct option is D

w=16(Q),   η=wQ=16 1−T2T1=16       −(1) 1−(T2−62)T1=13       −(2)solve (1) and (2) we get 1−(T2)T1+62T1=13 16+62T1=13 16=62T1 T1=372K=372-273=99°C 1−(T2)T1=16 (T2)T1=1-16=56 T2=56372=310K=310-273=37°C

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