A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62∘C, the efficiency of the engine is doubled. The temperatures of the source and sink are
A
80°C, 37°C
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B
95°C, 28°C
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C
90°C, 37°C
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D
99°C, 37°C
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Solution
The correct options are C 90°C, 37°C D 99°C, 37°C Initially η=(1−T2T1)=WQ=16 ....(i) Finally η′=(1−T′2T1)=(1−(T2−62)T1)=1−T2T1+62T1 =η+62T1 ...(ii) it is given that η′=2η. Hence solving equation (i) and (ii) ⇒T1=372K=99∘C and T2=310K=37∘C