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InterviewSolution
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A reversible heat engine A(based on carnot cycle ) absorbs heat from a reservoir at 1000 K and rejects heat to a reservoir at `T_(2)`. A second reversible engine B absorbs, the same amount of heat as rejected by the engine A, from the reservoir at `T_(2)` and rejects energy to a reservoir at 360 K. If the efficienices of engines A and B are the same then the temperature `T_(2)` is .A. 680 KB. 640 KC. 600 KD. none |
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Answer» Correct Answer - D For heat engines A: `eta= (1000- T_(2))/(1000) rArreta_(B) = (T_(2) - 360)/(T_(2))` `eta_(A) = eta_(B)` `So(1000-T_(2))/(1000)=(T_(2) - 360)/(T_(2)) = T_(2) = sqrt(360 xx 1000)=600 K` |
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