A reversible Carnot engine converts 1/6th of heat into input works. When the temperature of the sink is reduced by 62˚C then the efficiency is doubled. The temperature of the source and sink is:
Correct answer: D. 99˚C, 37˚C
- A. 80˚C, 37˚C
- B. 99˚C, 30˚C
- C. 99˚C, 25˚C
- D. 99˚C, 37˚C
Explanation
The efficiency of the engine = η = Work / Qinput.As W = ⅙ QinputAlso η = 1 - T2/T1 T1 and T2 are the temperatures of source and sink.⅙ = 1 - T2/T1T1 = 1.2 T2 (1)Now the sink temperature is reduced by 62oC, T2' = T2 - 62Also the new efficiency η' = 2 × ⅙ = 2/6η' = 1 - T2' /T12/6 = 1 - T2 - 62 / 1.2 T2T2 = 310 KT2 = 310 - 273 = 37o CFrom (1)T1 = 1.2 × 310 = 372K T1 = 372 - 273 = 990 CThis is numerical so it can only have one answer so option D is correct.
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