A reversible Carnot engine converts 1/6th of heat into input work. When the temperature of the sink is reduced by 62˚C and the efficiency is doubled then the temperature of the source and sink is:
Correct answer: D. 99oC , 37oC
- A. 80oC ,37oC
- B. 99oC , 30oC
- C. 99oC,25oC
- D. 99oC , 37oC
Explanation
Efficiency of Engine , = Work / QinputAs, W = 1 / 6 Qinput = 1 / 6Also, = 1 - T2 / T1WhereT1 and T2 are the temperatures of the source and sink respectively.So,1 / 6 =1 - T2 / T1T1 = 1.2 T2 ............ (1)Now the sink temperature is reduced by 62C, i.e. T2' = T2 - 62 ............ (2)Also the new efficiency is doubled, i.e. ' = 2 1 / 6 = 2 / 6 = 1 / 3 ...... (3)Substituting (1), (2) and (3)' = 1 - T2' / T11 / 3 = 1 - (T2 - 62) / 1.2 T2T2 = 310 KThus T2 = 310 - 273 = 37CFrom (1), T1 = 1.2 T2T1 = 1.2 x 310 = 372 Ki.e T1 = 372 - 273 = 99C
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