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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About Energy Losses and Efficiency
Efficiency compares useful output energy or power with the total input, usually expressed as a percentage. The topic covers energy losses through friction, heating, resistance and sound, along with power ratings and Sankey diagrams. Efficiency is not the same as conservation of energy, because lost energy is transferred rather than destroyed.
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