A reversible carnot engine converts 1/6th of heat into input work. When the temperature of the sink is reduced by 62oC, 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
Efficiency = T2-T1/T2 = ⅙ By rearranging we can get 6T2-6T1 = T2 T2 = 1.2T1T2 is the source and T1 is the sink. When the sink temperature is reduced by 62, the efficiency gets doubled. T2-(T1-62)/T2 = 2 x ⅙ = 2/6 By rearranging this equation we get 6T2-6T1+372 = 2T2 4T2 - 6T1 + 372 =0 Substitute the value of T2 from the first equation 4(1.2T1)-6T1+372 1.2T1 = 372 = 310K = 37˚C T2= 1.2 x 310 = 372K = 99˚C
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Thermal equilibrium means that bodies in contact have the same temperature and no net heat flows between them. The topic distinguishes heat from temperature, uses the zeroth law of thermodynamics, and covers heat transfer, specific heat capacity, thermal expansion and the conditions for reaching equilibrium.
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