Question

A reversible carnot engine converts 1/6^th of heat into input work. When the temperature of the sink is reduced by 62^oC, then the efficiency is doubled. The temperature of the source and sink is:

Options

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 = 1.2T1

T2 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

This option suggests that the temperatures of the heat source and the heat sink are 80°C and 37°C, respectively.

However, based on our calculations, the temperature of the heat source (Th ) is found to be 372 K, which is approximately 99°C, and the temperature of the heat sink (Tc ) is 310 K, which is not equivalent to 37°C.Therefore, option A) does not match the calculated temperatures and is incorrect.

This option suggests that the temperatures of the heat source and the heat sink are 99°C and 30°C, respectively.

However, based on our calculations, the temperature of the heat source (Tℎ ) is found to be 372 K, which is approximately 99°C, and the temperature of the heat sink (Tc ) is 310 K, which is not equivalent to 30°C.Therefore, option B) does not match the calculated temperatures and is incorrect.

This option suggests that the temperatures of the heat source and the heat sink are 99°C and 25°C, respectively.

However, based on our calculations, the temperature of the heat source (Tℎ ) is found to be 372 K, which is approximately 99°C, and the temperature of the heat sink (Tc ) is 310 K, which is not equivalent to 25°C.Therefore, option C) does not match the calculated temperatures and is incorrect.