Physics - Heat & Thermo Question with Solution | TestHub

The efficiency of a reversible engine (Carnot engine) is given by:

$$\eta =1-\frac{T_C}{T_H}$$

Initially, the engine converts one-sixth of the heat input into work, so the initial efficiency is ${\eta }_1=\frac{1}{6}$.

Thus,

$$\frac{1}{6}=1-\frac{T_C}{T_H}$$

$$\frac{T_C}{T_H}=1-\frac{1}{6}=\frac{5}{6}(1)$$

When the temperature of the sink is reduced by $62^{\circ }\mathrm{C}$, the new sink temperature is $T_C^{\prime }=T_C-62$.

The efficiency is doubled, so ${\eta }_2=2{\eta }_1=2\times \frac{1}{6}=\frac{1}{3}$.

The new efficiency is:

$${\eta }_2=1-\frac{T_C^{\prime }}{T_H}=1-\frac{T_C-62}{T_H}$$

$$\frac{1}{3}=1-\frac{T_C-62}{T_H}$$

$$\frac{T_C-62}{T_H}=1-\frac{1}{3}=\frac{2}{3}(2)$$

From equation (1), $T_C=\frac{5}{6}{T}_H$. Substitute this into equation (2):

$$\frac{\frac{5}{6}{T}_H-62}{T_H}=\frac{2}{3}$$

$$\frac{5}{6}-\frac{62}{T_H}=\frac{2}{3}$$

$$\frac{62}{T_H}=\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{4}{6}=\frac{1}{6}$$

$$T_H=62\times 6=372K$$

Now, calculate $T_C$ using equation (1):

$$T_C=\frac{5}{6}{T}_H=\frac{5}{6}\times 372=5\times 62=310K$$

Convert the temperatures from Kelvin to Celsius:

Source temperature $T_H=372-273=99^{\circ }\mathrm{C}$

Sink temperature $T_C=310-273=37^{\circ }\mathrm{C}$