A Carnot reversible engine convert 1/6 of heat input into work. When the temperature of the sink is reduced by 62 K, the efficiency of Carnot’s cycle becomes 1/3. The temperature of the source and sink will be

The efficiency of heat engine is given by

$\eta =\frac{W}{Q}=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}$

Where, $T_1$  is temperature ofsource and $T_2$  is temperature of sink.

Given, ${\eta }_1=\frac{1}{6},{\eta }_2=\frac{1}{3}$

$\therefore \frac{1}{6}=\frac{T_1-T_2}{T_1}$ $\Rightarrow \frac{T_2}{T_1}=\frac{5}{6}\left(i\right)$

and$\frac{1}{3}=\frac{T_1-\left(T_2-62\right)}{T_1}$

$\Rightarrow \frac{T_2}{T_1}-\frac{62}{T_1}=\frac{2}{3}\left(ii\right)$

Solving Eqs. (i) and (ii), we get$\frac{5}{6}-\frac{62}{T_1}=\frac{2}{3}$

$\Rightarrow T_1=372K\text{and}{T}_2=310K$