A Carnot engine has an efficiency of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$. When the temperature of the sink is reduced by ${62}^0 C$ its efficiency is doubled. The temperatures of the source and the sink are, respectively,

As $\eta =1-\frac{T_2}{T_1}$

According to first case

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

According to second case

$\frac{T_2-62}{T_1}=1-2\times \frac{1}{6}=\frac{2}{3} \left(ii\right)$

Equating equations (i) and (ii), we get

$$\begin{gathered} \frac{5}{6}=\frac{2}{3}+\frac{62}{T_1} \\ {T}_1=62\times 6=372 K=372 - 273={99}^{0}C \end{gathered}$$

$T_{\sin k}=\frac{5}{6}\times 372=310K={37}^{0}C$