An engine has an efficiency of 1/6. When the temperature of sink is reduced by $62°\mathrm{C}$, its efficiency is doubled. Temperatures of source and sink are

$\begin{array}{l}\text{From }\eta =1-\frac{{\text{T}}_2}{ {\text{T}}_1},\\ \Rightarrow \frac{{\text{T}}_2}{ {\text{T}}_1}=1-\eta =1-\frac{1}{6}=\frac{5}{6}\dots \text{ (i) }\\ \text{In 2nd case : }\frac{{\text{T}}_2-62}{ {\text{T}}_1}=1-{\eta }^{\text{'}}=1-\frac{2}{6}=\frac{2}{3}\dots \mathrm{..}\text{ (ii)}\\ \text{Using }\left(\text{i}\right),{\text{T}}_2-62=\frac{2}{3} {\text{T}}_1=\frac{2}{3}\times \frac{6}{5} {\text{T}}_2=\frac{4}{5} {\text{T}}_2\\ \Rightarrow \frac{1}{5} {\text{T}}_2=62,\\ \Rightarrow {\text{T}}_2=310 \text{K}=310-273={37}^{°}\text{C}\\ \therefore {\text{T}}_1=\frac{6}{5} {\text{T}}_2=\frac{6}{5}\times 310=372 \text{K}=372-273={99}^{°}\text{C}\end{array}$