where \(T_1\) is the temperature of the source and \(T_2\) is the temperature of the sink.

\(\therefore \frac{1}{6} = 1 - \frac{{{T_2}}}{{{T_1}}}\,\,\,or,\,\,\,\frac{{{T_2}}}{{{T_1}}} = \frac{5}{6}\)         \(...(i)\)

where the temperature of the sink is decreased by \({62^ \circ }C\,\left( {or\,62\,K} \right),\) efficiency becomes double.

Since, the temperature of the source remains unchanged

\(\therefore 2 \times \frac{1}{6} = 1 - \frac{{\left( {{T_2} - 62} \right)}}{{{T_1}}}\,\,or,\,\,\frac{1}{3} = 1 - \frac{{\left( {{T_2} - 62} \right)}}{{{T_1}}}\)

\(or,\,\frac{2}{3} = \frac{{{T_2} - 62}}{{{T_1}}}\,\,or,\,2{T_1} = 3{T_2} - 186\)

\(or,\,\,2{T_1} = 3\left[ {\frac{5}{6}} \right]{T_1} - 186\)        \([using  (i)]\)

\(\therefore \,\,\left[ {\frac{5}{2} - 2} \right]{T_1} = 186\,\,or,\,\,\frac{{{T_1}}}{2} = 186\)

\(or,\,\,{T_1} = 372\,K = {99^ \circ }C.\)