\(or,\,\,\frac{{{T_2}}}{{{T_1}}} = 1 - \eta = 1 - \frac{{40}}{{100}} = \frac{3}{5}\)
\(\therefore {T_1} = \frac{5}{3} \times {T_2} = \frac{5}{3} \times 300 = 500\,K.\)
\(Increases\,in\,efficiency = 50\% \,of\,40\% = 20\% \)
\(New\,efficiency,\eta ' = 40\% + 20\% = 60\% \)
\(\therefore \,\,\frac{{{T_2}}}{{{T_1}'}} = 1 - \frac{{60}}{{100}} = \frac{2}{5}\)
\({T_1}' = \frac{5}{2} \times {T_2} = \frac{5}{2} \times 300 = 750\,K.\)
\(Increases\,in\,temperature\,of\,source = {T_1}' - {T_1}\)
\( = 750 - 500 = 250\,K.\)