An engine has an efficiency of $$\frac{1}{6}.$$ When the temperature of sink is reduced by $${62^ \circ }C,$$  its efficiency is doubled. Temperature of the source is

Solution :
Efficiency of engine is given by $$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
$${T_2} =$$  temperature of sink
$${T_1} =$$  temperature of source
$$\therefore \frac{{{T_2}}}{{{T_1}}} = 1 - \eta = 1 - \frac{1}{6} = \frac{5}{6}\,......\left( {\text{i}} \right)$$
In other case,
$$\eqalign{ & \frac{{{T_2} - 62}}{{{T_1}}} = 1 - \eta = 1 - \frac{2}{6} = \frac{2}{3}\,......\left( {{\text{ii}}} \right) \cr & {\text{or}}\,\,{T_2} - 62 = \frac{2}{3}{T_1} = \frac{2}{3} \times \frac{6}{5}{T_2}\,\,\left[ {{\text{Using}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr & {\text{or }}\frac{1}{5}{T_2} = 62 \cr & \therefore {T_2} = 310\;K \cr & = 310 - {273^ \circ }C = {37^ \circ }C \cr} $$
Here, $${T_1} = \frac{6}{5}{T_2} = \frac{6}{5} \times 310$$
$$\eqalign{ & = 372\,K = 372 - 273 \cr & = {99^ \circ }C \cr} $$