An engine takes heat from a reservoir and converts its $$\frac{1}{6}$$ part into work. By decreasing temperature of sink by $${62^ \circ }C,$$  its efficiency becomes double. The temperatures of source and sink must be

Solution :
If $${T_1}$$ is temperature of source and $${T_2}$$ the temperature of sink, the efficiency of engine
$$\eqalign{ & \eta = \frac{{{\text{Work done}}\left( W \right)}}{{{\text{Heat taken}}\left( {{Q_1}} \right)}} = 1 - \frac{{{T_2}}}{{{T_1}}} \cr & \therefore 1 - \frac{{{T_2}}}{{{T_1}}} = \frac{1}{6}\,......\left( {\text{i}} \right) \cr} $$
When temperature of sink is reduced by $${62^ \circ }C,$$  then temperature of sink
$$\eqalign{ & {{T'}_2} = {T_2} - 62 \cr & \therefore \eta ' = 1 - \frac{{{{T'}_2}}}{{{T_1}}} \cr} $$
As according to question efficiency becomes double
So, $$\eta ' = 2\eta = \frac{2}{6} = \frac{1}{3}$$
$$\therefore \frac{1}{3} = 1 - \frac{{{T_2} - 62}}{{{T_1}}}\,......\left( {{\text{ii}}} \right)$$
From Eq. (i) $$\frac{{{T_2}}}{{{T_1}}} = \frac{5}{6}\,......\left( {{\text{iii}}} \right)$$
From Eq. (ii) $$\frac{{{T_2} - 62}}{{{T_1}}} = \frac{2}{3}\,......\left( {{\text{iv}}} \right)$$
Dividing Eq. (iii) by Eq. (iv)
$$\eqalign{ & \frac{{{T_2}}}{{{T_2} - 62}} = \frac{5}{4} \cr & \Rightarrow 4{T_2} = 5{T_2} - 310 \cr & \Rightarrow {T_2} = 310\,K \cr} $$
and from Eq. (iii), we have
$$\frac{{310}}{{{T_1}}} = \frac{5}{6} \Rightarrow {T_1} = 372\,K$$
Hence, $${T_1} = 372\,K = 372 - 273 = {99^ \circ }C$$
and $${T_2} = 310\,K = 310 - 273 = {37^ \circ }C$$
As kinetic energy of a gas depends on its atomicity.