A Carnot engine operating between temperatures $${T_1}$$ and $${T_2}$$ has efficiency $$\frac{1}{6}.$$ When $${T_2}$$ is lowered by $$62\,K$$  its efficiency increases to $$\frac{1}{3}.$$ Then $${T_1}$$ and $${T_2}$$ are, respectively

Solution :
$$\eqalign{ & {\eta _1} = 1 - \frac{{{T_2}}}{{{T_1}}} \Rightarrow \frac{1}{6} = 1 - \frac{{{T_2}}}{{{T_1}}} \Rightarrow \frac{{{T_2}}}{{{T_1}}} = \frac{5}{6}\,......\left( {\text{i}} \right) \cr & {\eta _2} = 1 - \frac{{{T_2} - 62}}{{{T_1}}} \Rightarrow \frac{1}{3} = 1 - \frac{{{T_2} - 62}}{{{T_1}}}\,......\left( {{\text{ii}}} \right) \cr} $$
On solving Eqs. (i) and (ii)
$${T_1} = 372\,K\,{\text{and}}\,{T_2} = 310\,K$$