A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by $62 \mathrm{K}$, the efficiency of Carnot's cycle becomes 1/3. What is the sum of the temperature (in kelvin) of the source and the sink?

The efficiency of heat engine is given by

$\mathrm{\eta }=\frac{W}{Q}=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}$

where $T_1$ is temperature of source and $T_2$ is temperature of sink.

Given, ${\mathrm{\eta }}_1=\frac{1}{6}, {\mathrm{\eta }}_2=\frac{1}{3}$

$\therefore \frac{1}{6}=\frac{T_1-T_2}{T_1}$ ...(i)

and $\frac{1}{3}=\frac{T_1-(T_2-62)}{T_1}$ ...(ii)

Solving Eqs. (i) and (ii), we get

${ T}_1=372 \mathrm{K}$

and $T_2=310 \mathrm{K}$

$T_1+T_2=310+372=682$