The efficiency of Carnot engine

$\mathrm{\eta }=\frac{\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{p}\mathrm{u}\mathrm{t}}{\mathrm{H}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{p}\mathrm{u}\mathrm{t}}=\frac{W}{Q_H}$

or $\mathrm{\eta }=\frac{Q_H-Q_L}{Q_H}=1-\frac{Q_L}{Q_H}$

Also we can show that

$\frac{Q_L}{Q_H}=\frac{T_L}{T_H}$

$\therefore \mathrm{\eta }=1-\frac{T_L}{T_H}$

where $T_L$ is temperature of sink and $T_H$ is temperature of hot reservoir.

According to question

$\frac{1}{5}=1-\frac{T_L}{T_H}$ ...(i)

$\mathrm{a}\mathrm{n}\mathrm{d} \frac{1}{3}=1-\frac{T_L-50}{T_H}$ ...(ii)

From Eq. (i)

$\frac{T_L}{T_H}=\frac{4}{5}$

$\Rightarrow T_H=\frac{5}{4}{T}_L$

Substituting value of $T_H$ in Eq. (ii), we get

$\frac{1}{3}=1-\frac{T_L-50}{\frac{5}{4}{T}_L}$

$\mathrm{o}\mathrm{r} \frac{4(T_L-50)}{5T_L}=\frac{2}{3}$

$\mathrm{o}\mathrm{r} T_L-50=\frac{2}{3}\times \frac{5}{4}{T}_L$

$\mathrm{o}\mathrm{r} T_L-\frac{5}{6}{T}_L=50$

$\therefore { T}_L=50\times 6=300 \mathrm{K}$