Explanation

(c) The efficiency of Carnot engine
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{Work\ output}{Heat\ input} = \frac{W}{Q_{H}}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{Q_{H} - Q_{L}}{Q_{H}} = 1 - \frac{Q_{L}}{Q_{H}}$
Also we can show that
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{Q_{L}}{Q_{H}} = \frac{T_{L}}{T_{H}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \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)
$and\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{3} = 1 - \frac{T_{L} - 50}{T_{H}}$ …(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
$\frac{1}{3} = 1 - \frac{T_{L} - 50}{\frac{5}{4}T_{L}}$
$or\ \ \frac{4(T_{L} - 50)}{5T_{L}} = \frac{2}{3}$
$or\ \ \ T_{L} - 50 = \frac{2}{3} \times \frac{5}{4}T_{L}$
$or\ T_{L} - \frac{5}{6}TL = 50$
$\therefore\ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ T}_{L} = 50 \times 6 = 300\ K$