Question

A reversible heat engine converts one- fourth of the heat input into work. When the temperature of the sink is reduced by $52\text{K}$, its efficiency is doubled. The temperature in Kelvin of the source will be_______.

Answer 1

$\eta =\frac{W}{Q_{\text{in}}}=\frac{1}{4}$
$\frac{1}{4}=1-\frac{T_2}{T_1}$
[For reversible heat engine]
$\frac{T_2}{T_1}=\frac{3}{4}$
When the temperature of the sink is reduced by $52\text{K}$ then its efficiency is doubled.
$\frac{1}{2}=1-\frac{(T_2-52)}{T_1}$
$\frac{(T_2-52)}{T_1}=\frac{1}{2}$
$\frac{T_2}{T_1}-\frac{52}{T_1}=\frac{1}{2}$
$\frac{3}{4}-\frac{52}{T_1}=\frac{1}{2}$
$\frac{52}{T_1}=\frac{1}{4}$
$T_1=208\text{K}$