Question:
A reversible heat engine converts one- fourth of the heat input into work. When the temperature of the sink is reduced by $52 \mathrm{~K}$, its efficiency is doubled. The temperature in Kelvin of the source will be
Solution:
(208)
$\because \mathrm{n}=\frac{\mathrm{W}}{\mathrm{Q}_{\mathrm{in}}}=\frac{1}{4}$
$\frac{1}{4}=1-\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$
$\frac{T_{1}}{T_{2}}=\frac{3}{4}$
When the temperature of the sink is reduced by $52 \mathrm{k}$ then its efficiency is doubled. $\frac{1}{2}=1-\frac{\left(T_{1}-52\right)}{T_{2}}$
$\frac{T_{1}-52}{T_{2}}=\frac{1}{2}$
$\frac{T_{1}}{T_{2}} \frac{-52}{T_{2}}=\frac{1}{2}$
$\frac{3}{4}-\frac{52}{T_{2}}=\frac{1}{2}$
$\frac{52}{T_{2}}=\frac{1}{4}$
$\mathrm{T}_{2}=208 \mathrm{k}$