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:
$\eta=\frac{1}{4}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$
$\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{3}{4}$
$\frac{T_{2}-52}{T_{1}}=\frac{1}{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.