A reversible heat engine converts $1 / 6$ th of heat it absorbs at source into work. When the temper

Question

A reversible heat engine converts $1 / 6$ th of heat it absorbs at source into work. When the temperature of the sink is reduced by $62^{\circ} \mathrm{C}$, its efficiency is doubled. Temperature of source is\

Professor PC Thomas Classes · Answer

$\eta=\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~T}_{1}} \quad \frac{1}{6}=\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~T}_{1}}$\ $6 \mathrm{~T}_{1}-6 \mathrm{~T}_{2}=\mathrm{T}_{1}$\ $5 \mathrm{~T}_{1}=6 \mathrm{~T}_{2}----------$ (1)\ Temp. of sink reduced by $62^{\circ} \mathrm{C}$.ie. $\mathrm{T}_{2}$ becomes ( $\mathrm{T}_{2}-62$ )K and $\eta$ becomes double .ie. $\frac{1}{3}$ $$\begin{align*} & \frac{1}{3}=\frac{\mathrm{T}_{1}-\left(\mathrm{T}_{2}-62ight)}{\mathrm{T}_{1}} \ & 3 \mathrm{~T}_{1}-3 \mathrm{~T}_{2}+186=\mathrm{T}_{1} \ & 2 \mathrm{~T}_{1}=3 \mathrm{~T}_{2}-186  	ag{2}\ & 	ext { Solve (1) \& (2) } \end{align*}$$