A reversible engine converts one-sixth of the heat supplied into work. When the temperature of the sink is…
A reversible engine converts one-sixth of the heat supplied into work. When the temperature of the sink is reduced by $62^{\circ} \mathrm{C}$, the efficiency of the engine is doubled. The temperatures of the source and sink are
$99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
$80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
$95^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
$90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
Solution
The efficiency of engine, $\eta=\frac{T_1-T_2}{T_1}$ According to question, $2 \eta=\frac{T_1-\left(T_2-62\right)}{T_1}$
or $\quad \frac{1}{2}=\frac{T_1-T_2}{\left(T_1-T_2\right)+62}$
or
$62=\left(T_1-T_2\right)$
Only option (a) satisfies the condition.