An engine has an efficiency of $1 / 6$. When the temperature of sink is reduced by $62^{\circ} \mathrm{C}$, its efficiency is doubled. Temperatures of the source is:

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Aicharya · Accepted Answer

Efficiency of an engine, $\quad \eta=1-\frac{T_2}{T_1}$ where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink. $\therefore \quad \frac{1}{6}=1-\frac{T_2}{T_1} \quad$ or, $\quad \frac{T_2}{T_1}=\frac{5}{6}$ $\therefore \quad 2 \times \frac{1}{6}=1-\frac{\left(T_2-62\right)}{T_1}$ or, $\frac{1}{3}=1-\frac{\left(T_2-62\right)}{T_1}$ or, $\frac{2}{3}=\frac{T_2-62}{T_1}$ or, $2 T_1=3 T_2-186$ or, $2 T_1=3\left[\frac{5}{6}\right] T_1-186 \quad$ [using (i)] $\therefore \quad\left[\frac{5}{2}-2\right] T_1=186 \quad$ or, $\quad \frac{T_1}{2}=186$ or, $T_1=372 \mathrm{~K}=99^{\circ} \mathrm{C}$. ~

An engine has an efficiency of $1 / 6$. When the temperature of sink is reduced by $62^{\circ} \mathrm{C}$,…

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