The efficiency of a heat engine is 1 / 6 . Its efficiency double when the temperature of sink decrease by 62^ C , its efficiency doubles. Then, what is the temperature of source?

JIPMER 2018

JIPMER 2018 · Physics · Thermodynamics

The efficiency of a heat engine is 1 / 6 . Its efficiency double...

The efficiency of a heat engine is

1 / 6

. Its efficiency double when the temperature of sink decrease by

62^{\circ} \mathrm{C}

, its efficiency doubles. Then, what is the temperature of source?

B. $372 K$

\because

Efficiency of a heat engine,

\eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}

\quad \frac{1}{6}=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}

\frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{1}{6}=\frac{5}{6} \Rightarrow \mathrm{T}_2=\frac{5 \mathrm{~T}_1}{6}

...(i) According to question, it's efficiency is doubled when temperature of sink decreased by

62^{\circ} \mathrm{C}

\therefore \quad 2 \eta=1-\frac{\mathrm{T}_2-62}{\mathrm{~T}_1}

\Rightarrow \quad 2\left(\frac{1}{6}\right)=1-\frac{\left(\frac{5 T_1}{6}-62\right)}{T_1} \quad

[From Eq. (i)]

\begin{array}{rlrl}\Rightarrow & \frac{5 \mathrm{~T}_1}{6}-62 & =\mathrm{T}_1\left(1-\frac{2}{6}\right)=\frac{2 \mathrm{~T}_1}{3} \\ \text{ or } & \frac{5 \mathrm{~T}_1}{6}-\frac{2 \mathrm{~T}_1}{3} & =62 \Rightarrow \frac{\mathrm{T}_1}{6}=62 \\ \text{ or } & \mathrm{T}_1=372 \mathrm{~K}\end{array}