Question

A Carnot reversible engine converts $1/6$ of heat input into work. When the temperature of the sink is reduced by $62K$ , the efficiency of Carnot's cycle becomes $1/3$ . What is the sum of the temperature (in kelvin) of the source and the sink?

Answer 1

The efficiency of heat engine is given by
$\eta=\frac{W}{Q}=1-\frac{Q_{2}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}$
where $T_{1}$ is temperature of source and $T_{2}$ is temperature of sink.
Given, $\eta_{1}=\frac{1}{6}, \, \eta_{2}=\frac{1}{3}$
$\therefore \, \, \frac{1}{6}=\frac{T_{1} - T_{2}}{T_{1}}...\left(1\right)$
and $\frac{1}{3}=\frac{T_{1} - \left(T_{2} - 62\right)}{T_{1}}...\left(2\right)$
Solving Eqs. $\left(1\right)$ and $\left(2\right)$ , we get
$T_{1}=372 \, K$ and $T_{2}=310 \, K$
$T_{1}+T_{2}=310+372=682$