A Carnot engine operating between two reservoirs has efficiency...

A carnot engine works between two reservoirs, one hot reservoir and one cold reservoir.

The absolute temperature $\left({T}_{H}\right)$ of the hot reservoir can be calculated as follows-

$\begin{cases}{T}_{H} & = & 99 ° C \\ & = & 99 + 273K \\ & = & 372K\end{cases}$

The formula to calculate the efficiency \left(η\right)$of the engine is given by</p><p>$η = 1 - \dfrac{{T}_{C}}{{T}_{H}} . . . . . . . . . . . . . . . . . . . . . ( 1 )$</p><p>where,${T}_{C}$is the temperature of the cold reservoir.</p><p>Substituting the values of the known parameters in equation (1), we get</p><p>$\begin{cases}\dfrac{1}{3} & = & 1 - \dfrac{{T}_{C}}{{T}_{H}} \\ & ⇒ & \dfrac{{T}_{C}}{{T}_{H}} = \dfrac{2}{3} \\ & ⇒ & {T}_{C} = \dfrac{2}{3} \times 372K \\ & = & 248K\end{cases}$</p><p>The new efficiency of the engine \left(η '\right)$ is given by

$η ' = 1 - \dfrac{{T}_{C} + X}{{T}_{H}} . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )$

Substituting $\dfrac{1}{6}$ for $η '$ and the known values of the temperatures into equation (2), we get

$\begin{cases}\dfrac{1}{6} & = & 1 - \dfrac{{T}_{C} + X}{{T}_{H}} \\ ⇒ & \dfrac{5}{6} = \dfrac{{T}_{C} + X}{{T}_{H}} \\ ⇒ & 248 + X = 5 \times 62 \\ ⇒ & X = 310 - 248 \\ = & 62K\end{cases}$