A Carnot engine operating between two reservoirs has efficiency 1 3 . When the temperature of cold reservoir…

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-

$\begin{array}{rcl}{T}_H& =& 99°C\\ & =& 99+273 \mathrm{K}\\ & =& 372 \mathrm{K}\end{array}$

The formula to calculate the efficiency $\left(\eta \right)$ of the engine is given by

$\eta = 1-\frac{T_C}{T_H}.....................\left(1\right)$

where, $T_C$ is the temperature of the cold reservoir.

Substituting the values of the known parameters in equation (1), we get

$\begin{array}{rcl}\frac{1}{3}& =& 1-\frac{T_C}{T_H}\\ & \Rightarrow & \frac{T_C}{T_H}=\frac{2}{3}\\ & \Rightarrow & T_C=\frac{2}{3}\times 372 \mathrm{K}\\ & =& 248 \mathrm{K}\end{array}$

The new efficiency of the engine $\left(\eta \text{'}\right)$ is given by

$\eta \text{'}= 1-\frac{T_C+X}{T_H}.........................\left(2\right)$

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

$\begin{array}{rcl}\frac{1}{6}& =& 1-\frac{T_C+X}{T_H}\\ & \Rightarrow & \frac{5}{6}=\frac{T_C+X}{T_H}\\ & \Rightarrow & 248+X=5\times 62\\ & \Rightarrow & X=310-248\\ & =& 62 K\end{array}$