Find the relation between the efficiency of a reversible ideal heat engine and the coefficient of performance of the reversible refrigerator obtained by running the engine backward.

The engine is running backward.

First, we have to measure the efficiency of the ideal gas which is defined as the work the engine does per cycle divided by the energy it absorbs as heat per cycle. Then we can calculate the coefficient of performance for ideal gas. By comparing these two equations we can get the relation between the efficiency of the heat engine and the coefficient of performance of the reversible refrigerator.

Formulae:

The coefficient of performance of the Carnot refrigerator,$K=\frac{T_L}{T_H-T_L}$ (1)

The efficiency of the Carnot cycle,$\epsilon =\frac{T_H-T_L}{T_H}$ (2)

Where $T_H$ and $T_L$ are the temperatures of the high and low-temperature reservoirs respectively.

As we know that an engine is a device that operates in a cycle. The Carnot engine is an ideal engine that follows the cycle in fig. 20-8.

And the refrigerator is a device that is operating in a cycle. So, we can find the coefficient of performance.

From equation (2), we get the following relation:

$T_H-T_L=\epsilon T_H$

Substitute this equation in equation (1), we get that

$$\begin{gathered} K=\frac{T_L}{\epsilon T_H} \\ \epsilon K=\frac{T_L}{T_H} \end{gathered}$$

By adding and subtracting$T_H$in the numerator of the above equation, we can get the required relation as follows:

$$\begin{gathered} \epsilon K=\frac{T_L-T_H+T_H}{T_H} \\ =\frac{T_L-T_H}{T_H}+1 \\ =-\left(\frac{T_H-T_L}{T_H}\right)+1 \\ =-\epsilon +1 \end{gathered}$$

Therefore, we can write

$$\begin{gathered} K=-1+\frac{1}{\epsilon } \\ \frac{1}{\epsilon }=K+1 \\ \epsilon =\frac{1}{K+1} \end{gathered}$$

Hence, the required relation between efficiency and performance is $\epsilon =\frac{1}{K+1}$