Question

Prove that the coefficient of performance of a Carnot refrigerator is $K_{Carnot}=T_C/\left(T_H-T_C\right)$.

Short answer

The coefficient of performance of refrigerator is given by$COP=\frac{T_c}{T_h-T_c}$.

Step-by-step solution

Step : 1 Explanation

In the cycle of changes, as the working substance returns to its initial state, therefore, there is no change in its initial energy that is $∆E_{int}=dU=0\left(I\right)$

If $Q_c$is the amount of heat extracted per cycle from the cold reservoir at lower temperature $T_c$

$W=$work done per cycle on the system $-$the refrigerant.

If $Q_h$is the amount of heat released per cycle from the source at higher temperature $T_h$

The net amount of heat absorbed is given by $dQ=Q_c-Q_h\left(II\right)$

Work done by the system is given by $dW=-W\left(III\right)$

According to the first law of thermodynamics $dQ=dU+dW\left(IV\right)$

Step : 2 Calculation

Substituting the values of dU and dQ from the equation (I) and (II) in equation
$\begin{array}{r}\left(I\right)Q_c-Q_h=0-W\\ W=Q_h-Q_c\\ W=Q_c\left[\frac{Q_h}{Q_c}-1\right]\left(IV\right)\end{array}$

In Carnot Cycle

$\frac{Q_h}{Q_c}=\frac{T_h}{T_c}\left(VI\right)$

Substituting the value of$\frac{Q_h}{Q_c}$in equation (V)

$W=Q_c\left[\frac{T_h}{T_c}-1\right]$(VII)

Also, coefficient of performance is defined as the ratio of the quantity of heat removed per cycle to the energy spent per cycle to remove this heat.
$\text{COP}=\frac{Q_c}{W}\left(VIII\right)$

Substituting the values of W in above equation, the coefficient of performance is calculated
$\text{as}COP=\frac{T_c}{T_h-T_c}\text{.}$

Conclusion :

The coefficient of performance of a refrigerator is given by $COP=\frac{T_c}{T_h-T_c}$