91Ó°ÊÓ

Let us write the expression of the maximum efficiency for a heat engine

$e=1-\frac{T_{\text{cold}}}{T_{\text{hot}}}$

Here, $T_{\text{hot}}$is the temperature of hot reservoir and $T_{\text{cold}}$is the temperature of the cold reservoir.

Write the expression of the work done by an optimal heat engine

$W=Q_{\text{hot}}-Q_{\text{cold}}…..\left(1\right)$

Here, $Q_{\text{hot}}$is the heat extracted from hot reservoir and $Q_{\text{cold}}$is the heat transferred to the cold reservoir.

Write the expression of $Q_{\text{cold}}$from the second law of thermodynamics

$Q_{\text{cold}}=Q_{\text{hot}}\left(\frac{T_{\text{ovdd}}}{T_{\text{hot}}}\right)$

Substitute $Q_{\text{hot}}\left(\frac{T_{\text{cold}}}{T_{\text{hot}}}\right)$for $Q_{\text{cold}}$in equation (1)

$W=Q_{\text{hot}}\left(1-\frac{T_{\text{cold}}}{T_{\text{hot}}}\right)…\text{(2)}$

An engine which is more efficient than a Carnot engine can produce an equal amount of work by extracting lesser heat and transferring smaller waste heat.

The heat engine, in this case, produces the necessary work to run the refrigerator. But, the refrigerator extracts more heat than the waste heat produced by the heat engine.

The composite system of the two constitutes a system that extracts a net amount of heat from the cold reservoir without requiring any input work. Such a system cannot be made in reality as it violates the second law of thermodynamics.

Thus, it is verified that a heat engine whose efficiency is better than the ideal value can be hooked up to an ordinary Carnot refrigerator to make a refrigerator that requires no work input.