The figure shows a heat engine (HE) working between two reservoirs. The amount of heat (Q2) rejected by the heat engine is drawn by a heat pump (HP). The heat pump receives the entire work output (W) of the heat engine. If temperature, T₁ > T₃ > T₂, then the relation between the efficiency (η) of the heat engine and the coefficient of performance (COP) of the heat pump is

The correct answer is COP = η^-1.

Step-by-Step Explanation:

1. Analyze the Heat Engine (HE):
Based on the provided diagram, the heat engine operates between reservoirs at temperatures $T_1$ and $T_2$.
- Heat absorbed from the high-temperature reservoir ($T_1$) = $Q_1$
- Heat rejected to the low-temperature reservoir ($T_2$) = $Q_2$
- Work output produced by the heat engine = $W$

By the first law of thermodynamics for the heat engine:
$W=Q_1-Q_2$
Rearranging this gives:
$Q_1=W+Q_2$

The thermal efficiency ($\eta$) of the heat engine is defined as the ratio of work output to heat input:
$\eta =\frac{W}{Q_1}$

Substituting the expression for $Q_1$ into the efficiency formula:
$\eta =\frac{W}{W+Q_2}$

2. Analyze the Heat Pump (HP):
The heat pump receives the entire work output $W$ of the heat engine as its work input. It draws heat $Q_2$ from the reservoir at $T_2$ and rejects heat $Q_3$ to the reservoir at $T_3$.

By the first law of thermodynamics for the heat pump:
$Q_3=W+Q_2$

The Coefficient of Performance (COP) of a heat pump is defined as the ratio of the desired heating effect (heat rejected to the hot reservoir, $Q_3$) to the work input ($W$):
$\text{COP}=\frac{Q_3}{W}$

Substituting the expression for $Q_3$ into the COP formula:
$\text{COP}=\frac{W+Q_2}{W}$

3. Relate the Efficiency (η) and COP:
Comparing the equations for efficiency and COP:
$\eta =\frac{W}{W+Q_2}$
$\text{COP}=\frac{W+Q_2}{W}$

We can clearly see that the COP of the heat pump is the reciprocal of the thermal efficiency of the heat engine:
$\text{COP}=\frac{1}{\eta }={\eta }^{-1}$