Reversed heat engines (A-level only)

AQA·Physics·3.11.2.6

Reversed Heat Engines

A reversed heat engine uses work to transfer energy as heat from a colder space to a hotter space. This transfer is against the natural direction of heat flow, which is from hotter to colder, so the engine requires a work input.

In each cycle, the engine takes in heat $$Q_C$$ from the cold space and receives work $$W$$. It then delivers heat $$Q_H$$ to the hot space. These quantities are related by conservation of energy:

QH=QC+WQ_H = Q_C + W

The heat delivered to the hot space equals the heat taken from the cold space plus the work supplied to the engine. Here, $$Q_C$$, $$Q_H$$ and $$W$$ are positive amounts of energy transferred during one cycle.

Refrigerators and heat pumps

A refrigerator uses this process to remove heat from a cold space and keep it cool. Its desired effect is the heat removed, $$Q_C$$.

A heat pump uses the same principle to deliver heat to a hot space and keep it warm. Its desired effect is the heat delivered, $$Q_H$$, which includes both the energy taken from the cold space and the work supplied.

Coefficients of Performance of Reversed Heat Engines

A reversed heat engine uses work to transfer heat from a colder space to a hotter space. The coefficient of performance (COP) measures the desired heat transfer per unit of work input.

The desired transfer depends on the device’s purpose. A refrigerator is used to remove heat $$Q_C$$ from the cold space, whereas a heat pump is used to deliver heat $$Q_H$$ to the hot space.

COP in terms of heat and work

The heat delivered to the hot space comes from both the heat removed from the cold space and the work input $$W$$. Therefore, $$Q_H = Q_C + W$$, or $$W = Q_H - Q_C$$. Using the relevant desired heat transfer gives:

COPref=QCW=QCQH−QC,COPhp=QHW=QHQH−QCCOP_{\mathrm{ref}} = \frac{Q_C}{W} = \frac{Q_C}{Q_H-Q_C}, \qquad COP_{\mathrm{hp}} = \frac{Q_H}{W} = \frac{Q_H}{Q_H-Q_C}

For the same device operating between the same spaces, $$Q_H = Q_C + W$$. Dividing by $$W$$ shows that $$COP_{\mathrm{hp}} = COP_{\mathrm{ref}} + 1$$.

A COP can exceed one because the desired heat transfer is not supplied by the work input alone: the device also transfers heat taken from the cold space. A COP greater than one does not mean that energy is created.

Ideal reversible COP and temperature

For an ideal reversible device operating between a cold space at temperature $$T_C$$ and a hot space at temperature $$T_H$$, the heat transfers satisfy $$Q_C/Q_H = T_C/T_H$$. Substituting this relationship into the COP expressions gives:

COPref=TCTH−TC,COPhp=THTH−TCCOP_{\mathrm{ref}} = \frac{T_C}{T_H-T_C}, \qquad COP_{\mathrm{hp}} = \frac{T_H}{T_H-T_C}

The temperatures in these expressions must be absolute temperatures in kelvin, with $$T_H > T_C$$. These expressions give ideal reversible performance; real devices operating between the same temperatures have lower COPs.