# Heat pumps

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.14.

A heat pump is essentially a backwards air conditioner, in that it extracts an amount of heat ${Q_{c}}$ from the cold outside air using an amount ${W}$ of electrical energy and dumps an amount ${Q_{h}=Q_{c}+W}$ of heat into an inside room in order to heat it. We can analyze its performance using the same logic as for a refrigerator. The coefficient of performance in this case is the amount of heat dumped into the room divided by the energy (work) needed to achieve this, so

$\displaystyle \mbox{COP}=\frac{Q_{h}}{W} \ \ \ \ \ (1)$

From conservation of energy

$\displaystyle \mbox{COP}=\frac{Q_{h}}{Q_{h}-Q_{c}}=\frac{1}{1-Q_{c}/Q_{h}} \ \ \ \ \ (2)$

Thus the COP is always greater than 1.

In an ideal refrigerator (e.g. one working on a reversed Carnot cycle) the entropy gained in absorbing ${Q_{c}}$ is equal to the entropy lost in expelling ${Q_{h}}$, so

 $\displaystyle \frac{Q_{c}}{T_{c}}$ $\displaystyle =$ $\displaystyle \frac{Q_{h}}{T_{h}}\ \ \ \ \ (3)$ $\displaystyle Q_{c}$ $\displaystyle =$ $\displaystyle \frac{T_{c}}{T_{h}}Q_{h} \ \ \ \ \ (4)$

The upper limit on the COP is therefore

$\displaystyle \mbox{COP}_{max}=\frac{1}{1-T_{c}/T_{h}}=\frac{T_{h}}{T_{h}-T_{c}} \ \ \ \ \ (5)$

A heat pump is more efficient than a purely electric heater (that is, a heater that generates heat by electrical resistance rather than by transferring heat from outside to inside) since in an electric heater, the work ${W}$ is equal to ${Q_{h}}$, while in a heat pump, this work can be used to acquire an additional amount ${Q_{c}}$ of heat which is expelled into the room along with the electrical work ${W}$. In other words, with a heat pump you have ${Q_{h}=W+Q_{c}}$ while with an electric heater, you have only ${Q_{h}=W}$.