Refrigerators

Let us now move on to consider refrigerators. An idealized refrigerator is an engine which extracts heat from a cold heat reservoir (temperature $T_2$, say) and rejects it to a somewhat hotter heat reservoir, which is usually the environment (temperature $T_1$, say). To make this machine work we always have to do some external work on the engine. For instance, the refrigerator in your home contains a small electric pump which does work on the freon (or whatever) in the cooling circuit. We can see that, in fact, a refrigerator is just a heat engine run in reverse. Hence, we can immediately carry over most of our heat engine analysis. Let $q_2$ be the heat absorbed per cycle from the colder reservoir, $q_1$ the heat rejected per cycle into the hotter reservoir, and $w$ the external work done per cycle on the engine. The first law of thermodynamics tells us that

$$w + q_2 = q_1.$$ (364)

The second law says that

$$\frac{q_1}{T_1} + \frac{-q_2}{T_2} \geq 0.$$ (365)

We can combine these two laws to give

$$\frac{w}{T_1} \geq q_2\left(\frac{1}{T_2} - \frac{1}{T_1}\right).$$ (366)

The most sensible way of defining the efficiency of a refrigerator is as the ratio of the heat extracted per cycle from the cold reservoir to the work done per cycle on the engine. With this definition

$$\eta = \frac{T_2}{T_1 - T_2}.$$ (367)

We can see that this efficiency is, in general, greater than unity. In other words, for one joule of work done on the engine, or pump, more than one joule of energy is extracted from whatever it is we are cooling. Clearly, refrigerators are intrinsically very efficient devices. Domestic refrigerators cool stuff down to about $4^\circ$ C (277$^\circ$ K) and reject heat to the environment at about $15^\circ$ C (288$^\circ$ K). The maximum theoretical efficiency of such devices is

$$\eta = \frac{277}{288-277} = 25.2.$$ (368)

So, for every joule of electricity we put into a refrigerator we can extract up to 25 joules of heat from its contents.