# Air conditioners

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.13.

An air conditioner is an example of a refrigerator in which the cold reservoir is the room to be cooled and the hot reservoir is the outside atmosphere. On a hot day, the rate at which heat leaks into an air conditioned room from the outside is roughly proportional to the temperature difference ${T_{h}-T_{c}}$ between the outside and inside. In that case, the work done to remove an amount ${Q_{c}}$ of heat in time ${\Delta t}$ is

$\displaystyle W=Q_{h}-Q_{c}=Q_{h}-K\left(T_{h}-T_{c}\right) \ \ \ \ \ (1)$

where ${Q_{h}}$ is the heat expelled to the outside and ${K}$ is a constant.

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

 $\displaystyle \frac{Q_{c}}{T_{c}}$ $\displaystyle =$ $\displaystyle \frac{Q_{h}}{T_{h}}\ \ \ \ \ (2)$ $\displaystyle Q_{h}$ $\displaystyle =$ $\displaystyle \frac{T_{h}}{T_{c}}Q_{c}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle K\frac{T_{h}}{T_{c}}\left(T_{h}-T_{c}\right) \ \ \ \ \ (4)$

The work required to maintain a temperature of ${T_{c}}$ is therefore

 $\displaystyle W$ $\displaystyle =$ $\displaystyle K\frac{T_{h}}{T_{c}}\left(T_{h}-T_{c}\right)-K\left(T_{h}-T_{c}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{K}{T_{c}}\left(T_{h}-T_{c}\right)^{2} \ \ \ \ \ (6)$

Thus lowering the inside temperature by a small amount can have a large effect on the work required to maintain this temperature, and thus on the cost of running the air conditioner. For example, suppose the outside temperature is ${30^{\circ}\mbox{ C}=303\mbox{ K}}$ and the inside temperature is ${22^{\circ}\mbox{ C}=295\mbox{ K}}$. If we wish to lower the inside temperature by only one degree, the extra work required is

$\displaystyle \frac{W_{294}}{W_{295}}=\frac{295}{294}\frac{9^{2}}{8^{2}}=1.27 \ \ \ \ \ (7)$

We need to use 27% more power to achieve a single degree more cooling. This is one reason why it is much more economical to bear with a slightly higher indoor temperature on a hot day.

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