Cool it down

We seek to calculate the average rate of energy usage:

$\langle P_{fridge} \rangle = \frac{1}{T}\int\limits_0^{T} dt P_{pump}$

In order to calculate this, we have to find out how long it takes for the fridge to warm up by $\Delta T$ , as well as how long the pump needs to run for to remove the energy associated with the warming. First, let us calculate the time to warm, $t_{warm}$ .

The problem states that our refrigerator loses gains energy by convective heat transfer, i.e., the heat transfer between the room and the fridge (with the pump turned off) is proportional to the temperature difference between them:

$\dot{Q}_{fridge} \sim k_H \left(T_{room} - T_{fridge}\right)$

where $k_H$ is a coefficient describing the rate of heat transfer. This relation is known as Newton's law of cooling.

We can relate this rate of energy exchange to the rate of temperature increase. For ordinary solids, the relationship between the change in heat energy and change in temperature is given by $\Delta Q_{fridge} = m c_H \Delta T_{fridge}$ , or $\dot{T}_{fridge} = \frac{1}{m c_H}\dot{Q}_{fridge}$ . This leads to:

$\dot{T}_{fridge} \sim \frac{k_H}{mc_H}\left(T_{room} - T_{fridge}\right)$

Solving this relation yields an exponential approach to room temperature:

$T_{fridge}(t) = T_{room} + \left[T_{fridge}(0)-T_{room}\right]e^{-\gamma t}$

Here, the constant $\gamma$ , describing the rate of cooling is proportional to $\frac{1}{m c_H}\dot{Q}_{fridge}$ .

If the fridge starts out at $T_{fridge}(0) = T_{fridge}^0$ , then it will warm by $\Delta T$ at time $t_{warm}$ given by

$T_{fridge}^0+\Delta T = T_{room} + \left[T_{fridge}^0-T_{room}\right]e^{-\gamma t_{warm}}$

or

$\begin{aligned} t_{warm} &= \frac{1}{\gamma}\log\frac{T_{fridge}^0+\Delta T-T_{room}}{T_{fridge}^0-T_{room}} \\ &= mc_H \log\frac{T_{fridge}^0+\Delta T-T_{room}}{T_{fridge}^0-T_{room}} \\ &\sim mc_H \end{aligned}$

which shows that the time required to warm the refrigerator is proportional to the heat capacity of the solid multiplied by a factor containing the information about the temperature gap between the room and solid. Importantly, the time is linearly proportional to the heat capacity of the solid. In the case of a compound solid (made of several kinds of material), we replace $mc_H$ by $\sum\limits_i m_ic_H^i$ .

Now, let us calculate the time required by the pump to remove the added heat from the refrigerator.

If the fridge has a total heat capacity $\sum\limits_im_ic_H^i$ , it gains the heat $\Delta T \sum\limits_im_ic_H^i$ upon warming by $\Delta T$ . As the pump can remove $r_{pump}$ Joules per unit time, this removal will take $\frac{\Delta Q}{r}$ seconds, or

$\begin{aligned} t_{pump} &= \frac{\Delta T \sum\limits_im_ic_H^i}{r_{pump}} \\ &\sim \frac{\sum\limits_im_ic_H^i}{r_{pump}} \equiv \frac{M_{eff}c_H^{eff}}{r_{pump}} \end{aligned}$

We find that the pump runs for a time proportional to the total heat capacity of the fridge.

Since the time for warming and the time for cooling are both proportional to constants times the total heat capacity $M_{eff}c_H^{eff}$ , that implies that the average energy use is precisely the same in both cases.

$\begin{aligned} \langle P_{fridge} \rangle &= \frac{1}{T}\int\limits_0^T dt P_{pump} \\ &= \frac{r_{pump}\times t_{pump} + 0\times t_{warm}}{t_{pump}+t_{warm}} \\ &= \frac{r_{pump} M_{eff}c_H^{eff}/r_{pump}}{ M_{eff}c_H^{eff}/r + M_{eff}c_H^{eff} \log\frac{T_{fridge}^0+\Delta T-T_{room}}{T_{fridge}^0-T_{room}}} \\ &= \frac{r_{pump} /r}{ 1/r + \log\frac{T_{fridge}^0+\Delta T-T_{room}}{T_{fridge}^0-T_{room}}} \end{aligned}$

which is independent of the total heat capacity.

Therefore, it doesn't matter how the refrigerator is filled.