I've been told I should defrost my freezer to save energy, wiki, here and here for example, but none of the linked sites is a peer-reviewed paper explaining why (the wiki article doesn't even have references), and I don't find it obvious. I don't understand how the mechanism works, and I ask you for a good paper read on the subject or an explanation.
[Physics] How does defrosting your freezer save energy
everyday-lifefreezingtemperaturethermodynamics
Related Solutions
If the temperature is not much below freezing, the rate of heat transfer from your plants (and particularly from the earth around their roots) is low, if there is a lot of water present, the high heat of fusion means that it will take a long time to actually freeze much of it. So maybe the plant makes it through the night without too much damage.
Note that if it doesn't warm up enough the next day the second night will kill them because it starts close to freezing.
Heat Capacity
You are comparing two situations, and to formalize this, let's say that the freezer has a given volume, $V$ and that volume is broken down into air and other frozen materials. Then we have the situation that $V=V_{air}+V_{stuff}$. The heat capacity of the entire freezer between the two situations will be vastly difference because the volumetric heat capacity for air and ice (which is the most common item included in "stuff") is vastly different by a factor of about 1000. I'll use the Wikipedia notations, where specific heat capacity is $c$ ($\frac{J}{kg K}$) and volumetric heat capacity is $c \rho$. The volumetric heat capacity times volume is the heat capacity (denoted $C$) so for the entire freezer we have the following.
$$C = \left(V c \rho \right)_{air} + \left( V c \rho \right)_{stuff}$$
The heat capacity matters because should you insert some mount of heat into the freezer, $Q$, then after equilibrium is reached, the temperature will raise by:
$$\Delta T = \frac{Q}{C}$$
So if we don't consider the work of the freezer thermal cycle actively cooling the air, then the freezer with more stuff in it will cool the item faster because the temperature of the air+stuff changes less, so the $\Delta T$ in the Newton cooling model will be greater and it will cool faster and cool to a lower temperature.
Air Cooling
It is relevant to note that the air is actively cooled by the freezer systems and the stuff is not. This is interesting because it will have relevance to the heat transfer mechanisms (more later). Although this isn't entirely correct, we could assume that the cooling device takes some flow rate of air continuously, $\dot{m}$ (kg/s), and cools with some enthalpy change, $\Delta h$ (J/kg). That enthalpy change removes heat from the system, but that doesn't change from one case to the other.
But that can't be the full picture. Heat is only removed from the freezer if the temperature drops below a certain value. So, another way to do this may be to assume that the cooler keeps the air at a certain temperature, but that may defeat the purpose of the problem. It is also possible that the thing being cooled, in fact, outpaces the cooling device and heats the air faster than it can be cooled. In that case the air temperature would rise above the programmed setpoint and stay there until the item being cooled started to approach the freezer temperature.
Heat Transfer Mechanisms
There are two mechanisms of heat transfer between the item you are freezing and other stuff in the freezer.
- Convection with the air
- Radiative heat transfer with other items and the walls
- Conduction with the bottom surface (same for both and I won't discuss)
Obviously, the 1st one transfers with the air and the 2nd one transfers with the stuff. I would like to propose several simplified models such that we can talk more specifically about predictions.
- The freezer cools the item faster than the active cooling is relevant
- The item quickly equilibrizes with the air, then the stuff and the active cooling then slowly lowers the temperature
- The air is kept at a constant temperature
As I've said already, number 1 clearly has the freezer with more stuff win. Number 2 could potentially go in the direction of the freezer with less stuff because of the larger heat capacity of air (which will only work for a small item), but the radiative heat transfer still advantages the case with more stuff. In number 3 the freezer with more stuff clearly wins unless the flow path for convection is restricted.
The exact one that will win does depend on the specific values for the system, so that's in the hands of the experimentalist. Generally though, the higher heat capacity and the ability to transfer heat through radiative transfer is likely to favor the freezer with more stuff in it.
Best Answer
Refrigerators and freezers work by running a really cold liquid through cooling pipes fitted in the cavity to be cooled. This flow (the compressor) is switched off when the set temperature is reached, the faster the set temperature is reached, the less energy the appliance uses.
Cold fluid at $T_c$ runs through the cooling pipes. The cavity to be cooled is at $T_f$. Now let's look at small area $A$ on the surface of a cooling pipe.
When the cooling pipe is clean (not frosted over) then Newton's cooling law tells us that the heat flux (amount of heat removed per unit of time) $\dot{q}$ through $A$ is:
$$\dot{q}_\textrm{clean}=hA(T_f-T_c)$$
Where $h$ is the heat transfer coefficient.
But when the surface is frosted over with porous ice, then:
$$\dot{q}_\textrm{frosted}=uA(T_f-T_c)$$
It can be shown that:
$$\frac{1}{u}=\frac{1}{h}+\frac{\theta}{k}\implies u=\frac{hk}{k+h\theta}$$
Where $\theta$ is the thickness of the frosty material and $k$ the thermal conductivity of the frosty material.
Because the frosty material is a poor conductor of heat ($k$ has a low value):
$$h>\frac{hk}{k+h\theta}$$
(Note that the frosty material isn't pure ice, it's highly porous ice that contains much entrapped air, thereby further lowering the $k$ value of the frost). And this means that, all other things being equal:
$$\dot{q}_\textrm{clean}>\dot{q}_\textrm{frosted}$$
Multiply this of course for the total surface area of the cooling pipes. So clean cooling pipes carry away the heat more quickly, resulting in the compressor running for shorter times to reach the set temperature. This saves energy, Note also how freezers that have been frosted over more (higher frost thickness $\theta$) perform worse.
A slightly more detailed approach:
$$\dot{q}_\textrm{clean}=u_1A(T_f-T_c)$$ $$\dot{q}_\textrm{frosted}=u_2A(T_f-T_c)$$ Here it can be shown that:
$$\frac{1}{u_1}=\frac{1}{h_1}+\frac{\theta_1}{k_1}+\frac{1}{h_2}$$ And: $$\frac{1}{u_2}=\frac{1}{h_1}+\frac{\theta_1}{k_1}+\frac{\theta_2}{k_2}+\frac{1}{h_3}$$
But here too, because the frost conducts heat poorly ($k_2$ is small):
$$u_1>u_2$$
So that clean pipes carry off the heat more quickly, all other things being equal.
Symbols used in this section:
$h_1$: convection heat transfer coefficient, cooling fluid to metal.
$h_2$: convection heat transfer coefficient, metal to air.
$h_3$: convection heat transfer coefficient, frost to air.
$k_1$: thermal conductivity, metal.
$k_2$: thermal conductivity, frost.
$\theta_1$: thickness, metal.
$\theta_2$: thickness, frost.