Given that everything else is equal (model of fridge, temperature settings, external temperature, altitude), over a given duration of having the door closed, does it require more electricity to cool an empty refrigerator AND maintain that temperature, than a full one?
Thermodynamics – Does an Empty Refrigerator Require More Power than a Full One?
everyday-lifethermal conductivitythermodynamics
Related Solutions
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.
Without doing the analysis, I would think that a cooling system is more effective at extracting heat from a warm container than from a cold one.
For a fridge, the effectiveness (or coefficient of performance) is $Eff=Q_c/W$ is the ratio of the heat removed from the cold source (the fridge) to the energy used for the purpose. It increases with the temperature of the cold source. This is actually the prime factor to be considered in the analysis.
If we assume that, for a given current temperature of its contents, and thus for a given coefficient of performance, the cooling capacity of the fridge (heat removed per second) is limited only by the power of its cooling engine (I do not know whether that is the case), then the heat pump will pump more heat per second when the fridge is warm.
Hence it is better to put all bottles at once and get the fridge warmer to have a maximum heat pumpimg capacity from the heat pump.
Heat sharing rate within the fridge may also be an important issue, but there are no data available to measure how important. If it is really low, thus leaving an important temperature gradient in the fridge, it may be useful to exchange the position of bottles, so as to have the warmer part of the load near the heat pump and have work with highest possible coefficient of performance.
Precise figures about the load do not matter very much. However, a load with large heat capacity will take longer to cool and will thus allow more time for heat sharing.
In the second part below, we prove formally that all bottles should be cooled at once, and we use the understanding to discuss the heat sharing issue in some more depth. The variability of the coefficient of performance with temperature is central to this analysis.
FORMAL STATEMENT AND PROOF
A refrigerator is a Carnot machine functionning as a heat exchanger, where we are interested in removing heat from the low-temperature reservoir, using work from an engine that provides compression.
The effectiveness, or coefficient of performance, noted here $Z$, is defined as $Z=Q_c/W$ where $W$ is the work provided and $Q_c$ is the amount of heat extracted from the cold source (the refrigerator) with that work. If we note $Q_h$ the amount of heat delivered to the hot source (outside the refrigerator), we have the equality $Q_h=W+Q_c$.
For an ideal Carnot cycle, we have $Z_{ideal}=Q_c/W=Q_c/(Q_h-Q_c)=T_c/(T_h-T_c)$ where $T_h$ and $T_c$ are the temperatures of the hot and cold source. (see http://en.wikipedia.org/wiki/Coefficient_of_performance).
Of course, the actual coefficient of performance $Z$ is less that the Carnot ideal. Short of knowing its specific, we will only assume that, like the ideal coefficient, it depends monotonically on the temperature $T_c$ of the cold source, the hot source (outside the refrigerator) being considered at constant temperature. Hence we only assume that the coefficient of performance $Z$ is a strictly increasing function of (cold source) temperature, i.e., such that $T_1< T_2\ \Rightarrow\ Z(T_1) < Z(T_2)$
We also assume that the mechanical power available for compression is invariant, i.e. does not depend on the temperature of the sources, at least within the range of temperatures considered.
Finally, we also assume that the heat capacity of the refrigerator itself is negligible, and that heat sharing within the refrigerator takes negligible time compared to cooling time so that the content may be considered to have uniform temperature. These later two assumptions will be discussed afterwards.
With the above assumptions, given two masses $m_1$ and $m_2$ to be cooled in the cold source, it is faster to cool booth simultaneously than to try to cool one first and later add the second one. It also consumes less energy.
PROOF
Cooling a mass $m$
Actually, the formulae above are about heat and work increments. This is necessary since $Z$ is temperature dependant, and temperature may vary. Also, since we intend to analyze the system from the point of view of the cold source, the heat increments are actually removed from that source and must be countd as negative.
So we can write $Z= -dQ_c/dW$, or $dQ_c/dW=-Z$. The power provided by the compressor is a constant $P=dW/dt$. Hence $dQ_c/dt= (dQ_c/dW)\times(dW/dt)=-Z\times P$.
On the other hand we know that removing heat reduces the temperature
according to the formula $\Delta Q=-cm \Delta T$, where $m$ is the mass
being colled and $c$ is the specific heat for the substance
constituting that mass.
Hence we have $dQ_c/dt= cm(dT_c/dt)$.
Combining the two formulae, we get $dT_c/dt=-ZP/cm$. But we cannot resolve this equation since $Z$ is an unknown function of $T_c$.
What we know is that $Z(T_c)$, $P$, $c$ and $m$ are strictly positive values. So $dT_c/dt$ is strictly negative. Hence $T_c$ will decrease with time. Since the function $Z(T_c)$ is a strictly increasing function, its value will also decrease with time, and hence the absolute value of the derivative $dT_c/dt$ will also decrease with time.
Hence the graphical representation of the evolution of the temperature will look like the red curve in the figure, where $T_0$ is the initial temperature at time $t_0$.
Cooling the same mass $m$ in two steps
If we consider cooling independently (in an identical refrigerator) another mass $m_1$, smaller than $m$, with initial temperature $T_0$ we get another curve, like the curve in blue in the left part of the figure up to point C, corresponding to the equation $dT_c/dt=-ZP/cm_1$. It is below the red curve because the smaller mass $m_1$ cools faster than $m$. Formally, if we draw a horizontal line like the line cutting both curves in A and B, this corresponds to the same temperature for both curves, hence to a common value of the coefficient of performance $Z$. Then $m_1< m\ \Rightarrow\ (dT_c/dt)_A<(dT_c/dt)_B$. Since this is true for any value of the temperature $T_c$, it confirms that the blue curve for $m_1$ decreases faster than the red curve for $m$.
Suppose now that at time $t_2$ the mass $m_1$ has been cooled to temperature $T_2$ corresponding to point C below the red curve. We add to $m_1$ another mass $m_2$ such that $m=m_1+m_2$, the mass $m_2$ being at the initial temperature $T_0$.
The mass $m_2$ being warmer than $m_1$ will share its heat with $m_1$ (in negligible time according to our hypothesis) so that both reach the temperature $T_1$ corresponding to point D, and pursue cooling.
At any time between $t_0$ and $t_2$, the temperature of $m_1$ (blue) is less than the temperature of $m$ (red). Hence the refrigerator works with a lower coefficient of performance $Z$ for $m_1$ than for $m$, and less heat has been removed from the refrigerator containing $m_1$ than from the refrigerator containing $m$ at time $t_2$. When we introduce the mass $m_2$ with $m_1$, the total heat introduced in the refrigerator is that of $m_1+m_2$ at temperature $T_0$. This is exactly the same as the heat introduced in the refrigerator containing $m$. Since less heat was removed from the $m_1+m_2$ refrigerator at time $t_2$, it is at a higher temperature that the $m$ refrigerator. Hence the point D is above the red curve.
The $m_1+m_2$ refrigerator now contains the same mass as the $m$ refrigerator. Hence it will follow an identical curve. But it is at temperature $T_1$ that was attained earlier, at time $t_1$ by the $m$ refrigerator. So the right part of the blue cooling curve for $m_1+m_2$, starting at point D is the same as the right part of the red coling curve for $m$ starting at point B, translated by a duration $\Delta t=t_2-t_1$.
Conclusion
The masses $m_1+m_2$ will always reach any temperature with a delay $\Delta t$ after the mass $m$ has reached it. Actual figures would be required to be more precise.
Given the problem, the cooling engine will be working at maximum power to get the fastest possible cooling in both cases. Then it is obvious that the faster solution is also the most economical energetically. This assumes either that the cooling is started just at the right time, or that the cooling power is reduced once the right temperature has been attained.
These results are based exclusively on our assumptions, independently of any actual figures. We will now discuss some of these assumptions.
DISCUSSION
Heat capacity of the refrigerator
We have assumed that the heat capacity of the refrigerator itself is negligible. We should however analyse its effect. We first note that the objective is to extract heat from a given number of bootles to bring them from $T_0=30^{\circ}C$ to $T_f=6^{\circ}C$. That corresponds to a precise amount $Q$ of heat to be removed, independently of the process used for that purpose.
If the refrigerator itself is initially at temperature $T_f$, it will at the start share heat with the mass to be cooled, thus warming up and cooling the bootles. But then it will have to be cooled down to $T_f$ again, so that it net contribution to the cooling process will be null, and the same amount $Q$ of heat has to be removed by the heat pump. However, by sharing heat at the beginning, it induces an early cooling, thus making the whole heat removal process operate at a lower temperature, hence with a lower coefficient of performance $Z$.
The net effect of the heat capacity of the refrigerator is thus to provide some early cooling, which can sometimes be considered an advantage, but at the expense of a lower effective coefficient of performance $Z$. These effects increase with the heat capacity of the refrigerator.
Note that if the initial temperature of the refrigerator is below the targeted final temperature $ T_f$, the difference multiplied by the heat capacity is a net contribution to the refrigeration process, though the loss on the coefficient of performance remains.
Hence it is better not to have anything else in the refrigerator, even already cooled, unless it is cooled to a much lower temperature than the final temperature $ T_f$ intended for the bottles.
Heat sharing rate
As we have seen from the previous discussion, the main objective is to remove a given amount $Q$ of heat, and the effectiveness of the removal decreases as temperature is lowered. If the rate of heat sharing within the refrigerator is small, the volume near the cooling system will cool faster, thus reducing the coefficient of performance, i.e. the rate of heat removal.
Hence, ensuring the best possible heat sharing can help in all circumstances. It should be noted that direct sharing between cylindrical bottles will be reduced to a minimum: just one line of contact. So, if space allows, it is probably preferable to help air circulate between the bottles. Keeping the bottles in vertical position will help if the refrigerator has grid shelves that let air through, rather than glass shelves. And, of course, the bottles must be unpacked.
Opening the door to quickly exchange bottles so that the warmer ones are placed near the cooling system will improve the coefficient of performance and reduce cooling time. It may have some cost in warming the refrigerator, but that is less important if the heat capacity of the load to be cooled is large (actual measurements would be useful).
Exchanging bootles will also avoid having to cool those close to the cooling system below the required temperature $ T_f$ in order to have all the bottles temperature at least as low as $ T_f$.
Best Answer
The two "no" answers you've already received are correct for all practical purposes.
In real-world cases there can be a difference though. The difference depends on when the refrigerator decides to cycle on and cool. If the fridge cycles on a timer or based on heat energy then there will be a difference due to the added heat capacity. The outside of the refrigerator will acquire heat due to conduction, convection, and radiation from external sources.
All heat transfer depends on $\Delta T$. The greater the difference in temperature between two systems the faster heat will flow. When you add heat energy to a full refrigerator, the system has greater heat capacity so the temperature changes more slowly and $\Delta T$ is greater than it would be in an empty refrigerator. If the refrigerator could keep the temperature absolutely constant at all times the difference would not matter. Because a real system cools and then stops cooling in discrete steps, a loaded refrigerator acquires heat from the environment slightly faster because it stays colder for longer.
The difference is so small though that I'm sure it couldn't be measured.
But, if the refrigerator only cycles on at a specific temperature there will be no difference. I wrote a numerical simulation to test this. Here is the plot:
The simulation is a simplified, idealized, purely numerical one, hence the lack of units. The Y axis is temperature and the X axis is time. The red curve is the loaded refrigerator and green curve is the unloaded one. For the simulation I made the loaded fridge have double the heat capacity.
Assuming the refrigerator only cycles on when the temp reaches the top of the graph then there is no difference between the two. Although twice as much heat energy must be put into the loaded refrigerator, the simulation shows that it takes exactly twice as long. That is, the average rate of heat flow is identical.
So, the answer to the question depends on how the refrigerator works.
If the refrigerator is time interval or heat energy interval based, a loaded fridge takes more energy to maintain a cold temperature.
If the fridge is purely thermostat based, there is no difference in energy consumption.