SUMMARY: Unless I'm missing something here or you're doing very odd things with your refrigerator, you'd at most save a couple dollars per year by keeping your fridge/freezer full. Moreover, stocking up on water (or other things) to fill up fridge/freezer space won't save you much at all unless you're keeping it stored there for a VERY long time, since it costs so much energy to cool the water in the first place.
There are valid reasons to do this:
- if you need to survive intermittent power outages (as TFD notes), having a lot of ice or food in the fridge will keep it cool longer
- a larger amount of food in the fridge/freezer will make it easier to chill or freeze new food more rapidly, which may sometimes aid in food safety/preservation
- similarly, having more food may help to minimize small temperature fluctuations when opening the door frequently, again perhaps aiding in food safety/quality in some cases
- if you have a very inefficient fridge that cools unevenly or is not well-insulated, having more food will prevent as much cycling on and off (though packing a fridge too full can also prevent it from cycling correctly)
All of these are perhaps good reasons to have a mild preference for keeping the fridge a little more full. But, from an energy perspective, there isn't a rationale for deliberately filling your fridge with excess food/water, since the energy required to cool any solids or liquids is usually many, many times the amount needed to cool air.
Also, if your main concern is cold air "falling out" of the fridge when the door is open, I'd suggest filling it with empty containers that have only air in them. They'll give you the benefit of not losing that excess cold air, but without the energy expenditure to cool a liquid you don't need to. (But again, the likely benefit is probably a couple dollars per year at most.)
I tried searching for some reliable stats, and though I find many, many sources which make this claim, I don't generally see actual numbers regarding energy savings or even a theoretical calculation to support the logic of the practice.
In fact, it sometimes seems to appear on "myths" lists of energy groups, such as here:
- MYTH: You can save energy by keeping your fridge full, closing it quickly, and regularly cleaning the coils.
Actually, all three of these actions are not worth your trouble. In
the study done by Balsnik it was found:
Total use from ALL fridge door openings adds up to <50 kWh/yr, or about $5.
Putting water bottles in your fridge to keep it full adds up to <0.1 kWh/yr.
Cleaning coils – no actual savings found.
Or from this document (on ultralow temperature freezers' efficiency):
A full freezer takes less energy to operate: An
ostensible reason for this idea is that the thermal mass takes longer
to warm up, so the compressor doesn’t have to work as hard. Think
about it: while the contents take longer to warm up, it also takes
longer to cool down so the compressor works just as long every day.
The fundamental insulating factors of wall thickness and gasket
integrity do not change with a full or empty freezer, so why should it
make a difference to heat transfer? While cycle frequency will go
down, cycle duration will go up. The heat entering the cabinet will
not change. There is a nominal power spike at the beginning of each
compressor cycle, so more cycles could plausibly increase energy use a
little bit. Data has not been widely shared, so it remains in urban
legend status for now.
The logic of this last reference seems to address the question of whether an unopened freezer/refrigerator will be more efficient when it is full (as is sometimes claimed -- that the fridge will have to "work less hard" somehow). Obviously that doesn't make much sense, as this quote notes.
However, to judge this accurately, we'd need to take into account what happens when you open the fridge/freezer. Here's a report from an energy-savings group that tried a number of tests (including opening the door for various amounts of time). They concluded that top freezer models of fridges did use less energy when full, though they note in their analysis that it doesn't include the additional energy required to cool the extra food in the first place. But once that food is cool and the fridge is full, there is some energy benefit for top freezers. (How much is unknown, since their graphs don't have numbers.) For other types of freezer models, the results of the tests were mixed, so there was no clear benefit of an empty vs. full fridge. Their conclusion: "So our advice is to not worry about keeping the fridge full, and focus more on keeping the door closed."
For a theoretical perspective on fridge opening, let's try some reasonable assumptions:
The average fridge size in the U.S. is around 20 ft3. If we assume that the fridge isn't packed full and half of the air present is replaced by room temperature air when the door is opened, that would be roughly 10 ft3, or about 0.28 m3.
Using stats from here, we can calculate that cooling that 10 ft3 of air down by 20°C (e.g., from "room temperature" of about 25°C to 5°C) would require about 6.8 kJ of energy, or 0.0019 kWh. For a freezer of similar size, the temperature of the air would likely have to be lowered about 40°C, rather than 20°C, so these numbers would be doubled.
If we open the fridge door 20 times per day, over a year that would add up to about 13.8 kWh for a fridge with 10 ft3 of empty space, or 27.5 kWh for a freezer with a similar amount of empty space. The stats in the first quotation above estimate 50 kWh/yr for all fridge door openings, so the numbers seem to be in the right ballpark. Basically, it costs a couple dollars every year in energy loss for opening the fridge.
Now, suppose we instead loaded up that 10 ft3 with water instead of air. (This is a ridiculously large amount of water, but I'm using it to keep the occupied volume the same for a comparison.)
The amount of energy required to cool water from room temperature can be calculated similarly from these numbers. Cooling 10 ft3 of water by 20°C would require approximately 23,000 kJ. Freezing it to -15°C from 25°C would require about 120,000 kJ. (This number is significantly higher, due to the excess energy required to turn liquid water into solid ice.) The effect of adding large amounts of water was shown clearly in the study mentioned above, where the addition of 150 lbs. of room temperature water caused the fridge to spike to about 65°F and take almost a day and a half to return to normal temperature.
To put these numbers in a more useful form:
You'd have to cool down air in the fridge about 3500 times to "pay for" the amount of energy expended cooling the same volume of water.
You'd have to cool down air in the freezer about 9000 times to "pay for" the amount of energy expended in freezing the same volume of water.
UPDATE: As Joe rightly points out in comments, I assumed dry air here to simplify the calculations. But real kitchen air will be moist, and its effect is not insignificant. (I assumed the error would be less than 50% or so, but under reasonable assumptions, it's probably off by a factor of 1.5-3, depending on your kitchen humidity and how humid your fridge is.)
Anyhow, assuming we start with a relative humidity of 50% in the kitchen at 25°C, and we assume the refrigerator cools to 5°C in the fridge and -15°C in the freezer while maintaining 50% relative humidity at those temperatures (which would obviously require water vapor removal), here are some updated stats:
- You'd have to cool down air in the fridge about 1800 times to "pay for" the amount of energy expended cooling the same volume of water.
- You'd have to cool down air in the freezer about 5500 times to "pay for" the amount of energy expended in freezing the same volume of water.
[See calculations below for details.]
Basically, depending on how often you open your fridge and the room temperature, you will probably need to refrigerate water for at least several months before seeing any energy savings (at all). You'd likely need to keep (the same) water frozen for at least a year to get any energy savings. Even then, for reasonable quantities of water (e.g., a few gallons), it's unlikely that you'd save more than a few dollars per year in energy costs (and likely less).
A final note about full fridges: even supposing you manage to save a few cents per year with a full fridge, my practical experience tells me that I hold the door open much longer when the fridge is full than when it's nearly empty, since I often need to move stuff around or take things out temporarily to get things in the back. So would this theoretical savings ever actually materialize? I don't know.
For those who are interested, here is the "work" for the calculations above. I assume a volume of 10 ft3 = ~0.28 m3. Note that various approximations were used here to get a "ballpark" figure -- in particular, densities and specific heats were assumed to be constant over the temperature range, which might introduce a 5-10% error for the air calculations, and much less for the water calculations.
(1) Cooling (dry) air by 20°C
- 0.28 m3 of air × density of 1.205 kg/m3 at 20°C from table = 0.337 kg
- 0.337 kg × 20°C [same as 20 K] × specific heat of 1.005 kJ/(kg K) = 6.8 kJ
- 6.8 kJ ÷ 3600 = 0.0019 kWh
(2) Cooling (dry) air by 40°C
- Same weight of initial air
- 0.337 kg × 40°C × 1.005 kJ/(kg K) = 13.6 kJ
(3) Cooling water from 25°C to 5°C
- Same volume of 0.28 m3
- 0.28 m3 × density of about 1000 kg/m3 = 280 kg
- 280 kg × 20°C × specific heat of 4.18 kJ/(kg K) from table = 23400 kJ
- NOTE: Obviously one cannot and should not fill a home fridge with ~600 lbs. of water, but I used the same volume here to make the energy required for comparable volumes, since it's asserted that replacing air by an equivalent volume of water will make a difference.
(4) Cooling water from 25°C to -15°C
- Ice is less dense than water, so to achieve a final volume of 10 m^3, we must start with less water.
- 0.28 m3 × density of ice of 916.8 kg/m3 = 256 kg
- Cool to 0C: 256 kg × 25°C × specific heat 4.18 kJ/(kg K) = 26800 kJ
- Freeze: 256 kg × heat of freezing 334 kJ/kg = 85700 kJ
- Cool ice to -15°C: 256 kg × 15°C × specific heat of ice 2.108 kJ/(kg K) = 8100 kJ
- Total cooling energy: 120,700 kJ
(5) Cooling similar amount of water to air in fridge = 23400 kJ ÷ 6.78 kJ = about 3450 times greater
(6) Cooling similar amount of water to air in freezer = 120700 kJ ÷ 13.6 kJ = about 8900 times greater
(7) Cooling air at 50% relative humidity by 20°C:
- We get weight fractions of water vapor in air at 50% humidity from a Mollier diagram. Here x at humidity of 0.5 is about 0.0098 kg/kg at 25°C and about 0.0026 kg/kg at 5°C.
- We then follow the calculation of enthalpy (H) of moist air as found at Joe's link here.
- At 25°C: H = (1.005 kJ/kg°C)(25°C) + (0.0098 kg/kg)[(1.84 kJ/kg°C)(25°C)+(2501 kJ/kg)] = 50.1 kJ/kg
- At 5°C: H = (1.005 kJ/kg°C)(5°C) + (0.0026 kg/kg)[(1.84 kJ/kg°C)(5°C)+(2501 kJ/kg)] = 11.6 kJ/kg
- Delta H (change in enthalpy) = 50.1 - 11.6 = 38.5 kJ/kg
- Moist air is slightly less dense than dry air: using figures from here, moist air is about 1.199 kg/m3 at 20°C.
- Mass of air using above volume assumption of 0.28 m3 is 0.336 kg
- Energy required to cool = change in enthalpy × mass = 38.5 kJ/kg × 0.336 kg = 12.9 kJ
- Note that various numbers here may vary slightly over the temperature change, but as in Joe's link, we can assume they're constant enough that it won't affect the final answer by more than a few percent.
(8) Cooling air from 25°C to -15°C in freezer
- Using the Mollier diagram linked above, we get a weight fraction of approximately 0.00055 kg/kg for 50% humidity at -15°C
- Use similar calculations to above
- H at -15°C = -13.7 kJ/kg
- delta H from 25°C to -15°C = 63.8 kJ/kg
- using mass and density as above, the total energy required to cool is = 21.4 kJ
(9) We calculate the ratios as above, ending up with 1800 times more energy to cool an equivalent volume of water in fridge, and 5600 times more energy to freeze it.
(10) Relative humidity can vary in both the kitchen and fridge, so these calculations should only be taken as a ballpark figure, perhaps varying by a factor of 2-3 in either direction in extreme cases. Regardless, the amount of energy needed to cool even moist air is insignificant compared to that required to cool any liquid or solid food.