[Elepsis]

I'm not so sure this logic makes sense -- in fact I think it might be entirely backwards.

I've always been under the impression that though you pay a penalty for freezing things the first time you put them in, thereafter a full refrigerator (and freezer) is actually much more efficient because it's more difficult to keep air cold than it is to keep any variety of items that are more dense. If we're talking about leakage, too, air obviously exits the freezer much more often than its contents do.

[jacquesm]

reduce it to an absurdity:

If you imagine a giant freezer filled with ice it will eventually melt. This takes energy, a larger freezer will take more, an empty one will take less than a full one.

Counteracting the heat transferred through the freezer walls from the outside in will require an amount of energy proportional to the mass in the freezer and the leakage from the outside.

It takes energy to maintain that imbalance, the bigger the imbalance the faster the heat will transfer.

If the stuff in the freezer is light (air) then it will take less energy to keep it cold than if the stuff in the freezer is heavy (ice), because there simply is less leakage from air (a thermal insulator) to the walls than there is from ice (a pretty good thermal conductor).

Thermal conductivity is not very tightly coupled to mass but it usually is a good indicator, in the case of ice vs air it works pretty good.

So, it takes more energy to keep a freezer full of ice at a low temperature than a freezer full of air...

For an encore, which is heavier: A cubic meter of dry air vs a cubic meter of humid air ?

[derobert]

I believe the inefficiency of an empty freezer/fridge actually comes from it being inefficient to cycle the compressor. The low heat capacity of air leads to the compressor cycling often.

[sokoloff]

Most people will get the last question wrong.

Humid air is less dense (dry air heavier) because water has a lower molecular weight than the average molecule of dry air.

[swombat]

Air has a very low heat capacity. It doesn't take much energy to cool a cubic metre of air quickly (see air con, for example). On the other hand, cooling a cubic metre of water and other watery solids takes a load more energy.

In terms of keeping it cool, that probably depends on how good your freezer's insulation is. If it's really good, it probably doesn't take too much energy to keep things cool (air or otherwise). If it's not so good, you'll incur a constant "your freezer sucks" tax.

In the case being discussed, we're talking about freezing stuff every day, and letting it thaw in/near your bed during the night. So the cost would definitely be there.

[calambrac]

No doubt there's still an additional cost, I just don't think it would be high enough to make it more expensive than running an air conditioner, given that you're going to be running the freezer regardless (depending, of course, on how much we're talking about. I'm picturing a window unit vs. a window fan with a baking pan's worth of ice).

Any ideas on how to be more rigorous about thinking this through or running an experiment?

[swombat]

You could take a certain quantity of ice, figure out how long it takes to melt under a fan in hot weather, then calculate the energy required to get that quantity of water into ice (heat capacity equations), and you'd have a rough figure for energy spent per unit time.

Then compare that with the energy consumption of an air-con unit.

[derobert]

An experiment is easy to run. Obtain a Kill-a-Watt or other power meter, then measure (a) how much power the freezer uses over a day w/o freezing extra ice for fan-ice cooling, repeating several times; (b) how much power the freezer uses over a day with freezing extra ice for fan-ice cooling, repeating several times; (c) how much power the fan uses blowing over the ice; (d) how much power an AC uses to keep the room comfortable, repeated several times; (e) [bonus] how much power an AC set to a higher temp. + a fan blowing on you uses to keep the room comfortable, repeated several times.

You can then compare (b)-(a)+(c) vs. (d) or (e).

Ideally, you'd do these experiments in identical conditions (temp., sunlight, wind, humidity, etc.), of course.