100808-1933 EST
chris:
I am not in condition to answer your question with accuracy. Meaning I do not know where is some data.
Here is some ballpark idea and then I will tell you how to make a measurement.
I have two freezers, both old. One was bought in 1972 for about $350 is 18 cubic feet and has never been serviced. Runs very well in the garage even below zero. The second is newer to us, slightly smaller, maybe 14 to 16 cubic feet, and only cost about $35. It also works in cold weather. Both will cool to an average temperature of about 0 deg F. with an ambient of 85 degree F, and a duty cycle of around 50%.
The power consumption for each is about 400 W at the start of the cooling cycle and thru the cycle gradually drops to about 320 W. So average consumption in the summer is about 160 W if nothing is added and door is kept shut.
Put about 2 quarts of fresh blackberries in one freezer the other day and these might have required 0.2 KWH to cool to 0 deg F from 85 deg F.
How you can run a controlled experiment. Get a Kill-A-Watt EZ. Monitor your freezer for maybe 6 cool-warm cycles. Start monitoring at the start of a cool period and end at the end of a warm cycle. Probably do this over night. From this you get your average heat absorption from the outside to inside. Do not open the door over this period, do not put anything in the freezer for maybe 4 hours before the test.
You should know your average internal temperature and average ambient temperature. Hopefully the ambient will be fairly stable.
Ideally you need a thermocouple in the water you are going to freeze to monitor its temperature. But not essential.
Use a large known quantity of water in your test. Maybe 4 gallons in small containers.
It is my recollection it took more than a day to cool 1 gallon to 0 deg from room temperature. This was in a plastic gallon jug.
At the beginning of a cool period put your water in the freezer. Monitor the KWH used from this point until the water reaches the average of the freezer and end the KWH collection at the end of the warming period. This energy is the normal loss thru the box plus the additional energy to cool the water. You need the total time of this test, and the average power from the first test, meaning you also needed its time. Multiply the average power of the first test by the time of the second test, and subtract that calculated energy of the first test from the energy of the second test and this result is the energy to cool your known amount of water. Some errors result from opening the door.
When my dad was a boy they had an uncooled ice house in the town. In the winter ice was cut from the lake and placed in the heavily insulated ice house. So the costs were labor, the capital to build the ice house, taxes, and some miscellaneous expenses. He was in an area where there were still snow banks on the ground at the 4ft of July.
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