This is a great question... there's a lot to love about free air conditioning!

It certainly would be an easy system to build. All you need is a big insulated container (probably in the form of a hole in the ground) with some coiled tubes at the bottom. You would run a chilled water circuit from the container to a radiator inside the air conditioner (see How Air Conditioners Works for details). You would need a small pump to pump the water in the chilled water loop, but that's it.

So let's make a couple of assumptions:

• Let's assume that your air conditioner runs for 12 hours a day for 3 months out of the year.
• Let's assume that your house has a 5 ton air conditioner (60,000 BTUs)
• Let's assume that we can store the snow and ice with 50 percent efficiency. That is, over the course of the summer we will lose half of it to melting, inefficiencies in our system, etc.
To cool the house you therefore need:
60,000 BTUs/hr * 12 hours/day * 90 days = 64,800,000 BTUs
Multiplying by our 50 percent efficiency rating, let's call it 130 million BTUs.

If you have a gram of ice at 32 degrees Fahrenheit (0 degrees Celsius), it will absorb 80 calories of energy converting from ice to liquid water. There are 252 calories in a BTU. So we need 3.15 grams of water to absorb one BTU of heat. The assumption here is that we are going to rely on the phase change from ice to water to power the air conditioner. Once all the ice melts, the water will warm up quickly.

So we need:

130,000,000 BTUs * 3.15 grams/BTU = 409,500,000 grams of ice
That's about 410,000 liters of ice, or 410,000 kilograms (902,000 pounds) of ice that you must store to cool your house all summer. That's a cube 740 centimeters (24.26 feet) on a side. Very roughly speaking, you would have to dig a hole as big as your house and insulate it well, and then in the winter you would have to shovel it full of nearly a million pounds of ice. But if you do that, you can cool your house for free! (The value of the equivalent electricity to cool the house, at 7.5 cents per kilowatt-hour, would be about \$1,500.)