2020-06-24

My apartment has an air conditioner, but I've never used it. Instead, every summer, I keep bottles of water in the fridge and drink it to cool down. (It helps that LA only gets really hot several days a month, and my apartment is only hottest around 5pm every day.) I was curious whether/how much energy I am actually saving. The calculation is simple: how much energy does it take to cool all the air in my apartment, versus chilling a bottle of water in the fridge?

At a rough estimate, my apartment has an area of 700 ft^{2}, and let's say 10 ft tall ceilings, for 7000 ft^{3} or 1.98×10^{8} cm^{3}. Assuming I like the room at 73F, and that it's 90F outside, the temperature drop is 90F - 73F = 305.372K - 295.928K = 9.444K. Finally, the specific heat of air in "typical room conditions" is 0.00121 J/cm^{3}K. Multiplying all of these together, we get

1.98×10^{8} cm^{3} × (305.372 - 295.928)K × 0.00121 J/cm^{3}K = 2,262,593.52 J

In other words, it takes just over 2 megajoules of energy to cool my apartment to 73F.

What about drinking chilled water instead? I have 1 liter Nalgene bottles in my fridge, which is kept at 4C. Going through the same calculation, we have 1000 cm^{3} of water, going from the same outside temperature of 90F = 305.372K to 4C = 277.15K. Water has a specific heat of about 4.1796 J/cm^{3}K. Multiplying all of these together, we get

1000 cm^{3} × (305.372 - 277.15)K × 4.1796 J/cm^{3}k = 117,956.6712 J.

This comes out to about 20 times less energy than air conditioning my apartment.