“Dedicated pool owners are dropping hundreds of pounds of ice into their 90-degree pools in hopes of some relief,” the Wall Street Journal reported last Thursday. “Only one problem: It doesn’t really work.” Why not? Assume a 15-by-30-foot pool, 6 feet deep. The water is 90 degrees Fahrenheit and you’d like to cool it to 80 degrees. How much ice would that take?
The pool holds just more than 20,000 gallons of water, all of which is 10 degrees too hot. To raise the temperature of one gallon of water by one degree Fahrenheit requires 2,100 calories. To cool a gallon by one degree requires getting rid of the same amount. To cool 20,000 gallons by 10 degrees means getting rid of 420,000,000 calories.
Ice cools by absorbing heat in two steps. First it melts, then the resulting water rises to the temperature of its surroundings. It takes about 36,000 calories to melt one pound of ice into 32-degree water. Each pound of ice produces about 0.12 gallons of water. Since it takes 2,100 calories to raise a gallon of water one degree, 0.12 gallons of water will absorb about 12,000 calories in the process of warming from 32° to 80°. Taking both steps together, one pound of ice will absorb about 48,000 calories in the process of becoming 80° pool water.
So to lower the temperature of a 20,000 gallon pool of 90 degree water by 10 degrees, you would need about 8,750 pounds of ice. A ten-pound bag of ice costs around a buck, so cooling your pool with ice cubes would cost $875. (It would also add about 3 inches to the depth of the water.) And, of course, as long as the air around the pool and the bodies in it are warmer than 80 degrees, the water would immediately start getting warmer again.
For those pool owners who would like to personalize this calculation, here’s how: 1) Take your pool’s volume, in gallons. 2) Divide by 1,000. 3) Multiply by the number of degrees (Fahrenheit) you’d like to cool the water. 4) Multiply that number by 43.75. 5) Think again.
