How much ice to lower the temp of water?

Question

I'd like to find out how much ice it would take to lower the temp of the water in my pool from 92 degrees to 82 degrees. My pool is approximately 13,500 gallons.

Answer 1

the answer is...

WAY more ice than you are willing to put in the pool!

look at it this way, 1350 gallonsof ice is one percent of the total water volume. if it cooled 1350 gallons of water to halfway between 32 and 92 degrees that would be 2700 galons of h2o at 62 degrees, which would get you to 86 degrees. if you replaced a full 20% of the water with ice, you would be at 80 degrees, but that is 2700 gallons of ice!!!!

Answer 2

Roughly speaking, one pound of ice will produce a 1-degree (Fahrenheit) drop in 44 pounds of water.

So, if you want to drop the temperature 10 degrees, you'll need 10 lbs of ice for every 44 lbs of water. That means 3,000 "gallons" of ice--I doubt you'll have enough room!

It's true that "cold" ice is slightly more effective than ice that is near its melting point. But usually the majority of the cooling effect comes from the melting of the ice, not from its initial coldness (unless it is REALLY cold).

Answer 3

10F drop = 5.56C drop (if you don't believe me, convert both 92 and 82 to C then do the subtraction)

To melt Ice at 0C, you will need 1 unit of ice for every 80 units of water that drops by 1C. (latent heat of fusion for ice is 80cal/g-C, 1 calorie drops the temperature of 1 gram of water by 1C).

After that, the water left by the ice will consume another 82 calories per gram to rise up to 82C

So (5.56/{80 + 82})(13,500gallons) (8.337lbs/gallon)

(5.56/162)(13,500gallons) (8.337lbs/gallon)
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>>>>>> [ ~3,863 pounds of ice ] <<<<<<<<<<
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Answer 4

The ice you intend to use could be at different temperatures , that makes the difference, you should introduce another variable to have your answer.