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Old 03-02-12, 06:22 AM   #4
Plantman
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Join Date: Feb 2012
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Here are some statistics I found online. I think the calculations are correct:

For water at its normal freezing point of 0 ºC, the specific heat of Fusion is 334 J g-1. This means that to convert 1 g of ice at 0 ºC to 1 g of* water at 0 ºC, 334 J of heat must be absorbed by the water. Conversely, when 1 g of water at 0 ºC freezes to give 1 g of ice* at 0 ºC, 334 J of heat will be released to the surroundings.

One joule is 0.239 calories.

0.239 calories x 334 J g-1 = 79.826 calories removed to form ice per gram or cubic centimeter of water
1 US gallon = 3,785.41178 cm3
3,785.41178 cm3 x 79.826 calories = 302174.28 calories to melt I gallon of ice
1 BTU=0.252164401 kilocalories or 252 calories

so 1 gallon releases 302174/252 or about 1200 BTU when it melts but there is no temperature change
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