MadSci Network: Chemistry |
The answer to this problem requires two quantities: the heat capacity of water (conveniently 1 calorie, or 4.184 joules, per gram per degree Celsius) and the heat of fusion of ice. It's complicated by the fact that I don't know whether you are talking Fahrenheit or Celsius, but we'll assume Celsius. The amount of heat given off by 28,000 gal of water cooling 15 degrees is easily calculated from the heat capacity of water. The amount of ice should be just enough to absorb that heat. If you want the whole assemblage (the ice and the 28,000 gal of water) to come to 80 degrees, you need to allow for the amount of heat energy required to (a) melt the ice and (b) bring the melted ice (e.g. water at 0 C) to 80 degrees. More accurately, you need to calculate how much ice (assumed to be at 0 degrees C) is needed to absorb the heat given off by the 28,000 gal. If you use the heat capacity of water AND the heat of fusion of ice, you can set up an equation in one variable (the mass of the ice). Naturally I don't want to deprive you of the fun of actually doing the calculation. Dan Berger MadSci Administrator
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