The Cold Drink
At a bar, there is a bucket containing ice, some of which has melted. A bartender gets an ice cube weighing 20 grams from the ice bucket and puts it into an insulated cup containing 100 grams of water at 20 degrees Celsius. Will the ice cube melt completely? What will be the final temperature of the water in the cup?
It takes 334 Joules to melt one gram of ice at 0C into water at 0C, so melting 20 grams of ice takes 20 x 334 = 6,680 Joules.
It takes 4.186 Joules to raise the temperature of a gram of water by 1C, so 4.186 x 100 x 20 = 8,372 Joules of heat energy available in the water. Since 8,372 > 6,680, all the ice will melt, leaving 1,692 Joules.
The final temperature of the water at equilibrium is therefore 1692 / (120 * 4.186) = 3.368C.
All this of course assumes an ideal insulator, as well as assuming that the ice was 0C. The way the question is phrased leads to the assumption that the ice has already started melting, and is therefore at 0C uniformly, but this is not guaranteed. The interior of the ice could still be below 0C while the exterior is melting at 0C. If this were the case, additional energy would be used to raise the temperature of the ice to 0C, and thus the equilibrium temperature of the water would be lower.
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