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 a perfectly 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?

The heat of fusion of water, as given in Wikipedia, is 79.72 cal/g. So during melting, the 20 g of ice absorbs 1594.4 calories (the small kind). The ice was at exactly 0° Celsius as it was already in a partially melted bucket.

The new water from the melted ice will also absorb some more calories: 20 calories per degree as there are 20 grams. Let the new temperature be x. The 100 g of old water will lose 100 calories per degree that it lowers below 20. The heat gained by the new ice/water must equal the heat lost by the old water:

1594.4 + 20*x = 100*(20-x) = 2000 - 100*x

120*x = 2000 - 1594.4 = 405.6

x = 405.6/120 = 3.38

Since the 2000 calories based on the 100 grams of water at 20°C is larger than the heat of melting of the ice, the ice does completely melt and the water at the end is at 3.38°C.

Posted by Charlie on 2013-04-04 14:12:22