If 25g of ice were to melt, 80*25 = 2000 calories would be removed from the Coke. To cool 350 g of Coke from 25 C to 0 C would required removing 8750 calories from the Coke. Therefore all of the ice will melt, and then heat up from 0C, while the Coke will be cooled to an intermediate temperature, T. To solve for that temperature, assume that all heat gained by the ice and water as it heats up equals the heat lost by the Coke.
80*25 + 80*1*(T-0) = 350*1*(25 - T)
2000 + 80T = 8750 -350 T
430T = 6750
T = 15.7 C
The heat of fusion of water is 80 cal/g. When 25 g of ice at 0ºC is placed in the contents of a 12 oz can of Coke (≈350 g) at 25ºC, what is the final temperature of the beverage?
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