How much energy does it take to turn a 15 gram pice of ice into 80F water?

Answer 1

I'm assuming that the ice is at 0°C.
80°F = 26.666°C.
79.7cal/g * 15g * 4.186J/cal = 5,004J fron ice to water.
1cal/g-deg * 15g * 4.186J/cal * 26.666deg = 1,674J from 0°C water to 26.666°C water.
Total is 5,004 + 1,647 = 6,651J


Doug

Answer 2

Hm, 400 calories to turn the water at 32 oF to 80 oF, but forget the calorie exchange from ice to water.

BTW you should be working in degrees centigrade/ celcius not farenheit if working with grams

1 calorie = 1 gram water ^ 1 oC