Heat of fusion of ice = 334 J/g Specific heat of water = 4.186 J/g*˚C

1st determine the amount of heat energy required to melt all the ice. The water’s temperature will decrease as the ice melts. Energy absorbed by ice = mass * Heat of fusion = 90 * 334 = 30,060 Joules

2nd determine the temperature of the water, after releasing the heat energy.

Energy released by water = Mass * Specific heat * ∆T Energy released by water = 670 * 4.18 * ∆T

670 * 4.186 * ∆T = 90 * 334 ∆T = 10.7˚ The water’s temperature decreased 10.7˚ as all the ice melted. Final temperature of water = 24-10.7 = 13.3˚ Now you have 665 grams of water at 16.4˚C and 80 grams of water at 0˚

3rd determine the final temperature of the water. 90 * 4.186 * (Tf – 0) = 670 * 4.186 * (13.3 – Tf) Divide both sides by 4.186 90 * Tf = 670 * (13.3 – Tf) 760 * Tf = 8911 Tf = 11.7˚