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Jun 19th, 2015

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 x Heat of fusion = 90 x 334 = 30,060 Joules

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

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

670 x 4.186 x ∆T = 90 x 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 670 grams of water at 13.3˚C and 90 grams of water at 0˚

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

Now, let’s do the entire problem at once to check the answer!

Heat absorbed by ice + heat absorbed by 0˚water = Heat released by 24˚ water

90 x 334 + 90 x 4.186 x (Tf – 0) = 670 x 4.186 x (24 – Tf) 30060 + 376.2 Tf = 67214.4 – 2800.6 Tf 3176.8 x Tf = 37154.4 Tf = 11.7˚C