Question

Two 20.0-g ice cubes at –16.0 °C are placed into 295 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

Answer 1

First, the 20-g of ice at -16 must warm to 0 (m x deltaT x Cp) then each gram must absorb 334 J/g to melt (m x Hf) , and finally the 20 grams will heat to the same final temp to which the 295-g of 25 degree water cools (m x deltaT x Cp).
1. (20-g x 16 oC x 2.027 J/g) + (20-g x 334 J/g) + [20-g x (Tf-0 oC) x 4.184 J/goC] = 295 x (25 - Tf) x 4.184 J/goC

2. 648.64 + 6680 + 83.68Tf = 30857 - 1234.28Tf

3. 1317Tf = 23528.36

4. Tf = 17.86 oC (Tf = final equilibrium temp)