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 40-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 40 grams will heat to the same final temp to which the 295-g of 25 degree water cools (m x deltaT x Cp).
1. (40-g x 16 ^oC x 2.027 J/g) + (40-g x 334 J/g) + [40-g x (Tf-0 ^oC) x 4.184 J/g^oC] = 295 x (25 - Tf) x 4.184 J/goC

2. 1412.48 + 13,360 + 167.36 Tf = 30857 - 1234.28 Tf

3. 1494.84Tf = 29622.72

4. Tf = 19.8^oC (Tf = final equilibrium temp)