Question

Four ice cubes at exactly 0 ∘C with a total mass of 53.0 g are combined with 140 g of water at 90 ∘C in an insulated container. (ΔH∘fus=6.02 kJ/mol, cwater=4.18J/g⋅∘C). If no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer 1

The ice cubes will do two things:
1) melt while at zero Celsius
2) heat up as a liquid to the final temperature
The 90 degree water will cool down to the final temperature.
The key is that the heat gained by the ice cubes is equal to the heat lost by the 90 C water cooling down.
q = (140 g) (90 - x) (4.18J/g⋅∘C) <--- that's the set up for the 90 degree water. The 'x' is the final temp.

heat gained by ice = heat lost by water = q = (m)(C)(∆T)
step 1: convert 6.02 kJ/mol to kJ/g = 6.02 kJ/mol x 1mol/18g = 0.334kJ/g =334 J/g
step 2: (53g)(334J/g) = 17702 J to melt the ice; now there is 53g water at 0˚C
step 3: 17702J + (53g)(4.18J/g/deg)(Tf-0) = (140g)(4.18J/g)(90-Tf)
17702 J + 221.5Tf = 60192 - 585.2 T2
806.7Tf = 42490
Tf = 52.67 degrees C