Question

A sample of steam with a mass of 0.532 g at a temperature of 100 ∘C condenses into an insulated container holding 4.25 g of water at 4.0 ∘C. (For water, ΔH∘vap=40.7 kJ/mol and Cwater=4.18 J/(g⋅∘C).)

Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer 1

Molar mass of water = 18.01532 g/mol (For water, ΔH^∘_vap=40.7 kJ/mol and C_water=4.18 J/(g⋅^∘C).)

(40.7 kJ/mol) x ((0.532 g H2O) / (18.01532 g H2O/mol)) = 1.2018 kJ = 1201.8 J lost by the steam

Supposing the heat lost by the steam is gained by the cold water:
(1201.8 J) / (4.18 J/g⋅^∘C) / (4.25g) = 67.65 ^∘C (an insulated container holding 4.25 g of water at 4.0 ^∘C)

change
4.0 ^∘C + 67.65 ^∘C = 71.65 ^∘C

Now you have two bodies of water, 0.532 g at 100^∘C, and 4.25 g at 71.65 ^∘C,

so find the final temperature using a weighted average:
((0.532g x 100^∘C) + (4.25g x 71.65^∘C)) / (0.532g + 4.25g) = (53.2 g/^∘C + 304.51g/^∘C ) / (4.782g) = 74.80^∘C

Assuming that no heat is lost to the surroundings.

Therefore, the final temperature of the mixture = 74.80^∘C