Question

12.78

A sample of steam with a mass of 0.514 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.50 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, 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

mass of steam = 0.514 g
molar mass of steam (H2O) = 18 g/mol
number of moles of steam = mass / molar mass
                                                       = 0.514/18
                                                       = 0.0286 mol

Heat lost by steam when it converts to water = n*ΔHvap
                                                                                             = 0.0286 * 40.7 KJ
                                                                                             = 1.162 KJ
                                                                                             = 1162 J

Let final temperature be T
total heat lost by steam = heat gained by 4.5 g water
1162 + 0.514* 4.18*(100-T) = 4.5 * 4.18 *(T-5)
1162 + 214.9 - 2.15*T = 18.81*T - 94.05
20.96*T = 1470.95
T = 70.2 oC
Answer: 70.2 oC