Please help chem question

Question

A 0.506 g sample of steam at 106.7 ∘C is condensed into a container with 5.39 g of water at 14.9 ∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 J g⋅ ∘C, the specific heat of steam is 2.01 J g⋅ ∘C, and Δ𝐻vap=40.7 kJ/mol.

Answer 1

These problems are best approached in a step-wise fashion.

The heat lost by the hot steam MUST EQUAL the heat gained by the cooler water.

Step 1: heat lost by steam going from 106.7º to 100º = q = mC∆T = (0.506 g)(2.01 J/gº)(6.7º) = 6.81 J

Step 2: heat lost by steam converting to water @ 100º = q = m∆Hvap = (0.506 g)(40.7 kJ/mol)(1 mol/18g) = 1.14 kJ = 1140 J

Step 3: heat lost by 100º water upon mixing = q = mC∆T = (0.506 g)(4.18 J/gº)(100-Tf) = 212 - 2.1Tf

Step 4: total heat lost by steam = 6.81 J + 1140 J + 212 - 2.1Tf

Step 5: heat gained by cooler water = q = mC∆T = (5.39 g)(4.18 J/gº)(Tf - 14.9) = 22.5Tf - 336

Step 6: Set heat lost to heat gained and solve for Tf (final temp)

6.81 J + 1140 J + 212 - 2.1Tf = 22.5Tf - 336

1359 + 336 = 22.5Tf + 2.1Tf

1695 = 24.6Tf

Tf = 68.9º