Phase transitions

Question

A 0.514 g sample of steam at 105.9 ∘C is condensed into a container with 4.93 g of water at 15.6 ∘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

Conservation of energy: the heat lost by the steam must equal the heat gained by the cooler water.

q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperature

For steam:

heat lost to go from 105.9º to 100º = q = mC∆T = (0.514 g)(2.01 J/gº)(5.9º) = 6.096 J

heat lost in phase change from steam to liquid @100º = q = m∆Hvap; since ∆Hvap is given in kJ/mol, we'll convert it to kJ / g. 40.7 kJ/mol x 1 mol / 18 g = 2.26 kJ/g. q = (0.514 g)(2.26 kJ/g) = 1.16 kg = 1160 J

heat lost to go from 100º to final temp = q = mC∆T = (0.514 g)(4.18 J/gº)(100-Tf) = 214.9 - 2.15Tf

Adding these up we have 6.096 J + 1160 J + 214.9 - 2.15Tf

For cooler water:

heat gained to go from 15.6º to final temp = q = mC∆T = (4.93 g)(4.18 J/gº)(Tf-15.6º) = 20.6Tf - 321.5

Setting heat lost by steam to heat gained by water, and solving for Tf, we have...

6.096 + 1160 + 214.9 - 2.15Tf = 20.6Tf - 321.5

22.75Tf = 1702.5

Tf = 74.8º

(Be sure to check all of the math)