A 0.579 g sample of steam at 103.5 ∘C is condensed into a container with 4.79 g of water at 14.3 ∘C. What is the final temperature of the water mixture if no heat is lost?

Question

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

Heat released by the steam MUST equal heat gained by the water

Take it in steps.

(1) heat released going from 103.5º to 100º

q = mC∆T = (0.579 g)(2.01 J/gº)(3.5º) = 4.07 J

(2) heat released going from steam @100º to liquid water @100º (phase change, no ∆ in temp)

q = m∆Hvap = (0.579 g)(1 mol/18 g)(40.7 kJ/mol) = 1.309 kJ = 1309 J

(3) heat released going from 100º to the final temp

q = mC∆T = (0.579 g)(4.18 J/gº)(100º - Tf) = 242 - 2.42Tf where Tf is the final temperature

ADDING these up for heat lost by original steam, we have...

4.07 + 1309 + 242 - 2.42Tf

For the water, we have

(1) heat gained by water

q = mC∆T = (4.79 g)(4.18 J/gº)(Tf - 14.3º) = 20.0Tf - 286

Setting heat lost by steam to heat gained by water, we have...

4.07 + 1309 + 242 - 2.42Tf = 20.0Tf - 286

Solving for Tf (final temperature)...

22.42Tf = 1841

Tf = 82.1ºC

(be sure to check the math)