A 0.557 g sample of steam at 104.3 ∘C is condensed into a container with 5.66 g of water at 14.0 ∘C.

Question

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

This is a step-wise process.

Step 1: Convert the steam @ 104.3º to steam at 100º

q = mC∆T = (0.557 g)(2.01 J/gº)(4.3º) = 4.81 J

Step 2: Convert the steam @ 100º to liquid @ 100º (phase change, no change in temp)

q = m∆Hvap = (0.557 g)(1 mol / 18 g)(40.7 kJ/mol) = 1.26 kJ = 1260 J

Step 3: Sum the heat lost by the steam

4.81 J + 1260 J = 1265 J

Step 4: Heat lost by the steam must equal the heat gained by the water

heat gained by water = q = 1265 J = mC∆T

1265 J = (5.66 g)(4.18 J/gº)(∆T)

∆T = 1265 J / (5.66 g)(4.18 J/gº)

∆T = 53.5º

Final temperature of the water mixture = 14.0º + 53.5º = 67.5ºC