Specific heat chem 2

Question

A 0.522 g0.522 g sample of steam at 107.0 ∘C107.0 ∘C is condensed into a container with 4.63 g4.63 g of water at 17.0 ∘C.17.0 ∘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,4.18  J g⋅ ∘C, the specific heat of steam is 2.01  J g⋅ ∘C,2.01  J g⋅ ∘C, and Δ𝐻vap=40.7 kJ/mol.

Answer 1

0.522 g steam @ 107.0C

condensed into 4.63 g H2O @ 17.0C

? final temp of water

Heat lost by steam must be equal to heat gained by water

The 2 equations we'll use are:

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

q = m∆Hvap and ∆Hvap = 40.7 kJ/mol x 1 mol/18 g x 1000 J/kJ = 2261 J/g

(1) heat lost by steam

heat to change steam at 107 to steam at 100 = q = mC∆T = (0.522g)(2.01J/gº)(7º) = 7.34 J

heat to change steam to liquid water at 100º (phase change) = q = m∆Hvap = (0.522g)(2261 J/g) = 1180 J

Total heat lost by steam = 7.34 J + 1180 J = 1187 J

(2) heat gained by water = q = mC∆T

q = (4.63 g)(4.18 J/gº)(∆T) where q is the heat lost by the steam

1187 J = (4.63 g)(4.18 J/gº)(∆T)

∆T = 61.3º

Final temperature of the water mixture = 17.0º + 61.3º = 78.3ºC