Question

Calculate the enthalpy change, ΔH, for the process in which 22.2 g of water is converted from liquid at 12.1 ∘C to vapor at

25.0 ∘C .

For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and Cs = 4.18 J/(g⋅∘C) for H2O(l).

Answer 1

We require 2 type of heat, latent heat and sensible heat

Sensible heat (CP): heat change due to Temperature difference

Latent heat (LH): Heat involved in changing phases (no change of T)

Then

Q3 = m*Cp wáter * (Tb – Tf)

Q4 = m*LH vap

Note that Tf = 0°C; Tb = 100°C, LH ice = 334 kJ/kg; LH water = 2264.76 kJ/kg.

Cp ice = 2.01 J/g°C ; Cp water = 4.184 J/g°C; Cp vapor = 2.030 kJ/kg°C

Then

Q3 = 22.2*4.184 * (25– 12.1) = 1198.21

Q4 = 22.2*44000/18 = 54266.6 J

QT = 1198.21+54266.6

QT = 55464.81 J = 55.45 kJ