Question

21. If a 2.50 mol sample of water is heated from -50 °C to 75 °C at a constant pressure of 1 atm, what is the change in enthalpy?

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

Q1 = m*Cp ice * (Tf – T1)

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tb – Tf)

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

Q1 = m*2.01 * (0 – -50)

Q2 = m*334

Q3 = m*4.184 * (75– 0)

now, substitute data

2.5 mol --> 2.5*18 = 45 grams ofw ater

Q1 = 45*2.01 * (0 – -50)

Q2 = 45*334

Q3 = 45*4.184 * (75– 0)

QT = 45*(2.01*(50) + 334 + 4.184*75)

QT = 33673.5 J

Note that since this is P = 1 atm, constant

Q = dH

dH = 33673.5 J = 33.67 kJ