Question

Calculate the amount of energy in kilojoules needed to change 189 g of water ice at − 10 ∘C to steam at 125 ∘C . The following constants may be useful:

Cm (ice)=36.57 J/(mol⋅∘C)

Cm (water)=75.40 J/(mol⋅∘C)

Cm (steam)=36.04 J/(mol⋅∘C)

ΔHfus=+6.01 kJ/mol

ΔHvap=+40.67 kJ/mol

Express your answer with the appropriate units.

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)

Q4 = m*LH vap

Q5 = m*Cp vap* (T2 – Tb)

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 – T1)

Q2 = m*334

Q3 = m*4.184 * (100 – 0)

Q4 = m*2264.76

Q5 = m*2.03* (T2 – 100)

QT = Q1+Q2+Q3+Q4+Q5

QT = m * (2.01 * (0 – T1) + 334 + 4.184 * (100 – 0) + 2264.76 + 2.03* (T2 – 100))

substitute T1 and T2

QT = 189 * (2.01 * (0 – -10) + 334 + 4.184 * (100 – 0) + 2264.76 + 2.03* (125– 100))

QT = 583633.89 J

Q = 583.634 kJ