Question

How much heat (in kJ) is required to warm 12.0 g of ice, initially at -10.0 ∘C, to steam at 111.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

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 = 12*2.01 * (0 – -10) = 241.2

Q2 = 12*334 = 4008

Q3 = 12*4.184 * (100 – 0) = 5020.8

Q4 = 12*2264.76 = 27177.12

Q5 = 12*2.03* (111– 100) = 267.96

QT = Q1+Q2+Q3+Q4+Q5 = 241.2+4008+5020.8+27177.12+267.96 = 36715.08J

QT = 36.715 kJ