Question

How much heat in kJ is required to warm 10.0g of ice, initially at -10.0oC, to steam at 110.0oC? The heat capacity of ice is 2.09 J/goC and that of steam is 1.84 J/goC.

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 = 10*2.09 * (0 – -10) = 209

Q2 = 10*334 = 3340

Q3 = 10*4.184 * (100 – 0) = 4184

Q4 = 10*2264.76 = 22647.6

Q5 = 10*1.84* (110– 100) = 184

QT = Q1+Q2+Q3+Q4+Q5 = 184+22647.6+4184+3340 +209

Qt = 30564.6 J = 30.56 kJ