Question

At 1 atm, how much energy is required to heat 69.0 g of H2O(s) at –12.0 °C to H2O(g) at 153.0 °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 = 69*2.01 * (0 – -12) = 1664.28

Q2 = 69*334 = 23046

Q3 = 69*4.184 * (100 – 0) = 28869.6

Q4 = 69*2264.76 = 156268.44

Q5 = 69*2.03* (153– 100) = 7423.71

QT = Q1+Q2+Q3+Q4+Q5 = 1664.28 + 23046 +28869.6+156268.44+7423.71 = 217272.03

QT = 217272.03 J

QT = 217.27 kJ