Question

At 1 atm, how much energy is required to heat 43.0 g of H2O (s) at -12.0 C to H2O(g) at 159.0 C?
?     kJ

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 = 43*2.01 * (0 – -12) = 1037.16

Q2 = 43*334 = 14362

Q3 = 43*4.184 * (100 – 0) = 17991.2

Q4 = 43*2264.76 = 97384.68

Q5 = 43*2.03* (159– 100) = 5150.11

QT = Q1+Q2+Q3+Q4+Q5 = 1037.16+14362+17991.2+97384.68+5150.11

QT = 135925.15 J = 135.93 kJ