Question

At 1 atm, how much energy is required to heat 65.0 g of H2O(s) at –18.0 °C to H2O(g) at 161.0 °C?

Answer 1

Enthalpy of fusion - Hf = 0.334 kJ/g = 334 J/g
Enthalpy of vaporization - Hvap = 2.26 kJ/g = 2260 J/g
Heat capacity of solid phase Cp,s = 2.11 J/ (g K)
Heat capacity of liquid phase Cp,l = 4.18 J/(g K)
Heat capacity of gaseous phase Cp,g = 2.08 J/(g K)

now you have few processes of heating ice to vapor:
1. heating ice to 0°C
delta T = 0°C - 24°C= 273K - 255K = 18K
Q1 = m * Cp,s * deltaT = 65g * 2.11 J/(g K) * 18K = 2468.7 J = 2.468 kJ
2. melting the ice
Q2= m * Hf = 65g * 334 J/g = 21710 J = 21.7kJ
3. heating water to 100°C
delta T = 100°C - 0°C = 373K - 273K = 100K
Q3= m * Cp,l * deltaT = 65g * 4.18 J/(g K) * 100K = 27170 J = 27.2 kJ
4. vaporization of water
Q4= m * Hvap = 65g * 2260 J/g = 146900 J = 146.9 kJ
5. heating of vapor
delta T = 161°C - 100°C = 434K - 373K = 61K
Q5= m * Cp,g * deltaT = 65g * 2.08 J/(g K) * 61K = 8247.2 J = 8.247 kJ


total heat is Q = Q1 + Q2 + Q3 + Q4 + Q5 = 2.468kJ + 21.7kJ + 27.2kJ + 146.9kJ + 8.247kJ = 206.515 kJ