Question

At 1 atm, how much energy is required to heat 51.0 g of H2O(s) at –22.0 °C to H2O(g) at 141.0 °C?

Answer 1

This is calculated in 5 stages.

1...mcΔT = Q: m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released.
To heat the ice from -22°C to 0°C (Δ T = 22°C)
= 51g x 2.1 J/g/°C x 22°C ΔT = 2356.2J = 2.3562 kJ.

2...mC = Q: mass x Latenr heat of melting.
To melt the ice at 0°C to water at 0°C.
= 51g x 334 J/g = 17034 J = 17.034 kJ.

3...mcΔT = Q: m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released
To heat the water from 0°C to boiling at 100°C
= 51g x 4.184 J/g/°C x 100°C ΔT = 21338.4 J = 21.3384 kJ.

4...mC = Q: mass x Latenr heat of vaporisation.
To vaporise the water to steam at 100°C
= 51g x 2,260 J/g = 115260 J = 115.260 kJ.

5...mcΔT = Q: m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released
To heat the steam from 100°C to 143°C
= 51g x 2.01 J/g/°C x 43°C ΔT = 4407.93 J = 4.40793 kJ

Total heat required = 2.3562 kJ + 17.034 kJ + 21.3384 kJ + 115.260 kJ + 4.40793 kJ = 160.3965 kJ.