Question

At 1 atm, how much energy is required to heat 49.0 g of H2O(s) at –22.0 °C to H2O(g) at 127.0 °C?

Answer 1

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)
= 49g x 2.1 J/g/°C x 22°C = 2,263J = 2.263 kJ.

2...mC = Q: mass x Latent heat of melting.
To melt the ice at 0°C to water at 0°C.
= 49g x 334 J/g = 16,366 J = 16.3 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
= 49g x 4.184 J/g/°C x 100°C = 20,501 J = 20.5 kJ.

4...mC = Q: mass x Latent heat of vaporisation.
To vaporise the water to steam at 100°C
= 49g x 2,260 J/g = 110,740 J = 110.7kJ.

5...mcΔT = Q: m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released
To heat the steam from 100°C to 127°C
= 49g x 2.01 J/g/°C x 27°C = 2778 J = 2.778 kJ

Total heat required = 2.263 kJ + 16.3 kJ + 20.5 kJ + 110.7 kJ + 2.778 kJ = 152.5kJ.