Question

At 1 atm, how much energy is required to heat 73.0 g of H2O(s) at –22.0 °C to H2O(g) at 121.0 °C?

Answer 1

H2O(s) (-22.0 C)................H20(s)(0 C)...............H20(l)(0 C)....................H2O(l)(100 C)...............H20(g)(0 C)..............H2O(121 C)

73 g is 73/18 = 4.05 moles

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)
= 73g x 2.1 J/g/°C x 22°C = 3.33726KJ .

2...mC = Q: mass x Latenr heat of melting.
To melt the ice at 0°C to water at 0°C.
= 73g x 334 J/g = 24.382 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
= 73g x 4.184 J/g/°C x 100°C = 30.5432 KJ

4...mC = Q: mass x Latenr heat of vaporisation.
To vaporise the water to steam at 100°C
= 73g x 2,260 J/g = 164.980 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 121°C
= 73g x 2.01 J/g/°C x 21°C =3.08133 KJ

Total heat required = 226.323 KJ