Question

At 1 atm, how much energy is required to heat 53.0 g of H2O(s) at –10.0 °C to H2O(g) at 139.0 °C?

Answer 1

This is calculated in 5 stages.

1.

Q = mcΔT : m = mass; c = specific heat of ice; ΔT = temp difference;

Q = Heat absorbed or Released.

To heat the ice from -10°C to 0°C (Δ T = 10°C)
= (53 g) x (2.1 J/g/°C) x (10°C) = 1,113 J

2.

Q = mC : mass x Laten heat of melting.
To melt the ice at 0°C to water at 0°C.
= (53 g) x (334 J/g) = 17,702 J

3.

Q =mcΔT = Q: m = mass; c = specific heat of water; ΔT = temp. difference;

Q = Heat absorbed or Released

To heat the water from 0°C to boiling at 100°C (Δ T = 100°C)
= (53 g) x (4.184 J/g/°C) x (100°C) = 22,175.2 J

4.

Q = mC : mass x Laten heat of vaporisation.
To vaporise the water to steam at 100°C
= (53g) x (2,260 J/g) = 119,780 J

5.

Q = mcΔT : m = mass; c = specific heat of steam: ΔT = temp. difference;

Q = Heat absorbed or Released

To heat the steam from 100°C to 139°C (Δ T = 39°C)
= (53 g) x (2.0 J/g/°C) x 39°C = 4,134 J

Total heat required = 1,113 J + 17,702 J + 22,175.2 J + 119,780 J + 4,134 J

                              = 164904.2 J

                               = 164.90 kJ