Energy required to heat water

Question

At 1 atm, how much energy is required to heat 53.0 g H2O(s) at −22.0 ∘C to H2O(g) at 119.0 ∘C? In KJ.

Answer 1

Energy is going to be required to

1) Raise the temperature of the ice

2) Melt the ice (at 0°C)

3) Raise the temperature of the water

4) Boil the water

5) Raise the temperature of the water vapor

1) Use q = mcΔT. (We're raising this to 0° since that's when we'll have a phase change)

q = 53g * 2.09J/g°C * (0°C - (-22°C) ) (2.09 is the specific heat of ice)

2) Use the heat of fusion of water. The heat of fusion of water is 334J/g. So the energy required for this is

334J/g * 53g

3) Use q = mcΔT

q = 53g * 4.18J/g°C * (100°C - (0°C) )

4) Use the heat of vaporization of water. The heat of vaporization of water is 2230J/g. So the energy required for this is 2230J/g * 53g

5) Use q = mcΔT

q = 53g * 2.09J/g°C * (119°C - (100°C) )

Add up all the energies. You'd also have to change to kJ