55,035.5 Joules
Heat of vaporisation is the amount of energy required to transform a mole of liquid water into gas at constant temperature.
Given the following parameters:
Mass of water = 24.40 grams
ΔHvap = 40.6 kJ/mol
Determine the moles of water
Mole = mass/molar mass
Mole = 24.40/18
Mole of water = 1.356moles
Determine the required heat required
Q = mole of water * ΔHvap
Q = 1.356moles * 40.6kJ/mole
Q = 55.0355kJ
Q = 55.0355 * 1000
Q = 55,035.5 Joules
Hence joules of heat needed to completely vaporise 24.40 grams of water at its boiling point is 55,035.5 Joules
