Question

Calculate the mass of water vapour produced when 9.0 x 10^3 kJ of energy is added to a large amount of liquid water at its boiling point. Please show your work. /4

Answer 1

enthalpy of vaporization for water = 40.66 kJ/mol

molar mass of water = 18 g/mol

So, heat of vaporization is 40.66 kJ for 18 g

Hence, enthalpy of vaporization = 40.66/18 = 2.259 kJ/g

Heat required to convert liquid water into vapors, at its boiling point, is given by:

or,    9.0 x 10³ kJ = m.(2.259 kJ/g)

solving for m, we get:

m = 3984.06 g = 3.984 kg

Therefore, 3.984 kg of water is produced when 9.0 x 10³ kJ of energy is added to a large amount of liquid water at its boiling point.