Question

Calculate the mass of water vapour that could be produced when (1.73x10^3)kJ of energy is added to an excess amount of liquid water at its boiling point.

Answer 1

The latent heat of vapourization of water(delta Hv )= 2260KJ/Kg (at 1atm).

Energy used to vapourize the "M" mass of water from it's boiling point (i.e. from 100 degree C)

E (kg)= M(kg) x delta Hv (in KJ/Kg).................(1)

So mass of water evoporate when E=1.73x103 KJ .................(2)

M= (E) / (delta Hv.)..................(3)

So,

M = (1.73x103KJ) / 2260 KJ/Kg........................(4)

solving equation (4) we get,

M= 0.765 Kg

M= 765 gm.