Question

The standard enthalpy change of vaporization of water is +40.7 kJ mol-1. An electric kettle is rated at 3kW (1kW = 1 kJ s-1). How long will it take to evaporate 1.0 kg of water at 100°C? Give the answer in minutes.

Answer 1

+40.7 kJ mol^-1 is same as per 18.02 g of water

We have 1000 g (1 kg)

Simple math gives us:

40.7 x 1000 / 18.02 = 2258.6 kJ

of energy needed to evaporate water

3 kW = 3 kJ / s or

3 kJ of energy is used per 1 second

2258.6 / 3 = 752.87 s or 12.5 min