How much energy is needed to vaporize 0.22 kg of water?

1 Answer
Apr 27, 2016

Answer:

#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#

Explanation:

Let's assume that the water is ready to be vaporized, i.e. it's already at the boiling point, and that we just need to put in enough energy to vaporize it. We need to know the enthalpy of vaporization for water (also called the latent heat of vaporization). This can be found from standard tables as:

#Delta H_(vap,water) = 2257 kJ//kg#

Now we just need to use this like a conversion factor to find the energy we are looking for knowing the mass of the water we are working with:

#E = Delta H_(vap,water) * m_(water) = (2257 kJ*0.22kg)/(kg) = 496kJ#