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

Apr 27, 2016

$E = \Delta {H}_{v a p , w a t e r} \cdot {m}_{w a t e r} = \frac{2257 k J \cdot 0.22 k g}{k g} = 496 k J$

#### 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:

$\Delta {H}_{v a p , w a t e r} = 2257 k J / k g$

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:

$E = \Delta {H}_{v a p , w a t e r} \cdot {m}_{w a t e r} = \frac{2257 k J \cdot 0.22 k g}{k g} = 496 k J$