# The latent heat of vaporization of water is 2260 J/g. How many kilojoules per gram is this, and how many grams of water will vaporized by the addition of 2.260*10^3 J of heat energy at 100°C?

Dec 4, 2015

$\text{2.26 kJ/g}$

#### Explanation:

For a given substance, the latent heat of vaporization tells you how much energy is needed to allow for one mole of that substance to go from liquid to gas at its boiling point, i.e. undergo a phase change.

In your case, the latent heat of vaporization for water is given to you in Joules per gram, which is an alternative to the more common kilojoules per mole.

So, you need to figure out how many kilojoules per gram are required to allow a given sample of water at its boiling point to go from liquid to vapor.

As you know, the conversion factor that exists between Joules and kilojoules is

$\text{1 kJ" = 10^3"J}$

In your case, $\text{2260 J/g}$will be equivalent to

2260 color(red)(cancel(color(black)("J")))/"g" * "1 kJ"/(1000color(red)(cancel(color(black)("J")))) = color(green)("2.26 kJ/g")

Now for the second part of the question. As you know,

$2260 = 2.26 \cdot {10}^{3}$

which means that

$2.26 \cdot {10}^{3} \text{J" = "2260 J}$

This is the latent heat of vaporization per gram of water, which means that adding that much heat to a sample of water will vaporize one gram of water at its boiling point.