# 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#?

##### 1 Answer

#### 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

#"1 kJ" = 10^3"J"#

In your case,

#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 * 10^3#

which means that

#2.26 * 10^3"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.