# Question fe189

Jun 15, 2016

$\text{7.0 kJ}$

#### Explanation:

A substance's heat of vaporization, $\Delta {H}_{\text{vap}}$, essentially tells you how much heat is required in order for one mole of said substance to go from liquid at its boiling point to vapor at its boiling point.

In your case, water is said to have a heat of vaporization of

$\Delta {H}_{\text{vap" = "42 kJ mol}}^{- 1}$

This tells you that in order to convert one mole of water from liquid to gas at its boiling point you must supply it with $\text{42 kJ}$ of heat.

Now, when water condenses it releases heat. More specifically, it releases the same amount of heat it absorbed when it evaporated at its boiling point.

$\text{liquid " stackrel(color(white)(+)color(blue)(DeltaH_"vap")color(white)(aaaaa))(->) " vapor}$

$\text{liquid " stackrel(color(white)(a)color(blue)(-DeltaH_"vap")color(white)(aaaa))(larr) " vapor}$

When water condenses, the heat of condensation is equal to

$\Delta {H}_{\text{cond" = -DeltaH_"vap}}$

The minus sign is used because it denotes heat released.

So, the first thing to do here is use water's molar mass to determine how many moles of water you have in that $\text{3.0 g}$ sample

3.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.1665 moles H"_2"O"

Since $1$ mole of water releases $\text{42 kJ}$ of heat when it condenses, it follows that $0.1665$ moles will release

0.1665 color(red)(cancel(color(black)("moles H"_2"O"))) * "42 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("7.0 kJ")color(white)(a/a)|)))#

The answer is rounded to two sig figs.

So, when $\text{3.0 g}$ of water condense, $\text{7.0 kJ}$ of heat are being released.