# Question 183ce

Jul 26, 2016

$\text{2.5 kJ}$

#### Explanation:

For starters, energy is not being absorbed when water condenses, it is being released.

Condensation is an exothermic process, which implies that when water goes from vapor at its boiling point to liquid at its boiling point, energy is being given off to the surroundings Now, in order to be able to solve this problem, you need to know the value of water's enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, which tells you how much heat is needed in order to convert $1$ mole of water from liquid at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$.

Likewise, the enthalpy of vaporization tells you how much heat is released when $1$ mole of water goes from vapor at ${100}^{\circ} \text{C}$ to liquid at ${100}^{\circ} \text{C}$.

In water's case, you have

$\Delta {H}_{\text{vap" = "40.66 kJ mol}}^{- 1}$

This value tells you that when $1$ mole of water evaporates at ${100}^{\circ} \text{C}$, $\text{44.066 kJ}$ of heat are being absorbed. Similarly, when $1$ mole of water condenses at ${100}^{\circ} \text{C}$, $\text{44.66 kJ}$ of heat are being released.

The first thing to do here is convert the sample of water from grams to moles by using water's molar mass

1 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.0555 moles H"_2"O"

So, if $\text{44.66 kJ}$ of heat are being given off when $1$ mole of water condenses at its boiling point, you can say that you have

0.0555 color(red)(cancel(color(black)("moles H"_ 2"O"))) * overbrace("40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_ 2"O")))))^(color(blue)(=DeltaH_ "vap")) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.5 kJ")color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you have one sig fig for the mass of water.

So, you can say that when $\text{1 g}$ of water condenses at ${100}^{\circ} \text{C}$, $\text{2.5 kJ}$ of heat are being released.