How much heat is required to boil 83.0 g of water at its boiling point?

Nov 25, 2015

$\text{206 kJ}$

Explanation:

Assuming that pressure is equal to $\text{1 atm}$, boiling water at its boiling point implies providing it with enough heat to turn it from liquid at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$.

The amount of heat needed to allow one mole of water to undergo this phase change is called the enthalpy chnge of vaporization, $\Delta {H}_{\text{vap}}$.

For water at ${100}^{\circ} \text{C}$, the enthalpy change of vaporization is equal to

$\Delta {H}_{\text{vap" = "40.66 kJ/mol}}$

http://www2.bren.ucsb.edu/~dturney/WebResources_13/WaterSteamIceProperties/EnthalpyOfVaporizationH2O.pdf

This tells you that in order to boil one mole of water at its boiling point, you need to provide it with $\text{40.66 kJ}$ of heat.

Your strategy now would be to use water's molar mass to determine how many moles of water you have in $\text{83.0 g}$

83.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.607 moles H"_2"O"

So, if you know how many moles of water you have, and how much energy is needed per mole, all you have to do is use a simple proportion to get

4.607color(red)(cancel(color(black)("moles"))) * "44.66 kJ"/(1color(red)(cancel(color(black)("mole")))) = "205.75 kJ"

Rounded to three sig figs, the number of sig figs you have for the mass of water, the answer will be

$q = \textcolor{g r e e n}{\text{206 kJ}}$