# Question 48544

Oct 26, 2016

$\text{16.7 kJ}$

#### Explanation:

The problem provides you with

• the enthalpy of vaporization for water, $\Delta {H}_{\text{vap" = "40.68 kJ mol}}^{- 1}$

This tells you how much heat is needed to convert one mole of water from liquid at its boiling point to vapor at its boiling point

• the enthalpy of fusion for water, $\Delta {H}_{\text{fus" = "6.01 kJ mol}}^{- 1}$

This tells you how much heat is needed to convert one mole of water from solid at its freezing point, i.e. from ice, to liquid at its freezing point

In your case, you want to know how much heat is needed to melt solid ice at its freezing point, so you will have to use the enthalpy of fusion.

The sample of ice was given to you in grams, so start by converting it to moles by using water's molar mass

50.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "2.7755 moles H"_2"O"

Now use the enthalpy of fusion as a conversion factor to determine how much heat would melt this sample of ice at its melting point, i.e. at ${0}^{\circ} \text{C}$

2.7755 color(red)(cancel(color(black)("moles H"_2"O"))) * "6.01 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("16.7 kJ")color(white)(a/a)|)))#

The answer is rounded to three sig figs.