# How much heat must be removed to freeze a tray of ice cubes if the water has a mass of 40.0 g? (The molar heat of fusion of water is 6.02 kJ/mol.)

Jun 11, 2017

$\text{13.4 kJ}$

#### Explanation:

The molar heat of fusion of water tells you the amount of heat

• needed to convert $1$ mole of water from solid at its freezing point to liquid at its freezing point
• given off when $1$ mole of water is converted from liquid at its freezing point to solid at its freezing point

You can thus say that when $1$ mole of liquid water goes from liquid at its normal freezing point of ${0}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$, $\text{6.04 kJ}$ of heat are being given off.

In other words, you have

• $\Delta {H}_{\text{fus" = -"6.02 kJ mol}}^{- 1} \to$ when going from liquid to solid

The minus sign is used here to denote heat given off.

• $\Delta {H}_{\text{fus" = +"6.02 kJ mol}}^{- 1} \to$ when going from solid to liquid

The plus sign symbolizes heat absorbed.

Now, as its name suggests, the molar heat of fusion uses moles, not grams, so start by converting the mass of ice to moles by using the molar mass of water

40.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "2.220 moles H"_2"O"

So, use the molar heat of fusion as a conversion factor to calculate the enthalpy change that corresponds to the freezing of $\text{40.0 g}$ of liquid water at its normal freezing point

2.220 color(red)(cancel(color(black)("moles H"_ 2"O"))) * overbrace((-"6.02 kJ")/(1color(red)(cancel(color(black)("mole H"_ 2"O")))))^(color(blue)(DeltaH_ "fus" color(white)(.)"for liquid to solid")) = -"13.4 kJ"

This means that when $\text{40.0 g}$ of water go from liquid at ${0}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$, $\text{13.4 kJ}$ of heat are being released, hence the minus sign.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{heat given off = 13.4 kJ}}}}$

The answer is rounded to three sig figs.