# 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.)

##### 1 Answer

#### Answer:

#### Explanation:

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

neededto convert#1# moleof water from solid at its freezing point to liquid at its freezing pointgiven offwhen#1# moleof water is converted from liquid at its freezing point to solid at its freezing point

You can thus say that when **mole** of liquid water goes from liquid at its normal freezing point of **given off**.

In other words, you have

#DeltaH_"fus" = -"6.02 kJ mol"^(-1) -># when going fromliquidtosolidThe minus sign is used here to denote

heat given off.

#DeltaH_"fus" = +"6.02 kJ mol"^(-1) -># when going fromsolidtoliquidThe 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

#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 **released**, hence the minus sign.

#color(darkgreen)(ul(color(black)("heat given off = 13.4 kJ")))#

The answer is rounded to three **sig figs**.