Question #04705

As you know, the molar enthalpy change of fusion, #DeltaH_"fus"#, is defined as the change in enthalpy when one mole of a substance undergoes a liquid to solid or a solid to liquid phase change at its melting / freezing point.

More specifically, the molar enthalpy change of fusion will tell you

• how much heat must be added to one mole of a substance at its melting point in order for a solid #-># liquid phase change to take place

• how much heat must be given off by one mole of a substance at its freezing point in order for a liquid #-># solid phase change to take place

In your case, you are interested in finding out the molar enthalpy change of fusion for the freezing of #"30.00 g"# of water at #0^@"C"#. This means that the molar enthalpy change of fusion must carry a negative sign, since it represents heat lost.

The equation you'll use here looks like this

#color(blue)(q = n * DeltaH_"fus")" "#, where

#q# - heat lost / absorbed
#n# - the number of moles of the substance
#DeltaH_"fus"# - the molar enthalpy change of fusion

Use water's molar mass to determine how many moles you have in that #"30.00-g"# sample

#30.00 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.6653 moles H"_2"O"#

No,w you know that #"10.00 kJ"# of heat are given off when this sample of water goes from liquid water at #0^@"C"# to ice at #0^@"C"#.

This means that the value of #q# will be

#q = -"10.00 kJ"#

Again, the negative sign is used to symbolize heat lost.

This means that you have

#DeltaH_"fus" = q/n#

#DeltaH_"fus" = (-"10.00 kJ")/"1.6653 moles" = color(green)(-"6.005 kJ/mol")#

The answer is rounded to four sig figs.

So, you can say that the molar enthalpy change of fusion for water will be equal to

• #DeltaH_"fus" = +"6.005 kJ/mol" -># when ice at #0^@"C"# melts to liquid water at #0^@"C"#
• #DeltaH_"fus" = -"6.005 kJ/mol" -># when liquid water at #0^@"C"# freezes to ice at #0^@"C"#