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
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
In other words, you have
DeltaH_"fus" = -"6.02 kJ mol"^(-1) -> when going from liquid to solidThe minus sign is used here to denote heat given off.
DeltaH_"fus" = +"6.02 kJ mol"^(-1) -> when going from solid to liquidThe 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
color(darkgreen)(ul(color(black)("heat given off = 13.4 kJ")))
The answer is rounded to three sig figs.