Calorimetry
Unit 7
Phases of Matter
Calorimetry
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Allows us to calculate the amount of energy
required to heat up a substance or to make a
substance change states.
Molar Heat of Fusion — The heat
absorbed by one mole of a substance when
changing from a solid to a liquid.
For water, it = 6.0 kiloJoules/mole
Heat of solidification is opposite of heat of
fusion (heat is released).
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Molar Heat of Vaporization — The heat
absorbed by one mole of a substance when
changing from a liquid to a gas.
For water, it = 40.7 kiloJoules/mole.
Heat of condensation is the opposite of heat
of vaporization (heat is released)
Heat Required For a Phase Change
Heat Absorbed or Released = q
q = (moles) x (Molar Heat Fusion/Vaporization)
Calculating Heat Required To
Change State
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Example #1: How much heat is needed
to melt 56.0 grams of ice into liquid (the
molar heat of fusion for ice is 6.0
kJ/mol)?
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56.0 g
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1 mole H2O
6.0 kJ =
18.0 g
1 mole
= 18.7 kJ will be absorbed
Example #2
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How much heat energy will be released
when 200grams steam condenses back to
a liquid water?
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Molar Heat
condensation
= 40.7kJ/mol
q = (moles) x (Molar Heat condensation)
200gram
1 mole
18gram
40.7 kJ
1 mole
= 452 kJ released
Heating a Substance with
No Phase Change
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Specific Heat Capacity--The amount
of energy required to raise one
gram of a substance one degree
Celcius.
Water’s Specific Heat (as a liquid)
= 4.184 Joules/gram oC
Energy to Change Temperature
Heat
Measured in Joules
Change in Temperature
Tfinal – Tinitial
In OCelcius
q = (mass) ( C ) ( T )
Mass
In grams
Specific Heat
Capacity
Example #3
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How much energy is needed to heat
80grams of water from 10oC to 55oC?
q=mC
T
= m C (Tfinal – Tinitial )
q = (80grams) ( 4.184 J/goC) (55oC – 10oC)
q = 15062 joules
divide by 1000 to get kilojoules
15062 J
1 kJ
1000J
=
15.06 kJ absorbed
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Example #4 -How much energy is needed to
change 150grams of ice from 0oC to 50oC?
This problem requires two steps. Since water is solid ice at
0oC, we need to melt the ice and then heat it up to 50oC.
Step 1 – Calculate heat required to melt 150grams ice
150g
1 mole
18grams
6.0 kJ
1 mole
=
50 kJ
Step 2 - Calculate heat required to heat liquid water from 0oC to 50oC
q = mC
T
= (150g)(4.184 J/goC)(50oC)
= 31380 J
 convert to kJ = 31.38kJ
*Add both heat values together for your final answer
50 kJ + 31.38kJ = 81.38 kJ heat absorbed.
Multiple Step Calorimetry Problems
Use
q = Moles x Molar Heat vap/fus
Vaporization
Gas
Heats
Liquid
Heats
Solid
Heats
melting
Use q = mC
T
* Add each individual energies (in kJ) together for total