Calorimetry

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Calorimetry
Temperature Change Problems
Unit 7 Honors Chemistry
Phase Change Problems

• How to use math to describe the movement of heat
  energy

2
Energy Conversions

• Heat is a specific type of energy that can be
  measured in different ways.
• The SI unit for heat is Joules
• 4.184 Joules 1 calorie (this will be given)
• 1000 calories 1 kilocalorie
• 1000 Joules 1 kiloJoule

3
Heat Conversions

• How many joules are in 130 calories?
• How many calories are in 50 Joules?

4.184 Joules
130 calories
543.92 Joules ( 540 J (Sig figs!)
1 calorie
1 calorie
50 Joules
11.95 calories
4.184 Joules
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Heat Conversions

• How many kilojoules are in 130 Calories?

4.184 Joules
1 kJ
130 Calories
1000J
1 Calorie
0.54 KiloJoules
5
Calorimetry

• 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 (Hf) The heat absorbed by
  one mole of a substance when changing from a
  solid to a liquid.
• For water, it 6.0 kiloJoules/mole
• or 334 Joules/gram (specific heat of fusion)
• Heat of solidification is opposite of heat of
  fusion (heat is released).

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• Molar Heat of Vaporization (Hv) The heat
  absorbed by one mole of a substance when changing
  from a liquid to a gas.
• For water, it 40.7 kiloJoules/mole.
• or 2260 Joules/gram (specific heat of
  vaporization
• Heat of condensation is the opposite of heat of
  vaporization (heat is released)
• Every pure substance will have a unique Molar
  heat of fusion (Hf) or vaporization (Hv)

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Heat Required For a Phase Change

• Heat Absorbed or Released q
• For Melting or Freezing use the following
• For Vaporization or Condensation use the
  following

q (moles) x Molar Heat Fusion
q (moles) x Molar Heat vaporization
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Calculating Heat Required To Change State

• 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)?
• 56.0 g 1 mole H2O 6.0 kJ
  18.0 g 1 mole
• 18.7 kJ will be absorbed

q (moles) x (Hf)
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Example 2

• How much heat energy in kJ will be released when
  200grams steam condenses back to a liquid water?
• Hv 40.7kJ/mol

q (moles) x (Hv)
200gram 1 mole 40.7 kJ
18gram 1 mole
452 kJ released Or -452kJ
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Heating a Substance with No Phase Change

• Specific Heat Capacity--The amount of energy
  required to raise one gram of a substance one
  degree Celcius.
• Waters Specific Heat (as a liquid)
• Cp 4.184 Joules/gram oC
• Every pure substance will have its own unique
  specific heat for every phase!

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Heating a Substance with No Phase Change

• When you see an increase in the temperature of a
  sample, the heat is being added to raise the
  temperature
• How much the temperature increases is based upon
  the heat capacity (Cp) and the mass of your
  sample
• The higher the heat capacity number, the longer
  it takes to heat a substance up and the longer
  the substance holds on to the heat.

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Energy to Change Temperature

• q (mass) ( Cp) ( T )

Change in Temperature Tfinal Tinitial In
OCelcius
Heat Measured in Joules
Specific Heat Capacity
Mass In grams
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Example 3
m

• How much energy is needed to heat 80 g of water
  from 10 oC to 55 oC?

Tfinal
Tinitial
q mCp?T m Cp (Tfinal Tinitial )
(55oC 10oC)
(80g)
( 4.184 J/g C)
q 15062 joules
Is the energy absorbed or released?
Absorbed, because temperature in increasing
Final Answer 15,062 J 15.06 kJ absorbed/
endothermic
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Example 4
m
How much energy is needed to cool 150 g of ice
from -2 oC to -55 oC?
Tfinal
Tinitial
q mCp?T m Cp (Tfinal Tinitial )
(-55oC -2oC)
(150g)
( 2.06 J/g C)
q - 16377 joules
Is the energy absorbed or released?
Released, because temperature in decreasing
Final Answer - 16377 J -16.3 kJ
released/exothermic
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Heat Problem Road Map
q (moles)Hv
q (moles)Hf
Gas Heats
Vaporization or Condensation
Liquid Heats
Melting or Freezing
q mCp?T
Solid Heats
Add each individual energies (in kJ) together
for total heat energy required for multistep
problems (up to 5 steps max!)
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• Example 5 -How much energy in kJ 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 6.0 kJ
50 kJ
18grams 1 mole
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.
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Calorimetry Formula Summary

• Phase Change
• Use Molar Heat constants
• Melting use q (moles) x (Hfusion)
• Vaporize use q (moles) x (HVaporization)
• No Phase Change
• Use specific heat capacity
• q (mass) ( Cp ) ( ?T )