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Calorimetry

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Temperature Change Problems Unit 7 Honors Chemistry Phase Change Problems Calorimetry How to use math to describe the movement of heat energy Energy Conversions Heat ... – PowerPoint PPT presentation

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Title: Calorimetry


1
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
4
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).

6
  • 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)

7
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
8
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)
9
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
10
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!

11
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.

12
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
13
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
14
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
15
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!)
16
  • 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.
17
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 )
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