Calorimetry Energetics

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CalorimetryEnergetics
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Heat versus Temperature

• Heat is energy that is transferred from one
  object to another due to a difference in
  temperature from hot to cold
• Temperature is a measure of the average kinetic
  energy of a body

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What must you measure to calculate the amount of
heat lost or gained by a substance?
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Factors Affecting Heat Quantities

• The amount of heat contained by an object depends
  primarily on three factors
• The mass of material
• The temperature
• The kind of material and its ability to absorb or
  retain heat.

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Units of Heat

• The heat required to raise the temperature of
  1.00 g of water 1 oC is known as a calorie
• The SI unit for heat is the joule.
• 1.00 calorie 4.184 Joules
• 1000 Joules 1 kJ

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Calorimetry

• Calorimetry involves the measurement of heat
  changes that occur in chemical processes or
  reactions.
• You need to measure
• The mass of the substance
• The temperature change
• The heat capacity of the material

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Heat Capacity and Specific Heat

• The heat capacity is a measure of a substances
  ability to absorb or release heat.
• The specific heat capacity of a material is the
  amount of heat required to raise the temperature
  of 1 gram of a substance 1 oC (or Kelvin)

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Specific Heat values for Some Common Substances
Substance c ( J g-1 K-1)
Water (liquid) 4.184
Water (steam) 2.080
Water (ice) 2.050
Copper 0.385
Aluminum 0.897
Ethanol 2.440
Lead 0.127
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Heat Calculations

• The heat equation may be stated as
• DQ m c DT
• where
• DQ Change in heat
• m mass in grams
• c specific heat capacity in J g-1 oC-1
• DT Temperature change

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Calorimeters
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Temperature Changes

• Measuring the temperature change in a calorimetry
  experiment can be difficult since the system is
  losing heat to the surroundings even as it is
  generating heat.
• By plotting a graph of time v temperature it is
  possible to extrapolate back to what the maximum
  temperature would have been had the system not
  been losing heat to the surroundings.

temperature graph vs time
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Heat Transfer Problem 1

• Calculate the heat that would be required an
  aluminum cooking pan whose mass is 400 grams,
  from 20oC to 200oC. The specific heat of
  aluminum is 0.902 J g-1 oC-1.

Solution DQ mCDT (400 g) (0.902 J g-1
oC-1)(200oC 20oC) 64,944 J
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Heat Transfer Problem 2

• What change in energy if 50.0 grams of water
  at 20.0oC is heated to 60.0oC? Assume that the
  loss of heat to the surroundings is negligible.
  The specific heat of water is 4.184 J g -1 oC-1

Solution DQ mCDT (50.0 g)
(4.184 J g-1 oC-1)(60.0 oC - 20.0oC)
Q 8368 8370 Joules
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Heat Transfer Problem 3

• What is the final temperature when 50 grams
  of water at 20oC is added to 80 grams water at
  60oC? Assume that the loss of heat to the
  surroundings is negligible. The specific heat of
  water is 4.184 J g-1 oC-1

Solution DQ (Cold) DQ (hot)
mCDT mCDT Let T final temperature
(50 g) (4.184 J g-1 oC-1)(T- 20oC)
(80 g) (4.184 J g-1
oC-1)(60oC- T) (50 g)(T- 20oC) (80
g)(60oC- T) 50T -1000 4800 80T
130T 5800 T 44.6 oC
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NaOH HCl ? H2O NaCl

• If 20.0 g of solid NaOH are added to 1000
  mL of a 0.500 M HCl solution, the temperature
  of the solution rises from 20.0 oC to 26.0 oC.
  The specific heat of the solution is 4.184 J g-1
  oC-1.
• Calculate the heat released by this reaction.
• Then calculate ?Hrxn (i.e., the heat released
  per mole of NaOH).

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Calculations

• Find moles of HCl and NaOH
• 0.20 grams/40.0 g/mole 0.50 moles NaOH
• 0.5 M x moles/1.0 liter ? 0.50 moles HCl
• Q c m DT
• Q 4.184 J g-1 oC-1 (1000 grams) 6 oC
• Q -25104 J (Energy is released)