Heat as Energy Transfer

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Heat as Energy Transfer

• Heat refers to energy that is transferred from
  one body to another because of a difference in
  temperature.
• It is measured in JOULES.
• A common unit for heat was the CALORIE. It was
  defined as the amount of heat necessary to raise
  the temperature of 1 gram of water by 1 degree
  Celsius.
• 1 Calorie 4.189 Joules.

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Specific Heat Calorimetry

• The amount of heat Q required to change the
  temperature of a system is found to be
  proportional too the mass m of the system and to
  the temperature change ?T, and can be expressed
  in the equation
• Q m c ?T
• where c is a quantity characteristic of the
  material called specific heat.

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Specific Heat Problem
Problem - How much heat is required to raise the
temperature of 20 kg of iron form 10oC to
90oC? Solution - Data - specific Heat for Iron c
450 J/kgoC Temperature Change ?T
(90-10) 80oC. Thus Q m c ?T 20 x 450 x 80
7.2 x 105 J. Therefore the amount of heat
required was 720 kJ.
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Law of Heat Exchange

• The law of heat exchange has its origins in the
  law of conservation of energy.
• If the system is completely isolated, no energy
  can flow into or out of it the heat lost by one
  part of the system is equal to the heat gained by
  the other part of the system.
• Heat lost heat gained.

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Heat Exchange Problem

• Problem - If 200 cm3 of tea at 95oC is poured
  into a 150 g glass initially at 25oC, what will
  be the final temperature of the mixture when
  equilibrium is reached, assuming that no heat
  flows to the surroundings?

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Solution Since tea is mainly water, its
specific heat is 4180 J/kg oC and its mass is its
density times its volume m ? V 1x103 x 200 x
10-6 0.20 kg Using the law of heat exchange
(and let T be the final Temperature of the
mixture) hat lost heat gained mtea ctea ?T
mcup ccup ?T 0.20 x 4180 x (95-T) 0.15 x 840 x
(T-25) 79 420 - 836T 126 T - 3150 T 85.8 oC
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Heat Exchange Problem

• We wish to determine the specific heat of a new
  alloy. A 0.150 kg sample of the alloy is heated
  to 540oC. It is then quickly placed in 400g of
  water at 10.0oC which is contained in a 200 g
  aluminum calorimeter cup. The final temperature
  of the mixture is 30.5oC. Calculate the specific
  heat of the alloy.

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Data Required for this problem to find Calloy

• Mass of alloy malloy 0.150 kg
• Temperature Change for alloy
• ?Talloy (540-30.5) 509.5oC
• Mass of water mwater 0.40kg
• Specific Heat of water cwater 4180 J/kg oC
• Temperature Change for water
• ?Twater (30.5 - 10) 20.5oC
• Mass of the calorimeter cup mcup 0.20kg
• Specific Heat of the cup ccup 900 J/kg oC
• Temperature Change for the cup
• ?Tcup (30.5 - 10.0)20.5oC

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Solution to the problem

• Heat Lost Heat Gained
• Heat lost by alloy heat gained by water and cup
• maca?Ta mwcw?Tw mccc?Tc
• 0.15 x ca x 509.5 0.4 x 4180 x 20.5 0.2 x 900
  x 20.5
• 76.4 ca (34 300 3700)
• ca 500 J/kg oC.
• Therefore the specific heat of the alloy was 500
  J/kg oC.

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Latent Heat

• The heat required to change the phase of a
  substance (eg from solid to a liquid) is known as
  Latent Heat.
• When the substance is changing phase, there is no
  change in the substances temperature.
• The formula is simply
• Q m L
• where Q is the heat (in Joules), m is the mass
  (in kg) and L is the Latent Heat (in J/kg)

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Latent Heat of Fusion Lv

• The Latent heat of fusion is the amount of heat
  required to change 1.0kg of a substance from the
  solid to the liquid state.
• For example, when 5.00kg of water freezes at 0oC,
  the water must release
• Q m L 5.00 x 3.33 x 105
• ? Q 1.67 x 106 J of energy

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Latent Heat of Vaporisation Lv

• The Latent heat of vaporisation is the amount of
  heat required to change 1.0kg of a substance from
  the liquid to the gaseous state.
• For example, when 5.00kg of water boils at 100oC,
  the water must absorb
• Q m L 5.00 x 22.6 x 105
• ? Q 1.13 x 107 J of energy

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Problem involving Latent Heats.

• Problem - How much energy does a refrigerator
  have to remove from 1.5kg of water at 20oC to
  make ice at -12oC?
• Data -
• mass of water mw 1.5 kg
• specific heat of water cw 4180 J/kgoC
• Temperature change of water ?Twater 20oC
• latent heat of water/ice Lf 3.33 x 105 J/kg
• mass of ice mice 1.5 kg
• specific heat of ice ci 2100 J/kgoC
• Temperature change of the ice ?Tice 12oC

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Solution
Q mwater cwater ?Twater m Lf mice cice
?Tice Q 1.5 x 4180 x 20 1.5 x 3.33 x 105
1.5 x 2100 x 12 Q 6.6 x 105 J Therefore 660 kJ
must be removed by the refrigerator in order to
freeze ice, initially at 20oC down to -12oC.
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Linear Thermal Expansion

• The change in length of a 1m length of a
  substance due to a temperature change of 1oC is
  called the coefficient of linear expansion (?).
• The formula to help find the change in length of
  a substance due to thermal expansion of solids is
  
• ?L Li ? ?T

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Linear Expansion Problem

• An electric company strung an aluminum wire
  between two piers 200m apart on a day when the
  temperature was 25oC. They strung it tight so
  that it would not sag. Find the length of the
  wire when the temperature fell to -25oC on a cold
  winters night? What might happen? What should
  have been done to prevent this from occurring?

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Volumetric Thermal Expansion

• The coefficient of volume expansion (?) is
  normally associated with liquids. Liquids take
  the shape of the container hence we are mainly
  interested in the volume changes of liquids with
  temperature.
• As with solids, the change in volume of liquids
  can be found using the formula
• ?V Vi ? ?T

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Volume Expansion Problem

• What would be the increase in the volume of 0.20L
  of acetone if it was heated form 10oC to 40oC?
• Solution
• ?V Vi ? ?T
• 14.87 x 10-4 x 0.20 x (40-10)
• 89.2 x 10-4 L
• 8.9 x 10-3 L