conservation of energy like mass, energy cannot be created or destroyed, just transferred from one...
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Conservation of EnergyLike mass, energy cannot be created or
destroyed, just transferred from one place to another within the system
The most common energy transfer is in the form of heat Transfer of heat is measured as a temperature
change Temperature is a measure of the average
kinetic energy of the particles in the substance
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Measuring Heat SI unit for heat is Joules
4.184 Joules = 1 calorie Calorie is the amount of heat it takes to raise 1
gram of water by 1 degree Celsius Specific heat
The amount of heat necessary to raise one gram of a substance by one degree Celsius
Unique for each substance The specific heat of water is 1 calorie/gºC Also 4.184 J/ gºC
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Kinetic TheoryAs the energy of the particles in a
substance increases Particles move faster Density decreases
As the energy of the particles in a substance decreases Particles move more slowly Density increases
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Calculating HeatHeat is calculated using the heat
equation Q = (m) (ΔT) (cp)
Q = heat M = mass ΔT = change in temperature
Tf – Ti (final temp – initial temp)
Cp = specific heat
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Example Problem How much heat is lost when a solid
aluminum ingot with a mass of 4110 g cools from 660.0 C to 25.0 C? q=(m) (T) (Cp) Mass is 4110g Initial temp is 660.0 C Final temp is 25.0 C Specific heat of aluminum is 0.903 J/gC Q = (4110) ( 660C – 25.0 C ) ( 0.903 J/gC) Q = (4110) ( 635C) ( 0.903 J/gC) Q = 2.35 x 106
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Energy and Change of State If the change in kinetic energy is large
enough this will result in a change of stateAs the amount of energy changes, the
temperature will increase or decrease until the boiling (heat increase) or freezing (heat loss) point is reached
At the freezing or boiling point two phases of matter can exist at the same temperature
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To make the change from one phase to another more energy will be absorbed (boiling or melting) or lost (condensing or freezing) without a corresponding change in temperature This is because this energy is used merely
to overcome the intermolecular forces maintaining one state and move to the new state
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Heat of vaporization Hv
Heat necessary to change from liquid to gas or from gas to liquid at the boiling/condensation point
Measurement of the amount of potential energy absorbed /lost when a change of state occurs between liquid and gas
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Heat of fusion Hf
Heat necessary to change from liquid to solid or from solid to liquid at the freezing/melting point
Measurement of the amount of potential energy absorbed /lost when a change of state occurs between liquid and solid
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Phase change graph
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Slanted sections of the graph represent change in kinetic energy M ΔTCp
Flat portions represent the change in potential energy necessary to change from one phase or state of matter to another – Hf or Hv; look at it again
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Phase change graph
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If one were to begin with a cold solid, and add heat, each step as heat is added must be calculated The solid warms up to the melting point
M ΔTCp (using the specific heat for the solid)
At the melting point, some liquid will begin to form the heat necessary to melt it completely is calculated by
mass of the solid times the heat of fusion
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The liquid increases temperature until reaching the boiling point calculated by M ΔTCp (using the specific
heat for the liquid) At the boiling point gas will begin to
form the heat necessary to vaporize it
completely is calculated by mass of the solid times the heat of vaporization
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Vapor may continue to increase in temperature calculated by M ΔTCp (using the specific
heat for the gas)The sum of all of these heats (from
initial temperature to final temperature) is used to calculate the heat gained or lost by the substance
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If a 26.5g block of ice at –15.0 ºC is
heated to water at 20.0 ºC, how much heat is used?
Ist heat the ice to its melting point M ΔTCp (26.5g)(0.0C – (-15.0C))(0.53cal/gºC) 210.7 cal
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Now melt the ice M(Hf) 26.5g (80 cal/g) 2120 cal
Now warm the water M ΔTCp 26.5g (20 ºC - 0 ºC)(1.00cal/g ºC) 530 cal
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Now add up the calories 210.7 cal + 2120 cal + 530 cal 2860.7 cal 2860 cal (2 sig fig.)
Now try one yourself
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How much heat is needed to heat 117g of water at 42 ºC to steam at 136 ºC
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117g (100 ºC - 42 ºC)(1.00 cal/g ºC) 6786 cal
117g (540 cal/g) 63180 cal
117(136 ºC - 100 ºC)(0.480 cal/g ºC) 2021.76 cal
6786 cal + 63180 cal + 2021.76 cal 71988 cal 7.20 x 104 cal
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How much heat does it take to raise the temp of 112g of ice at –4.0 ºC to 120 ºC?