Section 7.3

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Section 7.3Changes in State
Whats happening when a frozen ice pack melts?
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Change in State
The energy being put into the system is used for
breaking IMFs, not increasing motion
(temperature)
Breaking intermolecular forces requires energy
To melt or boil, intermolecular forces must be
broken
A sample with solid liquid will not rise above
the melting point until all the solid is
gone. The same is true for a sample of liquid
gas
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Melting

• When going from a solid to a liquid, some of the
  intermolecular forces are broken
• The Enthalpy of Fusion (Hfus) is the amount of
  energy needed to melt 1 gram of a substance
• The enthalpy of fusion of water is 80.87 cal/g or
  334 J/g
• All samples of a substance melt at the same
  temperature, but the more you have the longer it
  takes to melt (requires more energy).

Energy needed to melt 1 g
Energy needed to melt
Mass of the sample
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Example
Example Find the enthalpy of fusion of a
substance if it takes 5175 J to melt 10.5 g of
the substance.
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Example
Example Find the enthalpy of fusion of a
substance if it takes 5175 J to melt 10.5 g of
the substance.
H enthalpy (energy) m mass of sample Hfus
enthalpy of fusion
Hfus 493 J/g
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Vaporization

• When going from a liquid to a gas, all of the
  rest of the intermolecular forces are broken
• The Enthalpy of Vaporization (Hvap) is the amount
  of energy needed to boil 1 gram of a substance
• The Hvap of water is 547.2 cal/g or 2287 J/g
• All samples of a substance boil at the same
  temperature, but the more you have the longer it
  takes to boil (requires more energy).

Energy needed to boil 1 g
Energy needed to boil
Mass of the sample
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Example
Example If the enthalpy of vaporization of water
is 547.2 cal/g, how many calories are needed to
boil 25.0 g of water?
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Example
Example If the enthalpy of vaporization of water
is 547.2 cal/g, how many calories are needed to
boil 25.0 g of water?
H enthalpy (energy) m mass of sample Hfus
enthalpy of fusion
DH 1.37104 cal
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Changes in State go in Both Directions
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Going the other way

• The energy needed to melt 1 gram (Hfus) is the
  same as the energy released when 1 gram freezes.
• If it takes 547 J to melt a sample, then 547 J
  would be released when the sample freezes. DH
  will -547 J
• The energy needed to boil 1 gram (Hvap) is the
  same as the energy released when 1 gram is
  condensed.
• If it takes 2798 J to boil a sample, then 2798 J
  will be released when a sample is condensed.
  DH will -2798 J

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Example
Example How much energy is released with 157.5 g
of water is condensed? Hvap water 547.2 cal/g
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Example
Example How much energy is released with 157.5 g
of water is condensed? Hvap water 547.2 cal/g
H enthalpy (energy) m mass of sample Hfus
enthalpy of fusion
Since were condensing, we need to release
energyDH will be negative!
DH - 8.6104 cal
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Heating Curves
Heating curves show how the temperature changes
as energy is added to the sample
Boiling Condensing Point
Melting Freezing Point
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Going Up Down
Moving up the curve requires energy, while moving
down releases energy
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States of Matter on the Curve
Liquid gas Energy added breaks remaining IMFs
Liquid Only Energy added increases temp
Gas OnlyEnergy added increases temp
Solid OnlyEnergy added increases temp
Solid Liquid Energy added breaks IMFs
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Different Heat Capacities
The solid, liquid and gas states absorb water
differentlyuse the correct Cp!
Liquid Only Cp 1.00 cal/gC
Gas OnlyCp 0.48 cal/gC
Solid OnlyCp 0.51 cal/gC
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Changing States
Liquid gas Hvap 547.2 cal/g
Solid Liquid Hfus 80.87 cal/g
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Adding steps together
If you want to heat ice at -25C to water at
75C, youd have to first warm the ice to zero
before it could melt.
Then youd melt the ice
Then youd warm that water from 0C to your final
75
You can calculate the enthalpy needed for each
step and then add them together
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Example
Useful information Cp ice 0.51 cal/gC Cp
water 1.00 cal/gC Cp steam 0.48 cal/gC Hfus
80.87 cal/g Hvap 547.2 cal/g
Example How many calories are needed to change
15.0 g of ice at -12.0C to steam at 137.0C?
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Example
Useful information Cp ice 0.51 cal/gC Cp
water 1.00 cal/gC Cp steam 0.48 cal/gC Hfus
80.87 cal/g Hvap 547.2 cal/g
Example How many calories are needed to change
15.0 g of ice at -12.0C to steam at 137.0C?
Warm ice from -12.0C to 0C
91.8 cal
Melt ice
1213 cal
Warm water from 0C to 100C
1500 cal
Boil water
8208 cal
Warm steam from 100C to 137C
266 cal
Total energy 11279 cal
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Lets Practice
Useful information Cp ice 0.51 cal/gC Cp
water 1.00 cal/gC Cp steam 0.48 cal/gC Hfus
80.87 cal/g Hvap 547.2 cal/g
Example How many needed to change 40.5 g of
water at 25C to steam at 142C?
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Lets Practice
Useful information Cp ice 0.51 cal/gC Cp
water 1.00 cal/gC Cp steam 0.48 cal/gC Hfus
80.87 cal/g Hvap 547.2 cal/g
Example How many needed to change 40.5 g of
water at 25C to steam at 142C?
Warm water from 25C to 100C
3038 cal
Boil water
22162 cal
Warm steam from 100C to 142C
816 cal
Total energy 26016 cal