Section 7.3—Changes in State
What’s happening when a frozen ice pack melts?
The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature)
Change in State
To melt or boil, intermolecular forces must be broken
Breaking intermolecular forces requires energy
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
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).
fusHmH Energy needed to melt
Mass of the sample
Energy needed to melt 1 g
Example
Example:Find the enthalpy of
fusion of a substance if it takes 5175 J to melt 10.5 g of the
substance.
Example
fusHmH
H = enthalpy (energy)m = mass of sampleHfus = enthalpy of fusion
fusHgJ )5.10(5175
fusHg
J
5.10
5175Hfus = 493 J/g
Example:Find the enthalpy of
fusion of a substance if it takes 5175 J to melt 10.5 g of the
substance.
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).
vapHmH Energy needed to boil
Mass of the sample
Energy needed to boil 1 g
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?
Example
vapHmH
H = enthalpy (energy)m = mass of sampleHfus = enthalpy of fusion
gcalgH 2.547)0.25(
H = 1.37×104 cal
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?
Solid
Liquid
Gas
Melting
Vaporizing or Evaporating
Condensing
Freezing
Incr
easi
ng m
olec
ular
mot
ion
(tem
pera
ture
)
Changes in State go in Both Directions
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. H 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. H will = -2798 J
Example
Example:How much energy is
released with 157.5 g of water is condensed?
Hvap water = 547.2 cal/g
Example
vapHmH
H = enthalpy (energy)m = mass of sampleHfus = enthalpy of fusion
gcalgH 2.547)5.157(
H = - 8.6×104 cal
Example:How much energy is
released with 157.5 g of water is condensed?
Hvap water = 547.2 cal/g
Since we’re condensing, we need to “release” energy…H will be negative!
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Heating Curves
Melting & Freezing
Point
Boiling & Condensing
Point
Heating curves show how the temperature changes as energy is added to the sample
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Going Up & Down
+H
-H
Moving up the curve requires energy, while moving down releases energy
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
States of Matter on the Curve
Gas OnlyEnergy added
increases temp
Liquid & gasEnergy added breaks remaining IMF’s
Liquid OnlyEnergy added
increases temp
Solid & LiquidEnergy added breaks IMF’s
Solid OnlyEnergy added
increases temp
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Different Heat Capacities
Gas OnlyCp = 0.48 cal/g°C
Liquid OnlyCp = 1.00 cal/g°C
Solid OnlyCp = 0.51 cal/g°C
The solid, liquid and gas states absorb water differently—use the correct Cp!
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Changing States
Liquid & gasHvap = 547.2 cal/g
Solid & LiquidHfus = 80.87 cal/g
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Adding steps together
If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice to zero before it could melt.
Then you’d melt the ice
Then you’d warm that water from 0°C to your final 75°
You can calculate the enthalpy needed for each step and then add them together
Example:How many calories are
needed to change 15.0 g of ice at -12.0°C to steam at
137.0°C?
ExampleUseful information:Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C
Hfus = 80.87 cal/gHvap = 547.2 cal/g
TCpmH
fusHmH
vapHmH
Example:How many calories are
needed to change 15.0 g of ice at -12.0°C to steam at
137.0°C?
Example
gcalgH 2.5470.15
CCgcalgH 010000.10.15
CCgcalgH 12051.00.15
Useful information:Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C
Hfus = 80.87 cal/gHvap = 547.2 cal/g
TCpmH
fusHmH
CCCg
calgH 1000.13748.00.15
Warm ice from -12.0°C to 0°C
Melt ice
Warm water from 0°C to 100°C
Boil water
Warm steam from 100°C to 137°C
gcalgH 87.800.15
91.8 cal
1500 cal
1213 cal
8208 cal
266 cal
vapHmH
Total energy = 11279 cal
Example:How many needed to
change 40.5 g of water at 25°C to steam at 142°C?
Let’s PracticeUseful information:Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C
Hfus = 80.87 cal/gHvap = 547.2 cal/g
TCpmH
fusHmH
vapHmH
Example:How many needed to
change 40.5 g of water at 25°C to steam at 142°C?
Let’s Practice
gcalgH 2.5475.40
CCgcalgH 2510000.15.40
Useful information:Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C
Hfus = 80.87 cal/gHvap = 547.2 cal/g
TCpmH
fusHmH
CCCg
calgH 1000.14248.05.40
Warm water from 25°C to 100°C
Boil water
Warm steam from 100°C to 142°C
3038 cal
22162 cal
816 cal
vapHmH
Total energy = 26016 cal
