Section 7.3—Changes in
State
What’s happening when a frozen ice pack melts?
Change in State
To melt or boil,
intermolecular
forces must
be broken
Breaking
intermolecular
forces
requires
energy
The energy being
put into the
system is used
for breaking
IMF’s, not
increasing
motion
(temperature)
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).
H  m  H fus
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
Example:
Find the enthalpy of
fusion of a substance
if it takes 5175 J to
melt 10.5 g of the
substance.
H  m  H fus
H = enthalpy (energy)
m = mass of sample
Hfus = enthalpy of fusion
5175 J  (10.5 g )  H fus
5175 J
 H fus
10.5 g
Hfus = 493 J/g
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).
H  m  H vap
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
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  (25.0 g )  547.2 cal
H  m  H vap
H = enthalpy (energy)
m = mass of sample
Hfus = enthalpy of fusion
g
H = 1.37×104 cal
Increasing molecular motion (temperature)
Changes in State go in Both Directions
Vaporizing or
Evaporating
Liquid
Melting
Freezing
Solid
Gas
Condensing
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
Example:
How much energy is
released with 157.5 g of
water is condensed?
Hvap water = 547.2 cal/g
H  m  H vap
H = enthalpy (energy)
m = mass of sample
Hfus = enthalpy of fusion
Since we’re condensing, we need to “release” energy…H will be negative!
H  (157.5g )  547.2 cal
g
H = - 8.6×104 cal
Heating Curves
Heating curves show how the temperature changes as energy is added to the
sample
Melting &
Freezing
Point
Temperature
Boiling &
Condensing
Point
Heating curve of w ater
150
100
50
0
-50
Energy input
Going Up & Down
Moving up the curve requires energy, while moving down releases energy
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
States of Matter on the Curve
Liquid & gas
Energy added breaks remaining IMF’s
Liquid Only
Energy added
increases temp
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
Solid Only
Energy added
increases temp
Solid & Liquid
Energy added
breaks IMF’s
Gas Only
Energy added
increases temp
Different Heat Capacities
Liquid Only
Cp = 1.00 cal/g°C
The solid, liquid and gas states absorb
water differently—use the correct Cp!
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
Gas Only
Cp = 0.48 cal/g°C
Solid Only
Cp = 0.51 cal/g°C
Changing States
Liquid & gas
Hvap = 547.2 cal/g
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
Solid & Liquid
Hfus = 80.87 cal/g
Adding steps together
Temperature
Heating curve of w ater
150
100
50
0
-50
Energy input
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
Example:
How many calories are
needed to change 15.0 g of
ice at -12.0°C to steam at
137.0°C?
Useful information:
Cp ice = 0.51 cal/g°C H  m  Cp  T
Cp water = 1.00 cal/g°C
H  m  H fus
Cp steam = 0.48 cal/g°C
Hfus = 80.87 cal/g
H  m  H vap
Hvap = 547.2 cal/g
Example
Example:
How many calories are
needed to change 15.0 g of
ice at -12.0°C to steam at
137.0°C?
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
Useful information:
Cp ice = 0.51 cal/g°C H  m  Cp  T
Cp water = 1.00 cal/g°C
H  m  H fus
Cp steam = 0.48 cal/g°C
Hfus = 80.87 cal/g
H  m  H vap
Hvap = 547.2 cal/g

H  15.0 g  0.51cal  0 C  12 C
g
H  15.0 g  80.87 cal
1213 cal

H  15.0 g  0.48 cal
91.8 cal
g
H  15.0 g 1.00 cal  100 C  0 C
g
H  15.0 g  547.2 cal


1500 cal
8208 cal
g

gC

 137.0 C  100 C

266 cal
Total energy = 11279 cal
Let’s Practice
Example:
How many needed to
change 40.5 g of water at
25°C to steam at 142°C?
Useful information:
Cp ice = 0.51 cal/g°C H  m  Cp  T
Cp water = 1.00 cal/g°C
H  m  H fus
Cp steam = 0.48 cal/g°C
Hfus = 80.87 cal/g
H  m  H vap
Hvap = 547.2 cal/g
Let’s Practice
Example:
How many needed to
change 40.5 g of water at
25°C to steam at 142°C?
Warm water from 25°C to 100°C
Boil water
Warm steam from 100°C to 142°C
Useful information:
Cp ice = 0.51 cal/g°C H  m  Cp  T
Cp water = 1.00 cal/g°C
H  m  H fus
Cp steam = 0.48 cal/g°C
Hfus = 80.87 cal/g
H  m  H vap
Hvap = 547.2 cal/g

H  40.5g 1.00 cal  100 C  25 C
g
H  40.5g  547.2 cal
H  40.5 g  0.48 cal

22162 cal
g

gC
3038 cal

 142.0 C  100 C
 816 cal
Total energy = 26016 cal