Mr. Shields
Regents Chemistry
U07 L03
1
Phase Changes
Let’s review what we’ve learned previously about
PHASE CHANGES.
First … What is a phase
change?
A change from one state
of matter to another.
What does endothermic and exothermic mean?
Absorbs heat / Releases heat
2
Phase Change Overview
What are the names of the phase changes we’ve studied?
Gas
e
Energy Absorbed
Endothermic
a
Exothermic
c
Energy Released
Liquid
f
d
b
Solid
ENDOTHERMIC
Phase changes
EXOTHERMIC
Phase Chages
a.
d.
Deposition
b. Fusion
e.
Condensation
c.
f.
Solidification
Sublimation
Vaporization
3
Phase Changes
What actually happens to E as matter moves from
Solid  Liquid  Gas:
Energy is absorbed (endo)
Temp increases
KE increases
molecules move further apart
PE increases (Why?)
Remember … PE is a function of the position
of two bodies relative to one another
So … PE inc as molecular separation increases
4
Heating Curve
What happens when we keep adding energy to a solid ?
Solid becomes a liquid then the liquid
becomes a gas
The answer is NO!
15
10
Temp
As this energy is added
KE inc and so does Temp.
but does temp Uniformly inc.
over time?
20
5
0
-5
-10
-15
-20
Time
5
Heating Curve
So what does happen if the temperature does not
Uniformly increase?
What happens is described by what is called
the HEATING CURVE
6
Heating Curve
If we heat a solid it’s temperature increases
Steadily until we finally reach the temperature
at which the solid begins to melt
It begins to go thru a phase change
solid  Liquid
This phase change is called fusion
7
Solid/Liquid Phase Transition
As the solid begins to melt something
Unusual Happens.
As we continue to add heat to the solid
the temperature stops rising as the solid
continues to melt.
10
Temp
5
But why does this happen?
Melting Solid
0
-5
-10
Solid
-15
Heat added over Time
8
Solid/Liquid Phase Transition
Solids exist in a rigid, closely packed, highly
structured pattern
Liquids however have no such rigid structure.
As we reach the solids m.p. there is just enough energy to
begin overcoming the intermolecular forces between
molecules holding them together
in the solid state...
Molecules begin to separate
9
Heat of Fusion
Added additional heat energy goes into
Separating more and more molecules
As molecules move from solid to liquid the PE
Increases but since the temperature doesn’t rise
The KE remains constant.
The energy necessary to melt 1 GRAM of a
solid is called the … HEAT OF FUSION (Hf)
It is unique for every substance.
For water, the heat of fusion = 334 Joules/gram
10
Solid/Liquid Phase transition
Remember… heat ALWAYS flows from hot to cold
H C
Until the last piece of solid melts the temperature the
Solid/liquid mixture remains constant
Any excess heat in the liquid immediately flows
back into the colder solid
Once there’s no solid left all additional heat added
begins to increase the temperature of the liquid.
11
Solid/Liquid Phase transition
When heat is added what happens to KE, T & PE ?
Dual L/S Phase present
KE (T) is Constant
PE increase
Temp
Only a single solid (s) phase present
KE (T) increases
10
PE is Constant
5
S&L
Liquid
0
-5
Solid
-10
-15
Heat added over Time
Only a single liquid (l) phase present
KE (T) increases
PE is constant
Heating curve
12
Liquid/Vapor Phase transition
B.P
Once VP = Patm the liquid
Boils.
Temp
Any additional heat added
To the liquid inc. Temp. and 100
Vapor Pressure.
5
S&L
Liquid
0
-5
Solid
-10
-15
Heat added over Time
As in the transition from solid
To liquid, 2 phases are now present
& the Temp. of the boiling water remains constant
13
Heat of Vaporization
Molecules in the liquid phase form a close but
loosely organized structure 125
105
Gas
85
Temp
Molecules in the gas
Phase have no structure
And are widely separated.
L&G
65
45
25
5
-15
S&L
Solid
Liquid
Heat added over Time
To separate these molecules this much takes
lots of energy.
This energy is called the HEAT OF VAPORIZATION (Hv)
14
Heat of Vaporization
For water, Hv = 2,260 Joules/g
As long as the liquid is
boiling
T and KE will be
Constant
PE will Increase
as molecules
Move further apart
In the gas phase
M.P
Note Hv > Hf
The Heating Curve
B.P.
Liquid &
gas
Solid &
liquid
15
Summary
(b.p.)
Hv = 2,260 J/g
T const. / KE const. /PE inc
( 2 phases )
T inc. / KE inc / PE Const.
( 1 phase )
Hf =
334 J/g
(m.p.)
16
A Review: Specific Heat
Recall from our earlier discussions that …
The SPECIFIC HEAT of a substance is the amount
of heat required to raise the temperature of 1 g of the
material by 1 degree centigrade.
Each substance has it’s own unique specific heat…
The lower the specific heat the better the conductor
And ….
q = Cp x m x (Tf – Ti)
17
Specific Heat
One of the variables in Specific heat calculations
Involves Temperature
Change.
The Heating Curve Phase V
What phases of the
The heating curve
Involve changes in T?
Phase III Phase IV
Phase 1 Phase II
Phase 1, 3, 5
18
Specific Heat
It’s in these phases that specific heat calculations
are used to determine how much heat us needed
to raise the temperature
of the sample
The Heating Curve
Phase V
Phase III Phase IV
But how do we find
the heat necessary to
to fully melt or vaporize
a sample of matter
In regions where T does
Not change (phase II & IV)
Phase 1 Phase II
19
Melting/Vaporization Calculations
It’s even easier than specific heat calculations since
Temperature is not a variable.
In phase II:
Q= m x Hf
m=total mass
Hf=heat of fusion
The Heating Curve
Phase V
Phase III Phase IV
Phase 1 Phase II
In phase IV:
Q=m x Hv
20
Melting/Vaporization Calculations
Problem 1: How much heat is necessary to melt 100g of ice?
Problem 2: How much heat is necessary to vaporize this water?
Q = m x Hf
Q = 100g x 334 J/g
Q = 33,400 J
The Heating Curve
Phase V
Phase III Phase IV
Phase 1 Phase II
Q = m x Hv
Q = 100g x 2,260 J/g
Q = 226,000 J
21
Melting/Vaporization Calculations
Note that these calculations are true for all sorts of
matter, not just water.
For example here are some Hv as a function of the
intermolecular force:
H2O
NH3
H2S
F2
Hydrogen Bonding
Hydrogen Bonding
Dipole-Dipole
London Dispersion
2260 J/g
1276 J/g
553 J/g
155 J/g
22
Melting/Vaporization Calculations
Problem: What is the heat of vaporization of a liquid if it takes
6500 J to totally vaporize 8g?
Q = m x Hv
6500 J = 8g x Hv
Hv = 6500 / 8 = 812.5 J/g
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Heat of Fusion and Vaporization
from heating graphs
If heat is added at a constant rate over time to a specific mass
we can determine both Hf and Hv by noting both the initial onset
of melting or vaporization and the completion point.
24
Heat of Fusion and Vaporization
from heating graphs
For example in this graph the onset of vaporization occurs after
875 Joules have been added and is complete after 3100 J have
been added. Therefore
It took 3100-875 KJ to
Vaporize this sample.
If the mass of the
Sample is 300g then the
Hv = 2225KJ/300g
=7.417 KJ/g
The same can be done
To calculate the Hf
875 KJ
3,100 KJ
25
Heat of Fusion and Vaporization from
heating graphs
If we add heat to a sample at a fixed rate, say 422.4 J/min, we
Can then calculate both Hf and Hv if we know the mass
Of the sample.
In this example the water
starts to boil at 3.8 min
And is complete at 14.5 min
So we’ve added 422.4J/min
For 10.7min = 4520 J
Since the mass is 2 gram then
Hv. = 4520/2 = 2260 J
(Mass = 2 gram)
Time = 3.8min
1 2 3 4 5
Time (min)
Time = 14.5 min
10 11 12 132614
Cooling Curve
The opposite of a heating Curve is a
COOLING CURVE.
Brrrrr…
Since the sample is cooling it must be releasing heat.
As Temp decreases KE dec and as a sample goes from gas
To liquid to solid the PE must be decreasing.
Instead of Heat of vaporization and heat of fusion we have
the Heat of condensation and the Heat of solidification. They
Are equal to Hv and Hf but are opposite in value
-Hv = Hc
&
-Hf = Hs
27
Phase I:
T dec
KE dec
PE constant
Cooling Curve
Phase II: T constant
KE constant
PE dec
Hc = -Hv
100
80
Temp
Phase III: T dec
KE dec
PE constant
120
Phase IV: T constant
KE constant
PE dec
Hs = -Hf
Phase V: same as I and III
60
Gas
I
Gas & Liquid
II
III
IV
V
Liquid
Condensation
40
20
Liquid & solid
0
-20
solid
Solidification
-40
Time
Note: This is the mirror image of a heating curve
28
Cooling Curve Calculations
Problem: How much heat is released when 100g
Of water solidifies? How much heat is released when
100g of water condenses?
a)
Q = m x Hs (i.e. –Hf )
Q = 100g x -334 J/g
Q = -33,400 J (i.e. 33,400 joules are released)
The negative sign means this much heat is released (and not absorbed)
b)
Q = 100g x -2260 J/g = -226,000 J
29