Discusses macroscopic and microscopic properties of solids and liquids. **More good stuff available at: www.wsautter.com and http://www.youtube.com/results?search_query=wnsautter&aq=f

1. Copyright Sautter 2015

2. SOLIDS, LIQUIDS AND PHASE CHANGES
• What is the difference between a solid, a liquid and a gas?
• ENERGY STATE ! Solids are the lowest energy state,
liquids are intermediate and gases are the highest energy
state of matter.
• The energy differences which separate each state are called
the Heats of Phase Change. The Heat of Phase Change is
the energy required to change the state of a substance
without changing its temperature.
• The Heats of Phase Change are defined by the phase change
which occurs, for example, Heat of Melting (often called
fusion) is the heat needed to change a solid into a liquid at
its melting point. The Heat of Vaporization is the heat
needed to change a liquid into a gas at its boiling point.2

3. SOLIDS, LIQUIDS AND PHASE CHANGES
• When solids, liquids and gases are heated or cooled and
do not undergo phase change, the heated added or
removed from the substance can be calculated using the
equation: q = m x c x T
• The letter q represents heat is calories or joules. The
mass of the substance in grams is m. The letter c stands
for “specific heat” in calories per gram degree Celsius or
joules per gram degree Celsius. Specific heat indicated
how easy or difficult it is to change the temperature of a
substance. Large specific heats mean that it requires a
large amount of heat addition or removal to change
temperature. Small values mean that the substance
changes temperature readily.
• Delta T (T) is the Celsius temperature change of the
substance 3

4. HEAT IN JOULES
OR CALORIES
MASS IN
GRAMS
SPECIFIC HEAT
IN JOULES / G x 0C
OR CALORIES / G x 0C
TEMPERATURE
CHANGE IN 0C
CALCULATING HEAT QUANTITIES
(NO PHASE CHANGE OCCURS)
4

5. SOLIDS, LIQUIDS AND PHASE CHANGES
• Problem: What is the temperature of 15.0 grams of
iron when 250 joules of heat are removed from it?
(specific of iron = 0.450 j / g x 0C)
• Solution: q = m x c x T, q = 250 joules
m = 15.0 grams, c = 0.450 j / g x 0C
T = q / (m x c) = 250 / (15.0 x 0.450) = 37.0 0C
Since heat is removed from the sample the
temperature is lowered by 37.0 degrees, therefore the
answer is – 37.0 0C
5

6. SOLIDS, LIQUIDS AND PHASE CHANGES
• Problem: A 100 gram sample of aluminum at 100 0C is
placed in 200 grams of water at 25 0C. What is the final
temperature of the system? (c for Al = 0.900 j / g 0C and c
for water = 4.18 j / g 0C)
• Solution: Heat lost = Heat gained
(Conservation of Energy)
Al loses heat to the water since the Al is at the higher
initial temperature.
• mAl x cAl x TAl = mw x cw Tw,
T = temperature of system at equilibrium
100g(0.900 j /g0C)(1000C – T) = 200g(4.18 j /g0C)(T-250C)
• 9000 – 90T = 836T – 20900
926T = 29900, T = 32.3 0C 6

7. SOLIDS, LIQUIDS AND PHASE CHANGES
• When a substance undergoes phase change (change in
physical state) heat is added or removed from the
material however the temperature of the material
remains unchanged. All energy that is released or
absorbed during the phase change is used to change the
state of the substance. None is used to alter the
temperature.
• Only after the phase change is complete can
the temperature of the system rise or fall.
(Note the temperature plateaus on the next graph)
• The heat of phase change for different processes is
described by the physical change which is occurring for
example: Heat of Melting (sometimes called fusion),
Heat of Vaporization, Heat of Freezing, etc.
7

8. Time of Heating
T
E
M
P
E
R
A
T
U
R
E
0C
110
100
0
-10
All ice
Melting
starts
Melting
completed
Ice & water
All water
Boiling
starts
Water & steam
Boiling
completed
All
steam
Heating Curve for Water (-10 to 110 0C)
8

9. CALCULATING HEAT EXCHANGES DURING
PHASES CHANGES
• During a phase change the temperature of a substance
remains constant. Based on that observation, q = m c T
cannot be used to find heat gained or lost since T = 0.
• In order to find heat values during phase change we must
multiply the heat of phase change by the quantity of
substance involved. Heat of phase change is the quantity of
heat which is gained or lost when a substance changes state
without a temperature change.
• The heat of phase change may be expressed in calories or
joules per gram of substance or calories or joules per mole
of substance.
• Q = Heat of Phase Change x Amount of Substance 9

10. Heat of Melting
Heat of Vaporization
Heat of Condensation
Heat of Freezing
T
E
M
P
E
R
A
T
U
R
E
Time of Heating
HEATS OF PHASE CHANGE
DURING HEATING & COOLING
10

11. HEAT REQUIRED
FOR PHASE
CHANGE
HEAT OF MELTING (FUSION)
HEAT OF FREEZING
HEAT OF VAPORIZATION
HEAT OF CONDENSATION
HEAT OF SUBLIMATION
(ALL IN JOULES / GRAM,
CALORIES/ GRAM
OR PER MOLE
AMOUNT OF SUBSTANCE
CHANGING PHASE
(GRAMS OR MOLES)
CALCULATING HEAT GAINED
OR LOST DURING A PHASE
CHANGE
11

12. CALCULATING HEAT EXCHANGES
DURING PHASES CHANGES
• Problem: How much heat is needed to melt 200 grams of
ice at its melt point (0 0C) ? The heat of fusion for ice is
6019 joules per mole.
• Solution: Since a phase change is occurring without a
temperature change:
• qmelting = Heat of Melting x moles of ice melted
• Moles = grams / grams per mole, moles = 200 / 18 = 11.1
• qmelting = 6019 joules / mole x 11.1 moles = 66,878 joules
or 66.9 kilojoules.
12

13. CALCULATING HEAT EXCHANGES
DURING PHASES CHANGES
• Problem: How much heat is needed change 100 grams of ice at –10 0C
to steam at 110 0C. (c for ice = 2.06 j / g 0C, c for water = 4.18 j / g 0C,
c for steam = 2.02 j / g 0C, Heat of Fusion for ice = 6.02 Kj / mole,
Heat of Vaporization = 40.7 Kj / mole)
• Solution: qheating ice = m x c x T
• qheating ice = 100 x 2.06 x (0 – (-10)) = 2060 j
• qmelting = Heat of Melting x moles of ice melted
• qmelting = (6020 j / mole)(100 / 18 ) mole = 3.34 x 104 j
• qheating water = 100 x 4.18 x (100 – 0) = 4180 j
• qvaporization = (40.7 x 103 j / mole)(100 / 18) mole = 2.26 x 105 j
• qheating steam = 100 x 2.02 x (110 – 100) = 2020 j
• qtotal = 2060 + 3.34 x104 + 4018 + 2.26 x 105 + 2020 = 2.67 x 105 j
13

14. CALCULATING HEAT EXCHANGES
DURING PHASES CHANGES
• IMPORTANT POINTS IN CALCULATING HEAT QUANTITIES:
• (1) The specific heat of a substance is different for the same
substance in a different physical state. For example, as we have
seen, the specific heat of ice (2.06 j /g0C), that for water (4.18 j / g0C)
and steam (2.02 j / g0C) are different.
• (2) Heats of phase change are different for different changes
involving the same material. For example Heat of Fusion for ice is
6.02 Kj / mole while Heat of Vaporization for water is 40.7 Jj / mole.
Note that the Heat of Vaporization for a substance is always
noticeably larger than its Heat of Fusion.
• (3) Heat of phase change for a substance in the heating process
are equal but opposite in sign as compared for the reverse phase
change in the cooling process. For example, the Heat of Melting for
ice is 6.02 Kj / mole while the Heat of Freezing for water is – 6.02 Kj
/ mole. 14

15. TEMPERATURE AND PRESSURE EFFECTS
ON PHASE CHANGE
• Phase changes occur as:
• Melting or Fusion – solid to liquid (endothermic)
• Freezing or Solidification – liquid to solid (exothermic)
• Vaporization – liquid to gas (endothermic)
• Condensation – gas to liquid (exothermic)
• Sublimation – solid to gas (endothermic)
• Deposition – gas to solid (exothermic)
• All changes of physical state are affected by both temperatures and
pressures. Gas state is favored by low pressures and high
temperatures both allowing for free, random motion. Solids are
favored by low temperatures and high pressures both contributing
to an ordered low energy state. A phase diagram shows the
relationship between temperature, pressure and physical state
under a variety of conditions. 15

16. TEMPERATURE AND PRESSURE
EFFECTS ON PHASE CHANGE
• After a certain temperature (depending on the
substance) no amount of pressure can cause a gas to
form a liquid. This temperature is called the critical
temperature of the substance. The pressure which the
gas exhibits at this temperature is called the critical
pressure.This temperature – pressure point is called the
critical point.
• At a specific temperature and pressure all three phases
of a substance can coexist (solid, liquid and gas). This
point is called the triple point. At this point a very
slight change in temperature and / or pressure can
cause a solid to melt, a liquid to vaporize, a solid to
sublime, a liquid to freeze, etc. 16

17. P
R
E
S
S
U
R
E
TEMPERATURE
GAS
SOLID
LIQUID
PHASE DIAGRAM
TRIPLE
POINT
a
b
c
d
e
f
g
a = freezing
b = melting
c = vaporization
d = condensation
e= deposition
f = sublimation
g = critical point
17

18. P
R
E
S
S
U
R
E
GAS
SOLID
LIQUID
PHASE DIAGRAM
(for water)
a
b
Notice that moving
along the line
a-b, as pressure
rises, the solid
melts without a
temperature increase
TEMPERATURE
This behavior is
typical for
water but not
most substances
18

19. P
R
E
S
S
U
R
E
GAS
SOLID
LIQUID
PHASE DIAGRAM
(for most substances)
a
b Notice that moving
along the line
a-b, as pressure
rises, the solid
Will not melt
Increasing pressure
without a temperature
change will not
melt the substance
TEMPERATURE
This behavior is
typical for
most substances19

20. VAPOR PRESSURE OF LIQUIDS
• The measure of the tendency of a liquid to form a gas is
indicated by its vapor pressure. Liquids which evaporate
readily have high vapor pressures. The tendency of a liquid
to enter the gaseous state varies based on several factors:
• (1) the intermolecular forces which exist between the liquid
molecules. Factors such as polar character, London Forces
and hydrogen bonding (all of which have been discussed in
other programs) affect vaporization.As the strength of
intermolecular forces increases the vapor pressure of the
liquid decreases.
• (2) the Heat of Vaporization of a substance. Materials with
large Hvap have low vapor pressures.
• (3) the temperature of the liquid. As temperature rises the
liquid molecules increase their kinetic energies and the
tendency to form a gas increases. 20

21. 21
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