Unit States of Matter Solutions

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```					Unit 3 – Energy & Phase
Changes
Chapters: 13 & 15
Test: Friday, October 8th
Phases
• States of substances
are called phases when
they coexist as
physically distinct parts
of a mixture, such as ice
water.
– When energy is added to
or taken away from a
system, one phase can
change into another.
States of Matter – Phase Changes
• Melting/Freezing
changes between solid and liquid phases
• Vaporization/Condensation
changes between liquid and gas phases
• Sublimation/Deposition
changes directly between solid and gas phases
Heat (Thermal) Energy
Heat is a form of energy that flows between two
samples of matter because of their difference in
temperature

• Heat flows from hot to cold
• It can be absorbed or
released
• Measured in units of
calories or Joules
States of Matter – Phase Changes
• Endothermic phase
changes absorb
energy
– Sublimation
– Vaporization
– Melting
• Exothermic phase
changes release
energy
– Deposition
– Condensation
– Freezing
States of Matter – Phase Changes
Vaporization includes 2 processes:
• Evaporation
– Only occurs at the surface of the liquid
at room temperature
– Ex: Sweating cools the human body
• Boiling
– Occurs throughout the liquid
– Boiling point occurs when vapor
pressure = atmospheric pressure
– Ex: Boiling water
• Higher altitudes have lower
atmospheric pressure, so lower
boiling point
Phase Diagrams
• Temperature and pressure control the phase of a
substance.
• A phase diagram is a graph of pressure versus
temperature that shows in which phase a
substance exists under different conditions of
temperature and pressure.
– A phase diagram has three regions, each a
different phase and three curves that separate
each phase.
– The points on the curves indicate conditions
under which two phases coexist.
States of Matter – Phase Changes

Freezing
point      Boiling
point

Sublimation
point
States of Matter – Phase Changes

• Standard Pressure Line (1 atm)
• Normal Freezing Point
– Temp at which standard
pressure meets the solid-liquid
curve
• Normal Boiling Point                        Normal Freezing Point   Normal Boiling Point

STP
– Temp at which standard
pressure meets the liquid-
vapor curve
• Triple Point
– Pressure and temperature
where all three phases coexist
• Critical Point
– Pressure and temperature at
which a liquid can no longer
exist
States of Matter - Intermolecular Forces
• Intermolecular forces act between stable
molecules (not like bonding where forces act
within molecules)
• Weak intermolecular forces
– Low boiling points
– Most likely in gaseous state
• Strong intermolecular forces
– High boiling points
– Most likely in solid state
States of Matter - Vapor Pressure
• Vapor Pressure – pressure exerted by vapor
molecules above a liquid when dynamic
equilibrium is reached
• Dynamic equilibrium – 2 opposite processes
occurring at the same rate
– Ex: evaporation and condensation
• Vapor pressure depends on…
– Temperature (T V.P.)
– Strength of IMF (IMF V.P.)
• Volatile - how easily a fluid evaporates (high
vapor pressure)

Vapor       Vapor        Compare the vapor pressure of
water and alcohol. What do
Water       Alcohol
the arrows mean in these
diagrams?
States of Matter - Vapor Pressure
• Q: Why does water boil at a lower temperature at
higher altitudes?
States of Matter - Vapor Pressure

• A: Pressure is
lower, meaning less
energy required to
overcome
intermolecular
forces (less forces
“holding it together”)
States of Matter - Vapor Pressure
• Vapor pressure graphs
show vapor pressure
vs. temperature
• Standard pressure is 1
atm or 101.3 kPa –
indicated by the dark,
horizontal line
– Intersection of curved
lines and standard
pressure is the normal
boiling point
Heat of Fusion
• Heat of fusion (Hf) : the amount of energy needed
to change a solid into a liquid (melt)
– There is no temperature change during melting,
only an energy change.
• Heat of fusion can be determined by the following
equation: Q = mHf
– Q = heat required to melt (Joules - J)
– m = mass (grams - g)
–Hf = Latent heat of fusion
(Joules per gram - J/g)
Hfus Practice
• Calculate the heat of fusion when 60g of
HCl melts from solid to liquid at -114oC.
The latent heat of fusion for HCl is 54.6J/g.
– 3280 J or 3.28 kJ
Heat of Vaporization
• Heat of vaporization (Hv) : the amount of energy
needed to change a liquid into a gas (vaporize)
– There is no temperature change during
vaporization, only an energy change.
• Heat of vaporization can be determined by the
following equation: Q = mHv
– Q = heat required to vaporize
(Joules - J)
– m = mass (grams - g)
– Hf = Latent heat of vaporization
(Joules per gram - J/g)
Hvap Practice
• Calculate the heat of vaporization when
135g of Nitrogen vaporizes from liquid to
gas. The latent heat of vaporization for
nitrogen is 200J/g.
– 72.0 kJ
Heat of Fusion/Vaporization

Why do you think water requires more energy to change from liquid to gas?
Heat of Fusion/Vaporization:
Calculations
• Example: Calculate the heat of fusion when 60.0 g of
HCl melts from solid to liquid. The heat of fusion for HCl
is 54.6 J/g.

Q = 60.0 g * 54.6 J/g = 3280 J

• Example: Calculate the heat of vaporization when 135 g
of N2 vaporizes from liquid to gas. The heat of
vaporization for N2 is 200 J/g.

Q = 135 g * 200 J/g = 2.70 x 104 J
Specific Heat
• The specific heat of any substance is the amount
of heat required to raise the temperature of one
gram of that substance one degree Celsius.

Why will the
metal chairs
at the pool
get hot, while
the water
stays cool?
Specific Heat
• Each substance has its own specific heat
– Due to different composition and arrangement of
particles
• Water has an extremely high specific heat
(4.184 J/goC)
• Metals have low specific heats (<1 J/goC)

• Specific heat equation: Q = mcT
–   Q = heat absorbed or released (J)
–   c = specific heat (J/goC)
–   m = mass (g)
–   T = Tf - Ti change in temperature (oC)
Specific Heat Calculations

• Example: In the construction of skyscrapers, gaps must
be left between adjoining steel beams to allow for the
expansion and contraction of the metal. The temperature
of a 10.0 g sample of iron changes from 50.4°C to
25.0°C and releases of 114 J heat. What is the specific
heat of iron?

Q = mcT
114 = c * 10.0 * (50.4-25.0)
c = 0.449 J/goC
Measuring Heat
• Heat changes that occur during chemical
and physical processes can be measured
accurately and precisely using a
calorimeter.
• A calorimeter is an insulated device used
for measuring the amount of heat
absorbed or released during a chemical or
physical process.
Determining Specific Heat
• Suppose you put 125 g of water into a
foam-cup calorimeter and find that its
initial temperature is 25.6°C.
• You heat a 50.0-g sample of the
unknown metal to a temperature of
115.0°C and put the metal sample into
the water.
– Heat flows from the hot metal to the
cooler water and the temperature of
the water rises.
– The flow of heat stops only when
the temperature of the metal and the
water are equal.
Using Data from Calorimetry
• A piece of metal with a mass of 4.68 g absorbs
256 J of heat when its temperature increases by
182°C.
• What is the specific heat of the metal?
– Known
• mass of metal = 4.68 g metal
• quantity of heat absorbed, q = 256 J
• ∆T = 182°C
– Unknown
• specific heat, c = ? J/(g·°C)
Using Data from Calorimetry
• Substitute the known values into the equation.
• The calculated specific heat is the same as that
of strontium.
Heat of Fusion/Vaporization:
Graphs
Changing States and Temperature
TEMPERATURE ( C)

100

1. Heat solid to
melting point

0

q = (m) (Cp) (T2-T1)

HEAT (cal/g) OR TIME
Changing States and Temperature
TEMPERATURE ( C)

100

2. Melting solid
to liquid

0

q = (m) (Cp)

HEAT (cal/g) OR TIME
Changing States and Temperature
TEMPERATURE ( C)

100

3. Heat liquid
to boiling point

q = (m) (Cp) (T2-T1)
0

HEAT (cal/g) OR TIME
Changing States and Temperature
4. Change liquid
TEMPERATURE ( C)

to gas
100

q = (m) (Cp)

0

HEAT (cal/g) OR TIME
Changing States and Temperature
5. Heating
gas
TEMPERATURE ( C)

100

q = (m) (Cp) (T2-T1)

0

HEAT (cal/g) OR TIME
Heating/Cooling Curves
• During phase changes, there is no change in kinetic energy -
only potential energy increases!!!
– Liquid H2O at 0oC has more total energy than solid H2O at
0oC
– Gas H2O at 100oC has more total energy than liquid H2O at
100oC

Coke on ice - shows
3 phases of matter
Calculating Phase Changes

100
Temperature ( C)

0

*Sum of the Q’s =
total energy used

Heat (J) or Time (sec)
States of Matter - Phase Changes
Coke on ice - shows
3 phases of matter

• During phase changes, there is no change in kinetic energy - only
potential energy increases!!!
– Liquid H2O at 0oC has more potential energy than solid H2O at 0oC
– Gas H2O at 100oC has more potential energy than liquid H2O at 100oC

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