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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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