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Chapter 12 AIR The Sea We Live I

VIEWS: 3 PAGES: 89

  • pg 1
									                  CH104
        Chapter 2: Energy & Matter
                Temperature
            Energy in Reactions
                Specific Heat
             Energy from Food
               States of Matter
          Heating & Cooling Curves
CH104
        Chapter 2 Energy and Matter
                        2.2
                     Temperature




CH104
                                   2
                  Units of Measurement

                  Metric        Common
                   SI          Conversions
  Length        meter (m)      1 m = 1.09 yd
  Volume        liter (L)      1 L = 1.06 qt
  Mass          gram (g)       1 kg = 2.2 lb
  Temperature   Celsius (oC)   C = (F-32)/1.8
                Kelvin (K)      K = C + 273



CH104
            Temperature



        o            o
 180           100




CH104
                              Learning Check
   A. What is the temperature of freezing water?
     1) 0 °F        2) 0 °C     3) 0 K

   B. What is the temperature of boiling water?
      1) 100 °F     2) 32 °F           3) 373 K

   C. How many Celsius units are between the boiling
      and freezing points of water?
     1) 100          2) 180      3) 273
CH104
                                                  5
                                        Solution
   A. What is the temperature of freezing water?
     1) 0 °F        2) 0 °C     3) 0 K

   B. What is the temperature of boiling water?
      1) 100 °F     2) 32 °F           3) 373 K

   C. How many Celsius units are between the boiling
      and freezing points of water?
     1) 100          2) 180      3) 273
CH104
                                                  6
        Trivial Pursuit Question
• At what temperature does
  oF = oC?




   - 40 oF = - 40 oC

                   - 40 o


CH104
          Temperature conversion
    • Common scales used
          • Fahrenheit, Celsius and Kelvin.
              oF   = 1.8 oC + 32
              oC   = (oF - 32)
                        1.8
SI unit       K    = oC + 273
CH104
            Solving a Temperature Problem
A person with hypothermia has a
body temperature of 34.8 °C. What
Is that temperature in °F?

F = 1.8(C) + 32 °

F = (1.8)(34.8 °C) + 32 °
          exact tenth’s   exact


        = 62.6 ° + 32 °
        = 94.6 °F
         tenth’s
CH104
                                     9
        Practice: Temp Conversion

    • What is 75.0 º F in ºC?
         • ºC = (75.0 º F -32 º) = 23.9 ºC
                    1.8


    • What is -12 º F in ºC?
         • º F = 1.8 (-12) + 32 º F = 10 º C



CH104
                                  Learning Check
        The normal temperature of a chickadee is 105.8
        °F. What is that temperature on the Celsius scale?
        1) 73.8 °C     2) 58.8 °C        3) 41.0 °C




CH104
                                                   11
                                             Solution
        The normal temperature of a chickadee is 105.8
        °F. What is that temperature on the Celsius scale?
        1) 73.8 °C     2) 58.8 °C        3) 41.0 °C


           TC    =     TF – 32 °
                          1.8
                 =     (105.8 – 32 °)
                           1.8
                 =     73.8 °F = 41.0 °C
                         1.8 °        tenth’s place
CH104
                                                  12
                              Learning Check
    A pepperoni pizza is baked at 455 °F. What
    temperature is needed on the Celsius scale?
    1) 423 °C             2) 235 °C         3) 221 °C




CH104
                                               13
                       Solution
    A pepperoni pizza is baked at 455 °F. What
    temperature is needed on the Celsius scale?
    1) 423 °C             2) 235 °C         3) 221 °C



          TF – 32 ° = TC
           1.8

          (455 – 32 °) = 235 °C
              1.8       one’s place
CH104
                                               14
                              Learning Check
   On a cold winter day, the temperature is –15 °C.
   What is that temperature in °F?
   1) 19 °F          2) 59 °F         3) 5 °F




CH104
                                               15
                                            Solution
   On a cold winter day, the temperature is –15 °C.
   What is that temperature in °F?
   1) 19 °F          2) 59 °F         3) 5 °F

   TF   = 1.8TC + 32 °        Note: Be sure to use the
                              change sign key on your
                              calculator to enter the minus
   TF = 1.8(–15 °C) + 32      (–) sign.
                                      1.8 x 15 +/ – = –27
   °
      = – 27 + 32 °
      = 5 °F
         one’s place
CH104
                                                    16
        Temperatures




CH104
                       17
                                Learning Check
        What is normal body temperature of 37 °C in
        kelvins?
        1) 236 K      2) 310 K    3) 342 K




CH104
                                                18
                                          Solution
        What is normal body temperature of 37 °C in
        kelvins?
        1) 236 K      2) 310 K    3) 342 K



          TK   =     TC + 273
               =     37 °C + 273
               =     310. K
                    one’s place

CH104
                                                19
        Chapter 2 Energy and Matter
                   2.1
                  Energy




CH104
                                 20
                               Energy
        Energy = The capacity to cause change

                     Heat
                     Light
                     Wind

CH104
Potential Energy = stored Energy
                   (Has potential for motion)

        X



Kinetic Energy = Energy in motion
                 (Fulfilling its potential)




CH104
                           Learning Check
Identify each of the following as potential
  energy or kinetic energy.
A. roller blading
B. a peanut butter and jelly sandwich
C. mowing the lawn
D. gasoline in the gas tank



CH104
                                          23
                                   Solution
Identify each of the following as potential
  energy or kinetic energy.
A. roller blading      kinetic
B. a peanut butter and jelly sandwich potential
C. mowing the lawn kinetic
D. gasoline in the gas tank       potential



CH104
                                        24
                              Kinetic Energy
           KE = 1 mv2
                2           Which has more E?
    •Truck moving at 5 mph



        •Bicycle moving at 5 mph




CH104
                 Units of Measurement
                  Metric            Common
                   SI              Conversions
  Length         meter (m)       1 m = 1.09 yd
  Volume         liter (L)       1 L = 1.06 qt

  Mass           gram (g)        1 kg = 2.2 lb

  Temperature    Celsius (oC)    C = (F-32)/1.8
                  Kelvin (K)      K = C + 273

  Energy        calorie (cal)   1Kcal = 1000 cal = 1Cal
                Joule (J)       1 cal = 4.18 J
CH104
        Examples of Energy Values in Joules




CH104
                         Learning Check
  How many calories are obtained from a pat of
   butter if it provides 150 J of energy when
   metabolized?




CH104
                                        28
                                    Solution
  How many calories are obtained from a pat of
   butter if it provides 150 J of energy when
   metabolized?
        Given                  Need
        150 J   1 cal      =   36     cal
                4.184 J


   Conversion Factors:
1 cal      and 4.184 J
4.184 J         1 cal
CH104
        Chapter 2 Energy and Matter
                     2.3
                Specific Heat




CH104
                                30
                                      Energy
        Units of Energy =
         calorie       cal
         kilocalorie   kcal   1000 cal = 1 kcal

          Calorie      Cal    Cal = kcal

          joule        J      4.18 J = 1 cal

           British
                      BTU
         Thermal Unit

CH104
                               Energy
        calorie:
          •E to raise 1 g H20 by 1 oC
                         Specific Heat
          H 2O               1 cal
                   1oC
                            1g 1oC




CH104
                             Specific Heat
    E to raise Temp of 1g substance by 1 oC
            cal
            g oC




        H2O Al Sand Fe       Cu    Ag    Au
        1.00 0.22 0.19 0.11 0.093 0.057 0.031
 0oC = start                 Add 1 cal
CH104
                                     Specific Heat
      E to raise Temp of 1g substance by 1 oC
30o                                               Au
                cal                              0.031
20o                                        Ag
                g oC
                                      Cu 0.057
10o                            Fe
                                     0.093
                   Al Sand    0.11
1o                0.22 0.19
         H 2O
         1.00
     0oC = start                     Add 1 cal
 CH104
                                   Specific Heat
30o                Low SpHt; Heats quickly      Au
                                         Ag    0.031
20o
                                    Cu 0.057
10o                          Fe
                                   0.093
                 Al Sand    0.11
1o              0.22 0.19
         H 2O   High SpHt; Resists change
         1.00

 CH104
                               Specific Heat
                                Dehydrated person
                                  Body temp rises
                                           quickly

            Hydrated person
           Resists change in
              body temp

                           Heats quickly
                      Sand   Gets hot
   H2O Resists change 0.19
   1.00 Stays cold
CH104
                          Learning Check
A. For the same amount of heat added, a
  substance with a large specific heat
   1) has a smaller increase in temperature
   2) has a greater increase in temperature
B. When ocean water cools, the surrounding air
   1) cools 2) warms 3) stays the same
C. Sand in the desert is hot in the day and cool
   at night. Sand must have a
   1) high specific heat 2) low specific heat
CH104
                                    Solution
A. For the same amount of heat added, a
  substance with a large specific heat
   1) has a smaller increase in temperature
   2) has a greater increase in temperature
B. When ocean water cools, the surrounding air
   1) cools 2) warms 3) stays the same
C. Sand in the desert is hot in the day and cool
   at night. Sand must have a
   1) high specific heat 2) low specific heat
CH104
 Sample Problem:                Specific Heat
What is the specific heat of a metal if 24.8 g absorbs
 65.7 cal of energy and the temp rises from 20.2 C
 to 24.5 C?
                   cal     = 65.7 cal       = 0.62 cal
    SpHt =
                g oC        24.8g 4.3oC       1g 1oC

                       24.5oC
                                    DT = 4.3 Co
 m = 24.8g
                       20.2oC

CH104
    E = m DT SpHt
                              Specific Heat
    Sample Problem:
    How much energy does is take to heat
    50 g’s of water from 75oC to 87oC?

                      87oC
                                  DT = 12 Co
        m = 50g H2O

                      75oC            1 cal
                             SpHt =
                                      1g 1oC


CH104
    E = m DT SpHt
                            Specific Heat
    Sample Problem:
    How much energy does is take to heat
    50 g’s of water from 75oC to 87oC?
           m          DT    SpHt
        50g H2O   12 Co    1 cal   = 600 cal
                           1g 1oC to heat water




CH104
    E = m DT SpHt
                               Specific Heat
    Sample Problem:
A hot-water bottle contains 750 g of water at 65 °C.
If the water cools to body temp (37 °C), how many
calories of heat could be transferred to sore
muscles?

                     65oC
  m = 750g H2O                   DT = 28 Co

                     37oC
                            SpHt = 1 cal
                                   1g 1oC
CH104
    E = m DT SpHt
                                 Specific Heat
    Sample Problem:
A hot-water bottle contains 750 g of water at 65 °C.
If the water cools to body temp (37 °C), how many
calories of heat could be transferred to sore
muscles?


        m       DT        SpHt
  750g H2O     28 Co     1 cal     = 21000     cal
                        1g 1oC     from cool water

CH104
                         Learning Check
 How many kilojoules are needed to raise the
 temperature of 325 g of water from 15.0 °C to
   77.0 °C?




CH104
                                        45
    E = m DT SpHt                    Solution
 How many kilojoules are needed to raise the
 temperature of 325 g of water from 15.0 °C to
   77.0 °C?
        m       DT       SpHt
325g H2O       62 Co   4.184 J   1 KJ
                                          = 84.3 KJ

                       1g 1oC    1000 J

        77oC
                DT = 62 Co
        15oC
CH104
                                           46
        Chapter 2 Energy and Matter

                               2.4
                       Energy and Nutrition


               1 Cal     =   1000 calories
               1 Cal     =   1 kcal
               1 Cal     =   4184 J
               1 Cal     =   4.184 kJ


CH104
                                             47
                                    Energy
        Units of Energy =
           1 Cal    1000 cal   = 1 kcal
                    1 cal      = 4.18 J

            1 Cal   4184 J     = 4.184 kJ




CH104
                              Calorimeters
  A calorimeter
  • A reaction chamber
    & thermometer in
    H2O used to measure
    heat transfer
  • indicates the heat lost
    by a sample
  • indicates the heat
    gained by water



CH104
                  Energy from Food
                  4 kcal    17 kJ
                    g         g
  Carbohydrate

                  9 kcal    38 kcal
         Fat
                    g         g

        Protein   4 kcal    17 kcal
                    g         g

CH104
                         Energy from Food
    Sample Problem:
    How much energy in kcal (= Cal) would be
    obtained from 2 Tbl peanut butter
    containing 6 g carb, 16 g fat, & 7 g protein?
        6 g carb    4 kcal    = 24 kcal
                   1 g carb                = 196 kcal

        16 g fat    9 kcal    = 144 kcal   = 196 Cal
                    1 g fat

                   4 kcal                  = 200 Cal
    7 g protein             = 28 kcal
                1 g protein
CH104
                             Learning Check
  A cup of whole milk contains 12 g of carbohydrate,
  9.0 g of fat, and 9.0 g of protein. How many kcal
  (Cal) does a cup of milk contain? (Round to the
  nearest 10 kcal.)




CH104
                                              53
                                         Solution
  A cup of whole milk contains 12 g of carbohydrate,
  9.0 g of fat, and 5.0 g of protein. How many kcal
  (Cal) does a cup of milk contain? (Round to the
  nearest 10 kcal.)
    12 g carb 4 kcal          = 48 kcal
                 1 g carb                  = 149 kcal

    9 g fat     9 kcal     = 81 kcal     = 149 Cal
                1 g fat

               4 kcal                     = 150 Cal
5 g protein             = 20 kcal
            1 g protein
CH104
                                               54
        Chapter 2 Energy and Matter
                       2.5
            Classification of Matter




CH104
                                       55
        •                        Matter
            The stuff things are made of.
            Has Mass and takes up space.
                     (Air, water, rocks, etc..)

                   The amount of stuff (in g’s)
                        (Bowling Ball > Balloon)
                                Weight on earth.
                                 Pull of Gravity on matter.
CH104
                     Classification of matter
                       Matter

         Pure Substance             Mixture

Element       Compound

   Fe          FeS         Fe + S

        Mg     MgO         Mg + O2

CH104
                      Mixture          Mixtures
      Homogeneous               Heterogeneous
          (Solution)
   Uniform composition     Non-uniform composition


              Air
                                     Pizza

              Urine                  Fe + S
 Gasoline                Tea w/ice
                                     Sand
CH104
        Physical Separation of A Mixture
  Mixtures can be separated
  • involves only physical changes
  – Like Filtering & distilling




  Like when pasta and water are separated with a
  strainer
CH104
                           Learning Check
  Identify each of the following as a pure
    substance or a mixture.
  A. pasta and tomato sauce
  B. aluminum foil
  C. helium
  D. Air



CH104
                                    Solution
  Identify each of the following as a pure
    substance or a mixture.
  A. pasta and tomato sauce         mixture
  B. aluminum foil            pure substance
  C. helium                   pure substance
  D. Air                            mixture



CH104
                         Learning Check
   Identify each of the following as a
      homogeneous or heterogeneous mixture.
   A. hot fudge sundae
   B. shampoo
   C. sugar water
   D. peach pie



CH104
                                      64
                                  Solution
   Identify each of the following as a
      homogeneous or heterogeneous mixture.
   A. hot fudge sundae         heterogeneous
   B. Shampoo                  homogeneous
   C. sugar water              homogeneous
   D. peach pie                heterogeneous



CH104
                                       65
        Chapter 2 Energy and Matter
                       2.6
              States and Properties
                    of Matter




CH104
                 Elemental states at 25oC
H                            Solid                                         He

Li Be
                             Liquid                 B    C    N   O   F    Ne
                              Gas
Na Mg                                               Al   Si   P   S   Cl   Ar

K       Ca Sc   Ti    V   Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Rb Sr      Y    Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te                I    Xe

Cs Ba Ls Hf Ta            W Re Os     Ir   Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Ac
                     Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

                     Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
CH104
                                                                           6-2
                  Properties of matter
                     Physical State
   Property    Solid    Liquid      Gas
   Density     High       High         Low
                        like solids
                                      expands
   Shape       Fixed    Shape of
                        container     to fill
                                      container
   Compressibility
                Small     Small        Large
   Thermal
   expansion    Very                  Moderate
                small
                          Small
CH104
        Three States of Water




CH104
                        71
            Hydrogen Bonding of Water

        •




Frozen H2O:
Slow moving molecules
H-Bond in patterns
CH104
                            Learning Check
 Identify each as a S) solid, L) liquid, or G) gas.
 __ A. It has a definite volume but takes the
 shape of the container.
 __ B. Its particles are moving very rapidly.
 __ C. It fills the volume of a container.
 __ D. It has particles in a fixed arrangement.
 __ E. It has particles that are close together
 and are mobile.


CH104
                                      Solution
 Identify each as a S) solid, L) liquid, or G) gas.
 L A. It has a definite volume but takes the
 shape of the container.
 G B. Its particles are moving very rapidly.
 G C. It fills the volume of a container.
 S D. It has particles in a fixed arrangement.
 L E. It has particles that are close together
 and are mobile.


CH104
                       Physical Properties
  Physical properties
  • are observed or measured without changing the
  identity of a substance
  • include shape and color
  • include melting point and boiling point
            Physical Properties of Copper




CH104
                          Physical Change
 In a physical change,
 • the identity and
   composition of the
   substance do not
   change
 • the state can change
   or the material can
   be torn into smaller
   pieces

CH104
                                     76
        Examples of Physical Change




CH104
                               77
                     Chemical Properties
Chemical properties
• describe the ability of a substance to
  change into a new substance




CH104
                                           78
                        Chemical Change
During a chemical
  change,
• reacting substances
  form new substances
  with different
  compositions and
  properties
• a chemical reaction       Iron   Iron (III)
  takes place                        oxide
                             Fe
                                    Fe2O3
CH104
        Examples of Chemical Change




CH104
                               80
        Summary




CH104
            81
                           Learning Check
 Classify each of the following as a 1)
 physical change or 2) chemical change.
 A. ____ burning a candle
 B. ____ ice melting on the street
 C. ____ toasting a marshmallow
 D. ____ cutting a pizza
 E. ____ polishing a silver bowl

CH104
                                          82
                                       Solution
 Classify each of the following as a
 1) physical change or 2) chemical change.
 A. ____ burning a candle         2) chemical
 B. ____ ice melting on the street 1) physical
 C. ____ toasting a marshmallow 2) chemical
 D. ____ cutting a pizza         1) physical
 E. ____ polishing a silver bowl 2) chemical

CH104
                                          83
        Chapter 2 Energy and Matter
                    2.7
               Changes of State




CH104
                                  84
                              Changes of State
  Fast, far apart,
     Random                Vapor
    Vaporize          Boiling Pt       Condense
                                      Moderate, close,
                     Liquid              Random
                                       arrangement
                     Melting Pt =
        Melt         Freezing Pt        Freeze
   Slow, close,
       Fixed                  Solid
   arrangement
CH104
                        Changes of State
                    Vapor

                                Deposit
               Liquid            Frost
   Sublime
  Freeze Dry

                        Solid
CH104
  Heating Curve Specific Heat of Steam
                          0.48 cal to heat vapor
                          g oC

                                Heat of Vaporization
 Specific Heat of H2O           540 cal to vaporize water
                                 g
 1.00 cal to heat water
 g oC                            Heat of Fusion
                                 80 cal to melt ice
                                  g
              Specific Heat of Ice
              0.50 cal to heat ice
CH104         g oC
  Heating Curve


                  0.48 cal to heat vapor
                  g oC

                  540 cal to vaporize water
                   g
                  1.00 cal to heat water
                  g oC
                  80 cal to melt ice
                   g
                  0.50 cal to heat ice
                  g oC

CH104
Example: Calculate the total amount of heat needed
to change 500. g of ice at –10 oC into 500. g of steam
at 120oC.                   o
                   500g 20 C 0.48 cal   = 4800 cal
                              g  oC
                                      to heat vapor

                  500g 540 cal         = 270,000 cal
                         g           to vaporize water
                  500g 100 o C 1.00 cal = 50,000 cal
                               g o C to heat water

                  500g   80 cal        = 40,000 cal
                           g            to melt ice
                  500g 10 o C 0.50 cal = 2500 cal
                               g oC    to heat ice
                     367,300 cal = 3.67 x 105 cal
CH104
                          Learning Check
 A. A plateau (horizontal line) on a heating
 curve represents
       1) a temperature change
       2) a constant temperature
       3) a change of state

 B. A sloped line on a heating curve represents
       1) a temperature change
       2) a constant temperature
       3) a change of state
CH104
                                         90
                                    Solution
 A. A plateau (horizontal line) on a heating
 curve represents
       1) a temperature change
       2) a constant temperature
       3) a change of state

 B. A sloped line on a heating curve represents
       1) a temperature change
       2) a constant temperature
       3) a change of state
CH104
                         Learning Check
Use the cooling curve for water to answer each.
A. Water condenses at a temperature of
   1) 0 °C      2) 50 °C 3) 100 °C
B. At a temperature of 0 °C, liquid water
   1) freezes 2) melts 3) changes to a gas
C. At 40 °C, water is a
   1) solid          2) liquid 3) gas
D. When water freezes, heat is
   1) removed 2) added
CH104
                                        92
                                   Solution
Use the cooling curve for water to answer each.
A. Water condenses at a temperature of
   1) 0 °C      2) 50 °C 3) 100 °C
B. At a temperature of 0 °C, liquid water
   1) freezes 2) melts 3) changes to a gas
C. At 40 °C, water is a
   1) solid     2) liquid 3) gas
D. When water freezes, heat is
   1) removed         2) added
CH104
                                        93
                               Learning Check
  To reduce a fever, an infant is packed in 250 g of ice.
    If the ice (at 0 °C) melts and warms to body temp
    (37.0 °C), how many calories are removed?




CH104
                                                 94
                                           Solution
  To reduce a fever, an infant is packed in 250 g of ice.
    If the ice (at 0 °C) melts and warms to body temp
    (37.0 °C), how many calories are removed?

  37.0 °C           250g    37 o C 1.00 cal = 9,250 cal
                                   g o C to heat water
    STEP 2
                    250g    80 cal         = 20,000 cal
0.0 °C                        g             to melt ice
  STEP 1
                           29,000 cal = 2.9 x 104 cal


CH104

								
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