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					                           Report on the 2009 Tertiary Entrance Examination in

                                                   Physical Science
The TEE subject Physical Science was examined for the last time in 2009.

This report was written by the chief examiner. The opinions and recommendations expressed iare
not necessarily representative of, or endorsed by, the Curriculum Council.

The marking key appended to the report was prepared by the examining panel and modified as
appropriate at the pre-marking meeting. It is not intended as a set of model answers, and is not
exhaustive as regards alternative answers. It represents a standard of response that the examiners
deemed sufficient to earn full marks. Teachers who use this key should do so with its original
purpose in mind.

Candidature
                                            Year       Number    Number of
                                                       who sat   absentees
                                            2009         343        26
                                            2008         319        42
                                            2007         360        39

Summary
The 2009 paper provided a thorough coverage of the syllabus, examining both process and
content objectives. The format of the paper was the same as that used since 1986 and the
examiners attempted to make the questions as contextual as possible. The marking guide
provided clear and unambiguous direction to the markers and little reconciliation of marks was
necessary. Marks ranged from 9% to 93%. The mean was 58.43% (56.4% in 2008, 57.9% in
2007, 56.7% in 2006) with a standard deviation of 15.85% (14.41% in 2008, 14.27% in 2007).
Both close to desirable values of 60.0% and 15.0% the reliability of the paper was 0.7 (0.65 in
2008, 0.54 in 2007). The multiple-choice reliability was 0.83 (0.80 in 2008, 0.9 in 2007). Variations
in the difficulty of optional sections appear to be minor. The manipulation of formula, use of
scientific notation, appropriate units, and particularly an appreciation of significant figures seem to
have improved.

Comments on specific elements
Part A
Section 1 (Multiple-choice)
Mean 63.88 %, standard deviation 17.87%, correlation to exam total 0.83 (57.76%, 15.64%, 0.80 in
2008)

Section 2 (Short Answer and Problems)
Mean 58.41%, standard deviation 18.04%, correlation to exam total 0.97 (68.20%, 16.97%, 0.96 in
2008)




2009 examination report: Physical Science                                                                 1
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Section 3 (Comprehension)
Mean 54.71%, standard deviation 16.47%, correlation to exam total 0.71 (48.75%, 15.23%, 0.63 in
2008)

Part B
This year 94.2% (88.7% in 2008) of the candidates attempted the Water Option, 5.24% (7.8% in
2008) attempted the Sound and Light Option and 0.29% (3.4% in 2008) of candidates attempted
the Engines and Fuels Option.

Water (N = 323 (283 in 2008))
Mean 51.67%, standard deviation 15.78%, (39.56%, 17.39%,

Sound & Light (N = 18 (25 in 2008))
Mean 49.35%, standard deviation 18.03% (39.87%, 17.88% in 2008)

Engines and Fuels (N = 1 (11 in 2008))
Mean 26.67%, standard deviation 0.00% (31.21%, 16.71% in 2008)

Issues for the relevant committee to consider
Nil

Acknowledgements
The examining panel thanks all those who contributed to the development of the examination
materials and to the marking process.

Bruce Banyard
December 2009

2009 examining panel
Chief examiner: Bruce Banyard
Examiner: Michelle de Kok
Examiner: Brian Sanderson

Chief marker: Michelle de Kok




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                                            2009 TEE Physical Science

                                                       Marking key


Part A
Section 1: Multiple Choice
                                                       Question      Answer
                                                          1            d
                                                          2            c
                                                          3            b
                                                          4            a
                                                          5            d
                                                          6            a
                                                          7            c
                                                          8            a
                                                          9            b
                                                         10            d
                                                         11            a
                                                         12            d
                                                         13            a
                                                         14            c
                                                         15            c
                                                         16            c
                                                         17            b
                                                         18            d
                                                         19            b
                                                         20            b



Part A
Section 2: Short Answer and Problem Solving

1         Tin is relatively inert                                    1 mark
          Steel when exposed reacts                                  1 mark

2
                                  Property                 Explanation        Example (2)
Covalent molecular                Low boiling point        Small molecule     CO2, NH3
                                                           Secondary bond
                                                           need to be
                                                           broken
Covalent network                  High boiling point       Large network      Diamond, SiC
                                                           Primary bonds
                                                           need to be
                                                           broken
                                                                                             1 mark each box




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3
          (a)        Al(s) →                Al3+(aq)        +    3e-
          (b)        Pb2+(aq)               +       2e -
                                                            →    Pb(s)
          (c)        3 Pb2+(aq)             +       2 Al(s) →    2Al3+(aq)     +      3 Pb(s)
          (d)        Pb2+
                                                                                                  1 mark per part
4
          (a)        Cu(s) →         Cu2+(aq)             +      2e-
                        2+
                     Zn (aq)         +      2e-           →      Zn(s)                1 mk each
          (b)        Eq 1            -0.34 V,             Eq 2   -0.76 V       1 mk
                     -1.10 V                                                   1 mk
          (c)        No                                                               1 mk
          (d)        E0 is less than 0.                                               1 mk

5
                                 Lithium metal                         2,1
                                 Oxygen atom                           2,6
                                 Sulfide ion                           2,8,8
                                 Magnesium metal                       2,8,2
                                 Potassium ion                         2,8,8
                                                                                                      1 mk each
6         (a)        CaCO3                  →      CaO    +      CO2           1 mk correct formula
                                                                               1 mk balanced
          (b)        1 mol CaCO3 produces 1 mol CaO                            1 mk
                     100.1 g                          56.1 g
                     3500 kg                   x kg                            1 mk for these 2
                     X = 1960 kg                                               1 mk
          (c)        1 mol CaCO3 produces 1 mol CO2
                     100.1 g                   44.0 g
                     3500 kg                   x kg                            1 mk for these 3
                     X = 1540 kg                                               1 mk


     7          Methyl ethanoate




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                Methanal (formaldehyde)                                                    2 mks each


   8
 Molecule                       Electron dot diagram                   Name the shape      Solubility in water
                                                                                           @ 25°C
                                                                       Pyramidal           High

            NH3


                                                                       Straight            High
            HCl


                                                                       Straight            Slight


            CO2



                                                                                              1 mk each box
     9
     (a)        C5H12           +           8O2     →   5CO2   +      6H2O
     (b)        Ag+(aq)         +           Cl-(aq)     →      AgCl(s)
     (c)        2H+(aq)         +           MgO(s)      →      H2O(l) +     Mg2+(aq)
     (d)        Ba2+(aq)        +           SO42-(aq)   →      BaSO4(aq)
                                                                         1 mk correct formula, 1 mk balanced

     10
     (a)    c = 0.025 mol L-1        V=2L
            N = cV     =      0.025 x 2     =      0.050 mol
     (b)    V = 6.72 L @ STP         n = V/22.4    =      0.300 mol
     (c)    0.35 mol
     (d)    c = n/v    =      .35/2 =       0.175 mol L-1
                        -3
     50 to 500 1.68 x 10 mol                                                       1 mk each part




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     11




     12         (a)             15.20 15.30 16.00 15.30
                (b)             H+     +      OH-    →        H2O
                (c)             phenolphthalein, methyl orange
                (d)             15.30 mL
                (e)             15.30/1000*.11       =        1.68 x 10-3 mol
                (f)             Cleaning agent (HCl) diluted so did not need huge quantities of NaOH in
                burette
                (g)             mol of HCl              =        mol of NaOH in 20 mL
                                                        =        1.68 x 10-3 mol
                (h)             1.68 x 10-3 mol / .02   =        0.0842 mol L-1
                (i)             1.68 x 10-3 * 500/20    =        0.042 mol
                (j)             0.042/.05               =        0.084 mol L-1
                                                        (a) to (f) 1 mk each, rest 1 mk for setting out, 1 for ans.

     13         too many (note: if look at bauxite mining zero)
     14         (a)       0.5*m*v2       =      0.5*675*102 =                  3.38x104J
                                          4
                (b)       100/15*3.38x10 J      =      2.25 x 105J
                                                                               1 mk for setting out, 1 for ans.
     15         (a) Ep      =      mgh         =        1*9.8*3.7     =      36.3 J
                (b) Ek = 0.5*m*v2 =            Ep       =      36.3 m J
                v =         8.52 m s-1                                1 mk each step

     16         (a)




                                                                       50 Ω




                                                                       200 Ω


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                                                                       100 Ω           1 mk each bit
     (b)         P = VI = V2/R              2402/50       =      1150 W
                                                                                          1 mk choice, 1 mk ans
                       -1
     17          (a) R       =              R1-1+  R2-1   R = 2.5 Ω     1 mk for setting out, 1 for ans
                 (b) R = Σ R                = 7.5 Ω                            1 mk
                 (c) I = V/R                = 10/7.5      =      1.33 A 1 mk for setting out, 1 for ans

     18          (a) W = F*s                =      10010*80      =      8.01 * 105 J
                 (b) Ep = mgh               =      1001*9.8*20   =      1.96 * 105 J
                                                                                   1 mk for setting out, 1 for ans
                 (c) Working against friction.




     Part A
     Section 3: Comprehension

1          Systems will fail as it provides power to keep craft working.
2          (a)   H2(g) →            2H+(aq) +2e                 E0 = 0 V
           (b)       O2(g) + 4 H+(aq) + 4e– → 2 H2O()                  E0 = +1.23 V

3          O2(g) + 2H2(g)                   →      2H2O (l)             1.23 V

4          E = Pt = 200 000 * 3600 J = 7.20 x 108 J

5          1.23 V

6          eg        Solar cells            doesn’t use non-renewable energy sources OR non-polluting

7          Frequent stops to recharge, heavy (rmember: cf electric car NOT petrol)

8          Lighter, no moving parts in fuel cell, travels further.

8          Wasted in heat during production, heat during transmission.

9          Non-polluting (waste is water), cyclic process

10         Gas used to burn H2 is air, thus after burning O2 has combined with H2 to form H2O, and N2
           (lots) and CO2 (small) left from air.




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                                                          Part B

Part A
Physical Science 2009 Options                     Solutions

Water & Other Liquids
1     (a)




                                       covalent         1 shape, 1 electrons, 1 name




     (b)        bent




     (c)                                            1 diagram, 1 charges

     (d)        Hydrogen bonds


2    (a)       100 mL soln, add phenolphthalein, goes pink, add 6 mol L-1 H2SO4, until colour goes.
               Add 4 cm3 of acidified (NH4)0.5Mo7O24.4H2) ammonium molybdite + some acidified
               SnCl2. Stand 10 min, intensity of blue gives concn.
     (b)       Eutrophication algae grow, due to fertilizers such as phosphates, die, remove
               O2 from water, fish die.
3    (a)       Energy, cost, water sources (rainfall, aquifers, etc)
     (b)        Water salty (large amounts), use renewable energy, close to users.

4
                          Graph point                     Volume                   Density

           B        At 4°C                        W vol min                W density greatest

           C        From 4°C to 0°C               W v increases            W d decreases

           D        From 0°C to -4°C              I v decreases            I d increases




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5         (a)        Mix with water, see if it foams
          (b)        To precipitate salt from solution.

6
          (a)

    Primary       Grit removal & Sedimentation Water and solids sit and separate: scum & grease to top,
    Solids settle out

    Secondary Sedimentation contact between a suspension of micro-organisms and sewage causes
    formation of sludge.
    Sludge then interacts with bacteria and organics are digested.

    Tertiary – none other than allow water to percolate into soil and use soil filtering to remove nutrients.

    (b) Nutrients and pollutants into soil, then groundwater.

                                                   Sight & Light
    1     (a)        flash from pistol
          (b)        noise of pistol
          (b)        t=(.458+.467+.462+.465)/4 = 0.463 s         1 mk
                     V = s/t = 156/.463                          1 mk
                              =       337 m s-1                  1 mk
    2     (a)        v = λf           346    =     λ8            1 mk
                                      λ      =     43.2 m        1 mk
          (b)        ‘pulled’ by compression, pushed by compression
    3




    4     (a)        (i)   pitch increases
                     (ii)  pitch decreases
          (b)        As moves towards you waves reach you more quickly, so hear more.
                     As moves away from you waves reach you more slowly, so hear less.




2009 examination report: Physical Science                                                                       9
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   5      (a)        6
          (b)        0.30 m
          (c)        450 m s-1
          (d)




          (e)        (i)        Freq as heard by observer.
                     (ii)       mixture of harmonics
                     (iii)      Intensity as heard by observer

   6      (a)




   1 mk for type of mirror, 1 for position of mirror, 2 for rays, 1 for image
       (b)      virtual means that rays don’t cross and we cannot put image on a screen



                                                     Engines & Fuels

   1      (a)        Tem p change of water             Calculate increase in Q = mcΔT of water (use (b))
                     Change in fuel mass               Calculate energy content = Q/m using Q from (a)

          (b)        Measure mass of water             Calculate Q using ΔT from (a) Q = mcΔT

           (c)       retains heat within system (insulates radiation) burns chimney helps transfer (convection)
                     heat to water Doesn’t provide enough updraft 2 mks each

   2      (a)        The octane rating of a fuel mixture is the percentage of iso-octane and normal heptane.

          (b)        75




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   3      Petrol lower compression 7&:1 to 10:1, diesel’s higher to get fuel to ignite
          Ignite petrol with spark. This leads to diesel issues
                   Initial cost higher
                   Diesel dearer than unleaded
                   Engine heavier.
                   Less acceleration
                   Require more precision.
                   Larger batteries to start as need more current to turn over.
                   Acceleration less
                                                                                  1 mk for each point
   4      Use of waste materials, or renewable resources cf oil.                   2 mks
          Less pollutants eg sulphur oxides.                                       2 mks

   5      output 1200 kJ
          100/90 * 1200                                                          1 mk
          = 1330 kJ                                                              1 mk

   6      1 to 2 spark, increase temp, thus pressure increases                    2 mks
          2 to 3 energy stored in working fluid in cylinder converted into work (pushes piston down)
                                                                                  2 mks
          3 to 4 gases cooled as expanded, bottom                                 1 mk
          4 to 1 piston returns to top (compression), uses momentum of motor, expels waste
                                                                 1 mk




                                            End of marking key




2009 examination report: Physical Science                                                               11
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