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  • 7/29/2019 Chapter4 Enrichment(Answer)

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    JPN Pahang Physics

    Chapter 4: Heat (Teacher) Enrichment Module

    ANSWER

    Section A

    1. C

    2. D

    3. C

    4. C

    5. D

    6. D

    7. B

    8. C

    9. B

    10. B

    11. A

    12. A

    13. C

    14. D

    15. B

    16. A

    17. C

    18. B

    19. C

    20. D

    21. C

    Section B

    1

    (a)

    The quantity of heat energy required to increase the temperature of 1 kg of the substance

    by 1 or 1 K

    1

    (b)(i) The temperature of the soup in the metal pot is higher than the temperature of the soup

    in the clay pot.

    1

    (ii) The specific heat capacity of the metal pot is lower than the specific heat capacity of the

    clay pot.

    1

    (iii) An object with a lower value of specific heat capacity will

    increase temperature faster when heated by same amount of energy.

    The pot that has a lower specific heat capacity requires less heat to increase the

    temperature.

    1

    1

    (iv) The lower the specific heat capacity, the higher the increase in temperature. 1

    (c)(i) The temperature of the soup in the metal pot is lower than the temperature of the soup

    in the clay pot after 5 minutes move away from the burner.

    1

    (ii) An object with a lower value of specific heat capacity cools faster due to its

    lower amount of heat stored.

    1

    TOTAL 8

    2

    (a)

    There is no net flow of heat between two objects and they have same temperature 1

    (b) 45 0C 1

    (c) The heat from block P transfer to the water

    Until the rate of heat transfer between P and water become equal

    1

    1

    (d) E = mc

    = 0.3 x 900 x 55

    = 14 850 J

    1

    1

    TOTAL 63

    (a)

    Quantity of heat required to change 1 kg of a substance from solid to liquid without any

    change in temperature

    1

    (b) i) Cup B 1

    ii) The ice is exposed to the surroundings 1

    The ice absorb heat from the surroundings 1

    (c) Heat released = 0.2 x 4500 x (75.5 - 25) 1

    45450 J 1

    (d) i) mi x 3.36 x 105 + mi x 4200 (25 0) 1

    mi x 3.36 x 105 + mi x 4200 (25 0) = 45450 1

    0.103 kg 1

    ii) No heat loss from the coffee to surroundings 1

    TOTAL 10

    1

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    JPN Pahang Physics

    Chapter 4: Heat (Teacher) Enrichment Module

    4 (a)

    (i)

    (ii)

    h2greater // vice versa

    Temperature in Diagram 4.1 is lesser // vice versa

    1

    1

    (b) When the temperature increase, the hieght also increase

    When the temperature increase, the distance between molecules also increase

    When the distance between molecules increase, the volume also increase

    1

    1

    1(c) Charles Law 1

    (d) h1 increases 1

    (e) When force//pressure applied is decreases, the volume is increases at constant

    temperature

    1

    TOTAL 10

    Section C

    Question 1

    Part Mark Scheme Notes

    (a) 1 Degree of hotness of a substance(b)

    1

    2

    3

    4

    5

    Menyatakan perubahan suhu dengan tenaga

    Pada keadaan pepejal suhu bertambah jika tenaga

    bertambah

    Pada keadaan cecair, suhu bertamabh jika tenaga

    bertambah

    Menyatakan hubungan antara suhu takat lebur

    dengan tenaga dan penjelasan haba pendam

    Suhu tak berubah dengan pertambahan tenaga pada

    takat lebur

    Tenaga haba digunakan untuk meleburkan

    pepejal/menukarkan pepejal kepada cecair/ Tenaga

    digunakan untuk meleraikan ikatan molkeul pepejal

    kepada cecair

    Tenaga haba ini dinamakan haba pendam pelakuran

    (c)

    1

    2

    3

    4

    Menyatakan kaedah yang sesuai

    Masukkan ais ke dalam air kopi

    Kacau campuran ais dan air kopi

    Menjelaskan kepentingan kaedah

    Menyerap haba dari air kopi

    Haba dari air kopi dipindahkan lebih cepat dengan

    mengacau

    Terima lain-lain

    kaedah yang sesuai

    (d)

    1

    2

    3

    4

    5

    Menyatakan binaan paip dan penjelasankepentingannya

    Paip dibina berbentuk gelung mengelilingi kotak

    pembeku

    Supaya cecair penyejuk dapat menyerap haba lebih

    besar

    Menyatakan jenis bahan paip yang digunakan dan

    penjelasan pemilihan jenis bahan.

    Dibuat dari bahan yang mempunyai muatan haba

    tentu rendah

    Supaya dapat menyerap haba lebih cepat /pengalir

    haba yang baik

    Menyatakan jenis cecair yang sesuai danpenjelasan

    Terima semuacadangan ubahsuaian

    yang sesuai

    2

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    JPN Pahang Physics

    Chapter 4: Heat (Teacher) Enrichment Module

    6

    7 &8

    9

    10

    Cecair Freon

    Ia mudah mengembang menjadi wap pada suhu dan

    tekanan rendah

    Menyatakan ciri-ciri Keselamatan dan penjelasan

    Kotak hendaklah dibalut dengan bahan penebat

    Untuk mengelakkan renjatan jika berlaku kebocoran

    arus

    Pasangkan thermostat

    Supaya suhu dalam peti dikawal

    Menyatakan kedududukkan kotak dingin beku

    Kotak pembeku diletakkan di bahagian atas peti

    sejuk

    Menyatakan perolakan udara dalam peti sejuk

    Udara sejuk di bahagian atas akan turun ke bawah

    /pergerakan udara panas dan sejuk

    Jum 20

    2. (a) (i) Thermal equilibrium states that the net rate of heat flow from one medium to

    another medium and vice versa is zero// temperature for both medium is the same.

    1

    (ii) - Place the thermometer under the tongue

    - Heat transfer from the body to the mercury

    - The temperature rises // length of mercury increases

    - When the equilibrium state is achieves take the reading of thermometer.

    1

    1

    1

    1

    (b) (i) 100 C 1

    (ii) Heat supplied by metal block = heat gained by the water //m 1 c1 1 = m 2 c2 2

    0.25 x 900 x (100 ) = 0.5 x 4200 ( -29)

    = 27.74 C

    1

    1

    1

    (iii) No heat loss to the surrounding 1

    (c) Aspect Explanation

    Large numbers of fin blade - Increase surface area // release

    heat quickly // engine cools

    quickly

    Big size of fan Can suck more air // more air

    can be blow to the engineHigh boiling point Not easily to vaporize // the

    volume of liquid reduce

    slowly // takes a longer time to

    boil

    High specific heat capacity Takes a longer time to become

    hot // the rise in temperature is

    slow

    S is chosen Large numbers of fin

    blade,big size of fan, high

    boiling point, high specific

    heat capacity

    10Total 20

    3

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    JPN Pahang Physics

    Chapter 4: Heat (Teacher) Enrichment Module

    Section D

    Question 1 Marks Answer

    (a)(i) 1 V is directly proportional to T // V T

    (ii) 1

    1

    Show on graph with appropriate vertical and/or horizontal linecorresponding to 200 K.

    37 mm3

    (b)(i) 3 1. Drawing the gradient triangle

    Triangle size a minumum of 4 x 3

    2. Substitution

    (values from students triangle)

    3. Jawapan dengan unit betul

    0.1833 mm3 K- 1 ( Accept answers from 2 to 4 decimal places )

    (ii) 3 1. Explaining your working

    V = gradient x 3752. Substitution

    V = (0.1833) x 375

    3. Jawapan dengan unit betul

    68.74 mm3

    (c) 1

    1

    1

    Show on graph with appropriate vertical and/or horizontal line

    corresponding to 55 mm3

    300 K

    27 oC

    JUMLAH 12

    Question

    Number

    Answer Full

    Mark

    2 (a) The volume of the snack pack influenced by / effected by / depends on the

    atmospheric pressure of the surrounding. 1

    (b) For a fixed mass and at constant temperature of a gas, its volume

    will increase when the pressure decreases.

    or

    The volume of a gas increases when the pressure decreases.

    1

    (c)(i)

    To investigate the relationship between the volume and the pressure

    of a gas at fixed mass and temperature.

    1

    (ii) Manipulated variable :Volume of a gas, V

    Responding variable :Gas pressure,P

    Constant variable :Gas temperature,T / mass of a gas,m

    1

    1

    (iii)

    Bourdon gauge, a syringe with volume scale and rubber tube1

    4

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    JPN Pahang Physics

    Chapter 4: Heat (Teacher) Enrichment Module

    (iv)

    1

    (v)

    1. The piston of the syringe is pressed down slowly until the volumeof air inside is 120 cm3.

    2. The pressure of the air in the syringe is read from the Bourdon

    gauge.

    3. The experiment is repeated with volume of air inside the syringe

    at 100 cm3, 80 cm3 , 60 cm3 , 40 cm3 and 20 cm3.

    1

    1

    1

    (vi)

    Volume , V / cm3 Pressure , P /Pa

    120

    100

    80

    60

    40

    1

    (vii)

    1

    TOTAL

    12

    5

    Volume, V/cm3

    Pressure, P/Pa