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Physics Form 4 Answers

(d) Can 3 3 Q3 Q = CDq force per unit area hit on the wall
(e) (i) R 2 = 386 000 = 1 400 × (25 – 5) produces gas pressure.
T 27 2 = 28 000 J
= 7.89 × 10 km day Q4 c = Q Q2 According to Charles’ Law, the
13
3
–2
volume of gas increases with
3
(ii) R 2 = J 3 2 = 7.89 × 10 13 mDq temperature. This is because when
T 1 54 000 J
J = 42 890 km = 0.6 kg × 200°C a fixed mass of gas is heated, the
gas molecules acquired energy and
–1
(iii) v = 2pJ = 2p(42 890) = 450 J kg °C –1 increase their kinetic energy. Gas
T 24 × 60 × 60 mcDq molecules move faster and collide
= 3.12 km s Q5 t = P with the wall of the container more
–1
= 2 × 4 200 × 50 = 140 s frequent and this increases the
pressure of the gas. In order to keep
3 000
Chapter the pressure constant, the excess
4 Heat Q6 (a) Q 1 = mcDq pressure of the gas will produce a

= 0.6 × 4 200 × (100 – 27)
= 183 960 J force pushing the piston upward to
increase the volume of the gas.
Checkpoint 4.1 (b) Q 2 = mcDq Q3 The mass and temperature of the
= 0.4 × 900 × (100 – 27)
Q1 (a) Thermal equilibrium is the = 26 280 J gas are kept constant.
process of transferring heat Q4 This is because at 0°C, the gas
between two objects in thermal Checkpoint 4.3 molecules are still in a state of
contact until both are at the motion and therefore the pressure
same final temperature and Q1 The meaning is that to melt or still exist.
there is no net heat transfer freeze 1 kg of ice, the amount heat Q5 P 1 = 1.5 × 10 Pa; V 1 = 50 cm ;
5
3
5
between the two objects. absorbed or released is 3.34 × 10 J. V 2 = 30 cm ; P 2 = ?
3
(b) Hot coffee is cooled when ice Q2 (a) Q – mL = 0.2 × 3.34 × 10 5
cubes are added to it. = 66 800 J P 1 V 1 = P 2 V 2 → P 2 = P 1 V 1
5
Q2 • Net heat flow between P and Q (b) Q – mL = 0.4 × 2.26 × 10 6 = 1.5 × 10 × 50 V 2
is zero. = 904 000 J 30
• Temperature of P and Q is lower Q3 (a) Melting point = 75°C = 2.5 × 10 Pa
5
than 70°C but higher than 20°C. (b) t = 6 minute = 360 s
Q3 (a) Liquid suitable for X is mercury. m = 100 g = 0.1 kg Q6 P 1 = 100 kPa; T 1 = 270 K ;
(b) Melting for ice is 0 C; boiling Q = Pt = 100 × 360 T 2 = 324 K ; P 2 = ?
o
point is 100 C. = 36 000 J P 1 = P 2 → P 2 = 100 × 324
o
270
Q4 100°C → (16 + 9) = 25 cm L = Q = 36 000 T 1 T 2 = 120 kPa
m
0.1
Therefore, 16 × 100°C = 64°C = 3.6 × 10 J kg
–1
5
25 Q7 V 1 = 80 cm ; T 1 = 273 K ;
3
Q5 (a) 24°C Q4 (a) Q = mcDq + mL T 2 = 338 K ; V 2 = ?
(b) 17.2 cm = 1.5 × 4 200 × 30 + 1.5 × 80
(c) 2 cm → 0°C 3.34 × 10 5 V 1 = V 2 → V 2 = 273 × 338
and 16 cm → 68°C = 189 000 + 501 000 T 1 V 2 = 99.0 cm 3
Therefore, 14 cm : 68°C = 690 000 J
For increase of temperature of (b) Q = mL + mcDq
6
20°C, = 0.2 × 2.26 × 10 + 0.2 × 4 200 SPM Practice 4
l = 20 × 14 = 4.12 cm × (100 – 60) Objective Questions
= 452 000 + 33 600

68
= 485 600 J 1. B 2. B 3. C 4. C 5. B
Q5 Q = mcDq(ice) + mL(ice) + 6. C 7. A 8. D 9. A 10. D
Checkpoint 4.2 mcDq(ice) + mL(steam) 11. B 12. B 13. C 14. B 15. B
= 3.2 × 2 100 × 5 + 3.2 × 3.34 ×
Q1 (a) Mass – the larger the mass of 10 + 3.2 × 4 200 × 100 + 3.2 16. C 17. B 18. B 19. A 20. B
5
the object, the larger the heat × 2.26 × 10 6 21. D 22. B 23. C 24. D 25. A
capacity. = 9 678 400 J
(b) Shape – does not affect the Subjective Questions
heat capacity. Checkpoint 4.4
(c) Types of material – different Section A
types of materials have different Q1 Based on the kinetic energy of the 1. (a) (i) Thermal equilibrium
heat capacity. gas, the gas molecules are all in (ii) The change of oil
Q2 (a) Material T: The largest mass fast, random and continuous motion. temperature is greater than
The gas molecules are always in
water.
has the highest heat capacity. collision with each other and with (iii) The amount of heat energy
Q
(b) Specific heat capacity, c = mDq , the wall of the container. When a absorbed by oil and water is
means smaller mDq, the larger gas molecule collides with the wall the same.
specific heat capacity. Material and bounces back, the molecule (iv) The specific heat capacity of
P has the largest specific heat experiences a change in momentum. oil is smaller than water.
capacity due to the smallest mq The rate of change of momentum (v) PT = mcDq
value. from the rate of collisions with the 500 × (2 × 60) = 0.5 × c ×
container wall produces a force. The (97 – 25)
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ANS FOC PHYSICF F4 1P.indd 257 29/01/2020 2:04 PM