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OUP NSS PHYSICS AT WORK Book 1 solution(Heat and Gases)PHY_BK_ANS_1.pdf25Poxford
Text of NSS PHYSICS AT WORK Book 1 solution
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1 Heat and Gases Chapter 1 Temperature and Thermometers
New Senior Secondary Physics at Work Oxford University Press 2009
1
1 Temperature and Thermometers Practice 1.1 (p. 10) 1 B 2 D 3 A 4 Temperature is a measure of the degree of
hotness of an object. 5 (a) On the Celsius temperature scale, the
lower fixed point is the ice point (0 C) and the upper fixed point is the steam point (100 C).
(b) We can reproduce the lower and upper fixed points by using pure melting ice and pure boiling water at normal atmospheric pressure respectively.
6 (II), (IV), (V), (I), (III) 7 Let T be the temperature when the thread is
7.7 cm long.
0100
0T =
2.32.182.37.7
T = 30 C 8
The length of the mercury column at 100 C is
25 cm.
9 Let x be the column length for 37 C.
6216
x =
080037
x = 12.9 cm 10 According to the kinetic theory, all matter is
made up of particles. For solids, the particles are held in position by strong forces and so they have fixed shapes.
For liquids and gases, the particles are held by weaker forces and can move from one place to another. Therefore, they do not have fixed shapes.
Practice 1.2 (p. 16) 1 D 2 C 3 A 4 (a) Thermistor thermometer/ liquid-in-glass
thermometer (b) Resistance thermometer/ alcohol-in-glass
thermometer (c) Resistance thermometer (d) Liquid-in-glass thermometer/ infra-red
thermometer/ thermistor thermometer/ liquid crystal thermometer
5 (a) The curvature of the bimetallic strip. (b) It consists of a bimetallic strip which is
made up of two strips of different metals. The metals expand at different rates as they are heated. The different expansions of strips make the bimetallic strip bend one way. As a result, a particular curvature of the bimetallic strip represents a particular temperature.
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1 Heat and Gases Chapter 1 Temperature and Thermometers
New Senior Secondary Physics at Work Oxford University Press 2009
2
6 Let T be the temperature measured.
0100
0T =
351203580
T = 52.9 C Revision exercise 1 Multiple-choice (p. 19) 1 D 2 B 3 B 4 C
258025
R =
0100040
R = 47 units 5 B 6 C
0100
0T =
10901040
T = 37.5 C 7 A 8 A Conventional (p. 20) 1 Choose the ice point and the steam point as the
lower fixed point and the upper fixed point respectively. (1A)
Then divide the range between these fixed points into 100 equal divisions. (1A)
Each division is 1 C. (1A) The lower fixed point is taken as 0 C and the
upper fixed point is taken as 100 C. (1A) 2
01000x =
7.36.247.30.12
(1M)
x = 39.7 C (1A) 3 (a) Let T be the temperature when the length
of the alcohol column is 15.6 cm.
0100
0T =
2.44.182.46.15
(1M)
T = 80.3 C (1A)
(b) Let x be the length of the alcohol column at 30 C.
2.44.18
2.4
x =0100030
(1M)
x = 8.46 cm (1A)
4 (a) TN = TC 10033 (1M)
= 250 10033
= 82.5 N (1A) (b) TN = TC 100
33
=95 32FT 100
33 (1M)
= 32FT 6011 (1A)
5 (a)
(Correct labelled axis) (1A) (Correct points) (1A) (A smooth curve passing through all data
points) (1A) (b) 32 C (1A)
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1 Heat and Gases Chapter 1 Temperature and Thermometers
New Senior Secondary Physics at Work Oxford University Press 2009
3
6 (a) A rotary thermometer measures temperature by measuring the curvature of the bimetallic strip. (1A)
The bimetallic strip consists of two metal strips which expand at different rates when heated to cause a change in curvature of the strip. (1A)
If a strip with only one kind of metal is used, it would only expand but not bend when heated. (1A)
(b) Zinc (1A) Figure a shows that zinc expands more
when heated. Therefore, zinc corresponds to metal A which expands more as shown in Figure b. (1A)
7 (a) Water freezes at temperatures below the ice point and vaporizes at temperatures above the steam point. (1A)
The working range of a water-in-glass thermometer is much narrower than that of a mercury-in-glass thermometer. (1A)
(Or other reasonable answers) (b) Let T be the temperature measured when
the length of the column is 8.8 cm.
0100
0T =
8.34.168.38.8
(1M)
T = 39.7 C (1A) (c) Mercury is toxic. (1A) 8 (a) In a certain range of temperature, the
volume of mercury is proportional to the temperature. (1A)
(Or other reasonable answers) (b) Volume increased = 0.0748 0.0735 = 0.0013 cm3 (1M)
Length increased =area sectional cross
increased volume
=01.001.0
0013.0
= 13 cm (1M) Length of the mercury column = 13 + 3.6 = 16.6 cm (1A) 9 (a) To measure temperature (b) Any two from: (2A) Clinical thermometer has a smaller range. Clinical thermometer is more
accurate/reads to more (significant) figures/decimal places/more sensitive/has a narrower column.
Clinical thermometer can maintain reading/temperature.
Clinical thermometer has a kink/constriction/button to reset
Physics in articles (p. 22) (a) Energy of infra-red radiation emitted (1A) (b) It takes less time to obtain the results. (1A) (c) Doing experiment in laboratory (1A) (Or other reasonable answers) (d) It is less accurate. (1A) If it is used in a hospital, doctors may not be
able to determine the patients condition correctly and may miss noticing a dangerous situation instantly. (1A)
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1 Heat and Gases Chapter 2 Heat and Internal Energy
New Senior Secondary Physics at Work Oxford University Press 2009
1
2 Heat and Internal Energy Practice 2.1 (p. 26) 1 D 2 A 3 D 4 C 5 The internal energy of a body is the sum of the
kinetic energy and potential of all its particles. It is the total energy stored in the body.
Temperature is a measure of the degree of hotness of an object. When it increases, the kinetic energy, and hence the internal energy, of a body increases.
6 This statement is incorrect. Temperature accounts for the average kinetic energy of the particles in an object, while internal energy is the sum of the kinetic and potential energy of all the particles in an object. A drop of hot water has a higher temperature than water in an ocean, but the latter has more internal energy than the former since it contains much more particles.
Practice 2.2 (p.31) 1 C 2 B 3 A 4 D 5 A Time needed
=100021000420
= 210 s =60210 min = 3.5 min
6 B
P =tQ =
6052000 30
= 200 W
7 Heat is the energy transferred from one body to another as a result of a temperature difference, while internal energy is the energy stored in a body.
8 Energy transferred = Pt = 5 1000 30 60 = 9 000 000 J (= 9 MJ)
9 Power =tQ =
60151000900 = 1000 W
10 Power of the heater
=tQ =
60101000600 = 1000 W
Time needed =PQ =
10000001100 1 = 1100 s
11 The statement is incorrect. Heat always flows from a body with higher temperature to a body with lower temperature. However, an object having more internal energy does not mean that it has a higher temperature.
12 (a) B. It has a higher power and transfers more
energy to the water in a fixed time. (b) By E = Pt, for electric kettle A, E = 1500 5 60 = 450 000 J = 450 kJ for electric kettle B, E = 2000 5 60 = 600 000 J = 600 kJ (c) Boiling the same amount of water requires
the same amount of energy. Therefore, the kettles cost the same.
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1 Heat and Gases Chapter 2 Heat and Internal Energy
New Senior Secondary Physics at Work Oxford University Press 2009
2
Practice 2.3 (p.47) 1 C 2 B 3 D 4 C = mc = 5 480 = 2400 J C1 5 Copper has a higher temperature rise than
water. 6 Let T be the temperature of the soup after
5 minutes. By E = Pt = mcT, 200 5 60 = 0.5 3500 (T 20) T = 54.3 C 7 Let c be the specific heat capacity of the metal
block. Energy lost by the metal block = energy gained by the water bath m1c1T1 = m2c2T2 3 c (100 31.7) = 5 4200 (31.7 27) c = 482 J kg1 C1 The heat capacity of the metal block is
482 3 = 1450 J C1. 8 Let T be the final temperature of the mixture. Energy lost by the 80 C water = energy lost by the 30 C water m1c1T1 = m2c2T2 2 4200 (80 T) = 5 4200 (T 30) T = 44.3 C 9 Since water has a very high specific heat
capacity, it can absorb a lot of energy with only a small temperature rise. Hence water is suitable to be used as a coolant in motor cars and air-conditioners.
Revision exercise 2 Multiple-choice (p. 50) 1 B 2 A
3 B Let m be the mass of the water. m 4200 (35 20) = 2 480 (100 35) m = 0.990 kg 4 D 5 B 6 C 7 B 8 C Specific heat capacity of the liquid
=change etemperaturmass
nsferredenergy tra
= 1525260400
= 1200 J kg1 C1 9 D 10 (HKCEE 2002 Paper II Q20) 11 (HKCEE 2002 Paper II Q21) 12 (HKCEE 2007 Paper II Q10) 13 (HKCEE 2005 Paper II Q27) Conventional (p. 52) 1 P =
tQ
(1M)
= 605.1
4200109045.0
(1M) = 1680 W (1A)
2 c =Tm
Q (1M)
=5 3
6750
= 450 J kg1 C1 (1A) C = mc (1M) = 3 450 = 1350 J C1 (1A)
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1 Heat and Gases Chapter 2 Heat and Internal Energy
New Senior Secondary Physics at Work Oxford University Press 2009
3
3 Let T be the initial temperature of the iron sphere.
1.2 480 (T 15) = 3 4200 (15 12) (1M)
Mastering Nss Physics Book 3 Solution
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