nss physics at work book 1 solution

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

0

T

=2.32.18

2.37.7

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.

621

6

x

=080

037

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.



2

6 Let T be the temperature measured.

0100

0

T

=35120

3580

T = 52.9 C

Revision exercise 1 Multiple-choice (p. 19) 1 D

2 B

3 B

4 C

2580

25

R

=0100

040

R = 47 units

5 B

6 C

0100

0

T

=1090

1040

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 0100

0

x

=7.36.24

7.30.12

(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

0

T

=2.44.18

2.46.15

(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

=0100

030

(1M)

x = 8.46 cm (1A)

4 (a) TN = TC 100

33 (1M)

= 250 100

33

= 82.5 N (1A)

(b) TN = TC 100

33

=9

5 32FT

100

33 (1M)

= 32FT 60

11 (1A)

5 (a)

(Correct labelled axis) (1A)

(Correct points) (1A)

(A smooth curve passing through all data

points) (1A)

(b) 32 C (1A)



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

0

T

=8.34.16

8.38.8

(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)


(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)


(d) It is less accurate. (1A)

If it is used in a hospital, doctors may not be

able to determine the patient’s condition

correctly and may miss noticing a dangerous

situation instantly. (1A)

1 Heat and Gases Chapter 2 Heat and Internal Energy


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

=10002

1000420

= 210 s =60

210min = 3.5 min

6 B

P =t

Q=

605

2000 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 =t

Q=

6015

1000900

= 1000 W

10 Power of the heater

=t

Q=

6010

1000600

= 1000 W

Time needed =P

Q=

1000

0001100 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.



2

Practice 2.3 (p.47) 1 C

2 B

3 D

4 C = mc = 5 × 480 = 2400 J °C–1

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 C–1.

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

= 15252

60400

= 1200 J kg1 C1

9 D

10 (HKCEE 2002 Paper II Q20)




Conventional (p. 52)

1 P =t

Q (1M)

=

605.1

4200109045.0

(1M)

= 1680 W (1A)

2 c =Tm

Q

(1M)

=5 3

6750

= 450 J kg–1 °C–1 (1A)

C = mc (1M)

= 3 450

= 1350 J °C–1 (1A)



3

3 Let T be the initial temperature of the iron

sphere.

1.2 480 (T – 15) = 3 4200 (15 – 12)

(1M)

T = 80.6 °C (1A)

4 (a) Let T be the final temperature of the water.

2 450 (90 – T) = 3 4200 (T – 10)

(1M)

T = 15.3 °C (1A)

(b) Water has a high specific heat capacity. It

can absorb a large amount of heat from

engines without rising to a high

temperature. (1A)

5 Energy released

= 0.3 (80 – 65) 4200 (1M)

= 18 900 J (1A)

6 (a) The specific heat capacity of pottery is

greater than that of metal. (1A)

As the bowl and the mug have the same

mass, the heat capacity of the mug is

greater than that of the bowl. (1A)

For the same temperature rise, the energy

absorbed by the mug is greater than that

by the bowl. (1A)

To reach thermal equilibrium (same

temperature), more energy is transferred to

the mug than to the bowl. (1A)

(b) The soup in the metal bowl has a higher

final temperature. (1A)

7 (a) (i) Let T be the temperature of the

‘mixture’ before it is heated.

0.8 4200 (20 – T)

= 2 0.08 2400 (T – 2)

(1M)

T = 18.2 C (1A)

The temperature of the ‘mixture’

before heated is 18.2 C.

(ii) Energy needed

= (0.8 4200 + 2 0.08 2400)

(90 18.2)

= 269 000 J (1M)

Power of the stove

=605

000 269

(1M)

= 897 W (1A)

(b) Energy provided by the stove

= 897 60

= 53 820 J (1M)

Let c be the specific heat capacity of the

noodles.

0.5 (90 15) c = 53 820 (1M)

c = 1440 J kg1 C1(1A)

The specific heat capacity of the noodles

is 1440 J kg1 C1.

8 (a) By Q = mcT, (1M)

Energy gained by the water

= 0.45 4200 (35 15)

= 3.8 104 J (1A)

(b) (i) 3.8 104 J (1A)

(ii) 0.12 390 T = 3.8 104 (1M)

T = 812 C (1A)

(iii) 812 + 35 = 847 C (1A)

9 (a) (i) 0.75 hour (45 minutes) (1A)

(ii) The biggest temperature difference is

12 C (at 3:30 pm). (1A)

(b) E = mcT (1M)

= 10 4000 20

= 800 kJ (1A)

The water in the radiator takes 800 kJ to

raise its temperature by 20 C.

(c) The radiator loses energy to the

surroundings and so it needs energy

greater than that in (b). (1A)



4

(d) The student is incorrect. (1A)

From 7:00 pm to 8:00 pm, the temperature

drop of water is 22 C and that of oil is

33 C. Since the specific heat capacity of

water is twice that of oil, (1A)

by E = mcT, water gives out more energy

than oil. (1A)

10 (a) By E = Pt and P = VI, (1M)

E = VIt

= 12 4.2 5 60

= 15 120 J (1A)

(b) (i) By E = mcT, (1M)

15 120 = 0.8 c 19

c = 995 J kg–1 C–1 (1A)

The specific heat capacity of

aluminium is 995 J kg–1 C–1.

(ii) (1) Some energy is lost to the

surroundings. (1A)

(2) Wrap the aluminium block

with cotton wool. (1A)

Physics in articles (p. 54) 1 (a) When the water is stirred, the average

kinetic energy of the water particles

increases. (1A)

(b) By Q = mcT,

T =mc

Q (1M)

42005.0

1050

= 0.5 C

The temperature of the water increases by

0.5 C. (1A)

(c) The statement is correct. (1A)

A liquid with a lower specific heat

capacity will have a larger temperature

change for the same energy transferred to

it. (1A)

As a result, the temperature change can be

measured more accurately. (1A)

2 (a) The body temperature is higher than the

temperature of cold water, (1A)

so heat flows from his/her body to the

water. (1A)

(b) Water has a very high heat capacity. The

temperature of water rises very little after

it has absorbed a large amount of energy

from a human body. (1A)

Therefore, people keep on losing energy at

a high rate in cold water. (1A)

This may cause hypothermia and damage

important organs, causing

unconsciousness or even death. (1A)

(c) A high percentage of the mass of our body

is made up of water. (1A)

Water has a high specific heat capacity, so

the temperature of the body only changes

slowly when the temperature of the

surroundings changes. (1A)

1 Heat and Gases Chapter 3 Change of State


1

3 Change of State Practice 3.1 (p. 70) 1 B

2 A

Energy provided by the heater

= Pt

= 1000 (10 60)

= 600 000 J

Energy for heating up the water to 100 C

= mcT

= 0.5 4200 (100 20)

= 168 000 J

By E = ml,

Maximum amount of water boiled away

=l

E=

610262

000 168000 600

.= 0.191 kg

3 65 C

4 Incorrect

5 Energy required

= mlf + mcT

= 108 3.34 105 + 108 4200 (4 – 0)

= 3.51 1013 J

6 Energy needed

= mcT mlv

= 0.2 4200 (100 10) 0.2 2.26 106

= 5.28 105 J

7 Energy needed to change water at 0 C to

water at 100 C

= mcT = 1 4200 (100 – 0) = 420 000 J

Let m be the amount of 100 C steam needed.

Energy lost by steam

= energy taken up by water

m 2.26 106 = 420 000

m = 0.186 kg

8 A control is needed because ice absorbs

energy from the surroundings and melts at

room temperature.

If the energy absorbed from the surroundings

is ignored, by E = ml, the specific latent heat

of fusion of ice found would be smaller than it

should be.

9 Energy lost by the coke

= energy needed to melt the ice

+ energy needed to raise the temperature of

water from 0 C to T

0.3 5300 (25 T)

= 0.1 3.34 105 + 0.1 4200 T

T = 3.16 C

The final temperature is 3.16 C.

T is higher in reality.

10 Energy released by the juice

= mcT = 0.3 3850 (68 – 15) = 61 215 J

Energy needed to change 1 kg of ice at

0 C to water at 15 C

= mlf + mcT

= 1 3.34 105 + 1 4200 (15 – 0)

= 397 000 J

Minimum amount of ice needed

=000397

21561= 0.154 kg

11 Let c be the specific heat capacity of the

coconut milk.

Energy gained by the ice

= energy lost by the milk

0.17 3.34 105 = 0.2 c (70 – 0)

c = 4060 J kg–1 C–1

12 Energy released by cooling water at

20 C to 14 C

= mcT = 0.3 4200 (20 – 14) = 7560 J



2

Let l be the specific latent heat of fusion of

ice.

Energy absorbed by changing melting ice to

water at 14 C

= ml + mcT

= 0.02 l + 0.02 4200 (14 – 0)

= 0.02 l + 1176

Energy lost by water = energy gained by ice

7560 = 0.02 l + 1176

l = 3.19 105 J kg–1

13 Assume the mass, the temperature and the

specific heat capacity of hot drink are 0.3 kg,

50 C and 4200 J kg1 C1 respectively.

To cool the cup of hot drink, the water/ice

absorbs energy from the hot drink. If all the

0.2-kg ice at 0 C just melts to become water

at 0 C, it will absorb 3.34 105 0.2

= 6.68 104 J of energy. This can cool down

0.3-kg of hot drink by a temperature of

(42003.0

1068.6 4

=) 53.0 C. However, since the

initial temperature of the hot drink is 50 C

only, the drink will become 0 C.

On the other hand, the hot drink cooled by the

0-C water bath must have a final temperature

higher than 0 C. Therefore, ice can cool the

hot drink to a lower temperature in this case.

(One may get a different conclusion if

different assumptions on the mass, the

temperature and the specific heat capacity of

the cup of hot drink are made.)

14 (a) None

(b) c

(c) c

(d) lf

(e) lf

(f) lv, lf, c

(g) lv, c

(h) None

(i) lf

(j) lf

Practice 3.2 (p. 80) 1 A

2 C

3 C

4 When we get out of a swimming pool, water

on our skin absorbs energy from our bodies

and evaporates. Therefore, we lose energy and

feel cold.

If it is windy, the evaporation rate, and

therefore the rate at which our bodies lose

energy, will be higher. We would feel much

cooler.

5 On a humid day, the rate of evaporation is

lower. As a result, less energy is taken away

by evaporation. Therefore, the soup cools

down slower.

6 When vapour meets a cool surface, it will

condense on the surface and release energy.

(a) The glasses are cooler than the

surroundings when the person has just got

out of the car. Therefore, water vapour in

air condenses on them.

(b) The reason is similar to (a). The water

vapour in the steamy bathroom is at a

higher temperature than the glasses. And

the large amount of vapour enhances the

condensation.

7 (a) Assume that the latent heat of vaporization

is solely obtained from his body.

E = mlv

= 0.5 2.26 106

= 1.13 106 J (= 1.13 MJ)



3

The maximum amount of energy removed

is 1.13 106 J.

(b) 1.13 106 J is removed from his body in 1

hour. Therefore,

rate of cooling by sweating

=6060

1013.1 6

= 314 W

8 (a) Energy required

= mlv = 0.2 2.26 106 = 4.52 105 J

(b) By E = mcT,

decrease in temperature

=mc

E=

350050

000 452

= 2.58 C

9 Vaporization.

The refrigerating liquid absorbs energy in the

process of vaporization, so that the food is

cooled.

10 As glass A is half covered by a plastic sheet,

some water vapour that evaporates from the

hot water is trapped in the glass and the air

becomes humid. Therefore, the rate of

evaporation is slower in glass A and so the

water level in glass A drops more slowly than

that of glass B.

Revision exercise 3 Multiple-choice (p. 84) 1 B

2 A

3 D

4 B

5 B

6 B

7 B

8 C

Let m be the mass of steam and km be the

mass of ice.

Energy lost by steam = energy gained by ice

m 2.26 106 + m 4200 (100 50)

= km 3.34 105 + km 4200 (50 0)

k = 4.5

The ratio of the mass of ice to the mass of

steam is 4.5 : 1.

9 A

10 A






Conventional (p. 86) 1 Energy has to be removed from the water

= mcT + mlf (1M)

= 0.2 4200 30 + 0.2 3.34 105

= 92 000 J (= 92 kJ) (1A)

2 When snow melts, it absorbs latent heat of

fusion from its surroundings. (1A)

Therefore, the surrounding air loses energy

(1A)

and the air temperature drops. (1A)

3 (a) Melting point (1A)

(b) Room temperature (1A)

(c) (i) The average KE of the ice molecules

increases in this period. (1A)

(ii) The average PE of the ice molecules

increases in this period. (1A)

4 The water on a wet finger absorbs energy from

the finger and evaporates. The finger would

feel cold and the cooling effect increases in the

wind. (1A)



4

Therefore, from the side of the finger we feel

cooler, the wind direction can be told. (1A)

5 Let T be the final temperature of the ‘mixture’.

Energy lost by water = energy gained by ice

(1M)

0.5 4200 (30 T)

= 0.1 3.34 105 + 0.1 4200 T (1M)

T = 11.7 C (1A)

6 (a) Total energy released by the water

= mcwaterT + mlf + mciceT (2M)

= 3 4200 (30 – 0) + 3 3.34 105

+ 3 2060 [0 – (–5)] (1M)

= 1 410 900 J

= 1.41 MJ (1A)

(b) Effective power of the refrigerator

=t

E (1M)

=6060

0004101

= 392 W (1A)

7 Let T be the final temperature of the ‘mixture’.

Energy lost by water

= energy gained by ice (1M)

1 4200 (20 T)

= 0.5 3.34 105 + 0.5 4200 T (1M)

T = 13.2 C (1A)

Since the final temperature should be between

0 C and 20 C, the result 13.2 C shows that

not all the ice melts. Therefore, the final

temperature of the ‘mixture’ is 0 C. (1A)

8 (a) Temperature is higher on a sunny day.

(1A)

Therefore, particles move faster on

average and they can break free from

liquid more easily. (1A)

(b) On a windy day, particles in the vapour

are blown away. (1A)

This makes fewer particles in the vapour

ready to return to the liquid. As a result,

the rate of evaporation increases. (1A)

(c) Water vapour condenses to droplets on the

cold surfaces. (1A)

Since the temperature is lower than the

freezing point, the droplets solidify to

form frost. (1A)

9 When 0.1 kg of steam at 110 C condenses to

water at 100 C,

energy released

= mcT + mlv

= 0.1 2000 (110 – 100) + 0.1 2.26 106

= 228 000 J (1M)

Power supplied by the cooker

=t

E (1M)

=60

000 228

= 3800 W (1A)

10 (a) Energy needed

= mcT + mlv (1M)

= 0.5 4200 (100 – 25)

+ 0.5 2.26 106 (1M)

= 1 287 500 J

= 1.29 MJ (1A)

(b) Steam reaching the cover condenses to

water droplets, which may drip back to the

wok. (1A)

Thus the energy required is larger than the

result in (a). (1A)



5

11

(Slopes larger than those in the old curve)

(1A)

(Shorter horizontal line) (1A)

(Horizontal lines at the same height) (1A)

12 Energy absorbed by the liquid

= Energy supplied by the heater

= Pt

= 10 3 60

= 1800 J (1M)

Let l be the specific latent heat of vaporization

of the liquid.

E = mlv (1M)

1800 = 0.01 l

l = 180 000

= 1.8 105 J kg–1 (1A)

13 (a) Freezing water tends to warm the

surroundings. (1A)

(b) Since water has high specific heat capacity

and latent heat of fusion, (1A)

it can release a large amount of energy

before it freezes. Therefore, spraying

water on fruit trees can protect fruits from

freezing. (1A)

14 (a) In heating the water, the balance reading

remains unchanged at first. This shows

that the water temperature is below the

boiling point and keeps increasing. (1A)

The balance reading then drops. It shows

that the boiling point is reached. Water

temperature keeps steady. (1A)

(Temperature increases at first.) (1A)

(Curve remains steady afterwards.) (1A)

(b) Since the beaker has more water, the

initial balance reading is larger. (1A)

And by Pt = mcT, it will take a longer

time for the heater to boil the water. (1A)

Hence, the horizontal line of the new

curve is longer and is above the old curve.

After the water boils, the rate of

vaporization of water depends on the

power of the heater, which is unchanged.

(1A)

Hence, the slope of the curve remains the

same.

(Longer horizontal line above the old

curve, same slope) (1A)

0

balance reading / kg

time / s



6

15 (a) (i) When the water particles near the

surface absorb enough energy from

the surroundings, they can escape

into the space above the sea and form

water vapour. (1A)

(ii) The vapour is heated up by the sun

and it rises. (1A)

The temperature is lower in higher

altitude. Therefore, the vapour

condenses back to water droplets.

(1A)

Droplets accumulate and form larger

droplets. The larger droplets then fall

as rain. (1A)

(iii) Condensation of vapour releases heat.

(1A)

Surrounding air is warmed. (1A)

(b) If the temperature is low enough, water

droplets freeze to ice and fall as snow.(1A)

16 (a) (i) The air current brought by the fan

removes water molecules in the air

around the wick. (1A)

Hence, it is easier for the water in the

wick to evaporate. (1A)

(ii) Yes, it is true. (1A)

As relative humidity increases, the

rate of evaporation decreases. (1A)

It is harder for the water to evaporate

from the wick. (1A)

(b) (i) I would feel cooler. (1A)

This is because the water in the wick

absorbs energy from the surroundings

to vaporize. (1A)

(ii) I will feel warmer. (1A)

It would become more humid after

using the humidifier for a long time.

It is harder for sweat on our skin to

evaporate. (1A)

17 (HKCEE 2005 Paper I Q3)



Physics in articles (p. 90) 1 (a) By E = mlv (1M)

lv of water =m

E

=2

10853.4 6

= 2 426 500

= 2.43 106 J kg–1 (1A)

(b) The result is larger than the standard value.

This shows that more energy is needed to

vaporize the water. (1A)

Vapour of sweat can condense on the

clothes of an athlete and drip back to the

skin. (1A)

(c) The more humid the air, the lower the rate

of evaporation. (1A)

As a result, the efficiency of cooling by

sweating is lower. Therefore, marathon

runners feel hotter if the air is humid.(1A)

2 (a) They would evaporates gradually. (1A)

(b) Any two of the following: temperature /

wind speed / humidity. (2A)

(c) Vapour condenses on mirrors when we are

having showers.

Water condenses in a dehumidifier. (2A)


1 Heat and Gases Chapter 4 Transfer Process


1

4 Transfer Process Practice 4.1 (p. 100) 1 C

2 C

A: The fur is not a source of heat. It cannot

raise the temperature of the thermometer.

C: The fur slows down the rate of heat transfer

(from the bulb to the surroundings).

3 C

4

Material ExampleGood

conductor Good

insulatorSolid (metal)

copper

Solid (non-metal)

plastic

Liquid water Gas air

5 Metal is a good conductor. When we touch a

metal railing, it conducts energy away from

our hands quickly. However, wood is a good

insulator. When we touch a wooden railing,

only little energy is conducted away from our

hands. Therefore, a metal railing feels colder

than a wooden railing even if they have the

same temperature.

6 Goose-down in the jacket traps air. Since air is

a poor conductor of heat, energy cannot easily

escape by conduction.

7 In conduction, particles vibrate and transfer

energy when they collide with each other.

Since there is no particle in a vacuum,

conduction does not occur.

8 Since fat is a good insulator, it reduces heat

transfer from the body to the surroundings by

conduction. Hence, fat people feel warmer

than thin people in winter.

9 Since copper is a better conductor of heat than

stainless steel, it conducts heat faster and

avoids the accumulation of heat at a point.

This ensures even heating.

10 (a) Water is a poor conductor of heat. Heat is

conducted slowly from the top of the

boiling tube down to the ice.

(b) We cannot get the same result if we

replace the boiling tube by a metal tube.

Metal is a good conductor of heat. Heat is

conducted along the tube to the ice

quickly.

11 (a) Heat flows from inside to outside.

(b) Polyfoam is a poor conductor of heat. Heat

flows from inside through the polyfoam

container to outside slowly, so the food is

kept warm.

(c) In summer, the temperature outside the

container is higher than the temperature of

the cold food. Therefore, heat flows from

outside to inside. The polyfoam container

keeps the food cool by slowing the flow of

heat.

Practice 4.2 (p. 108) 1 A

2 A

3 B

4 The cover of the container reduces energy loss

by convection.

5 In a convection current, hot water rises and

cold water sinks. If the heating element of an

electric kettle is fixed at the top, water at the

bottom cannot be heated through convection.



2

6 In convection, particles of a fluid carry energy

from the hot region to the cold region and the

particles in the cold region move along the

convection current to the hot region to absorb

energy. Since no particles exist in a vacuum,

no convection current can be formed and

convection does not occur.

7 The warm air around the light bulb rises. The

opening lets the warm air flow out to prevent

overheating.

8 The air trapped in cotton and feathers can not

move around and so it reduces heat loss

through conduction and convection. On the

other hand, the air surrounding a hot pan can

move freely and it can carry the energy away

from the pan through convection.

Practice 4.3 (p. 119) 1 D

2 D

3 B

4 C

5 Heat travels from the sun to the earth by

radiation.

6 Since black paper is a better absorber of

radiation than white paper, snow under the

black paper melts faster than that under the

white paper.

7 It is very cold in the space. The spacesuit is

white in colour to reduce energy loss through

radiation.

If it were dull black, the astronaut would

radiate energy quickly and feel very cold. On

the other hand, the temperature of the side

facing the sun would increase rapidly.

8 Light-coloured surfaces are poor radiators of

heat. The hot air inside the hot air balloon

cools down more slowly. This saves fuel.

9 Thermometer A will have the highest reading.

Since dull black surfaces are good absorbers

of radiation, the blackened foil will absorb the

largest amount of radiation from the sun.

10 For keeping warm, John should wear the

transparent plastic coat on the outside. The

transparent plastic coat lets the radiation from

the sun pass through. The radiation is then

absorbed by the black coat, which is a good

absorber of radiation. The plastic coat also

reduces energy loss from the black coat to

outside by conduction and convection.

Revision exercise 4 Multiple-choice (p. 125) 1 C

2 B

3 D

4 B

5 C

6 A

7 C

8 D





Conventional (p. 126) 1 Black surfaces are good absorbers of radiation

and become hot under sunlight. (1A)

There is a risk of explosion of the fuel in the

tanks under high temperature. (1A)



3

2 Shiny surfaces are poor absorbers and poor

radiators of heat. (1A)

The shiny foil blanket reduces energy loss by

radiation and avoids the runner cooling down

too quickly. (1A)

3 Air is a poor conductor of heat. (1A)

The layer of air reduces the amount of energy

entering the food compartment. (1A)

4 (a) Aluminium (1A)

(b) Since aluminium conducts heat better than

stainless steel does, it conducts heat away

from the hot part faster. (1A)

This can reduce the accumulation of heat

and ensure even heating. (1A)

5 (a) The transparent plate allows sunlight to

pass through. (1A)

Furthermore, it reduces energy loss by the

panel through convection by trapping the

warm air between the plates. (1A)

(b) Since a black surface is a good radiation

absorber, (1A)

a lower plate with a black surface can be

heated up by radiation more quickly. (1A)

The surface should be dull in colour

because a dull surface is a better radiation

absorber than a shiny surface. (1A)

6 (a) Radiation from the sun can pass through

the car windows. (1A)

This warms up the air and other materials

inside the car. The warm air and materials

emit infra-red radiation which cannot

escape through the windows easily. (1A)

Moreover, since all the windows of the car

are closed, the car has no energy loss

through convection. (1A)

Hence, the temperature inside the car rises

drastically.

(b) The shield covering the car windows can

reflect sunlight away. This avoids the air

and materials inside the car being warmed

up. (1A)

7 (a) There is a layer of trapped air in a

double-glazed window. Since air is a

better insulator than glass, (1A)

it reduces heat loss by conduction. (1A)

(b) Since conduction cannot occur in a

vacuum, (1A)

a double-glazed window with a vacuum

between the panes can stop heat loss by

conduction. (1A)

Therefore, it performs better than that with

air. (1A)

8 (Design for preventing heat loss by conduction,

e.g. walls made of a good insulator or double

walls.) (1A)

(Design for preventing heat loss by convection,

e.g. airtight lid.) (1A)

(Design for preventing heat loss by radiation,

e.g. outside walls with light or silver colour.)

(1A)

(Appropriate drawing of the container) (1A)

(Appropriate labels of the components) (1A)



4

9 (a) Radiation (1A)

(b) The specific heat capacity of sea water is

much higher than that of sand. (1A)

For similar amount of energy lost to the

surroundings, the temperature drop of sea

water is mild, while that of sand is huge.

(1A)

(For teacher’s reference:

specific heat capacity of sand

= 835 J kg1 C1

Specific heat capacity of water

= 4200 J kg1 C1)

(c)

(Correct cold and warm regions) (2A)

(Correct movement of air) (1A)

10 (a) Radiation (1A)

(b) The fan speeds up the movement of air.

(1A)

This reduces the temperature difference at

different points. (1A)

Therefore, the food can be cooked more

evenly. (1A)

(c) As a good insulator, the vitreous enamel

reduces energy loss to the outside of the

oven. (1A)

(Or it prevents the outside of the oven

from being too hot to touch.)

(d) Since a shiny surface is a poor absorber of

radiation, (1A)

it reduces energy loss by radiation. (1A)

11 (a) (i) Radiation (1A)

(ii) Conduction (1A)

(b) This creates an environment similar to a

greenhouse for the heater. (1A)

Hence, it raises the temperature of the air

around the heater and allows the water

inside the heater to absorb more energy.

(1A)

(c) This is because black objects are good

absorbers of radiation. (1A)

(d) Copper is more suitable for making pipes.

(1A)

Energy is transferred from the pipes to

water by conduction and copper is a better

conductor than plastic. (1A)

12 (a) The glass plate traps the air inside and

reduces energy loss by convection. (1A)

Also, it absorbs most of the infra-red

radiation emitted by the warm objects

inside and reduces the energy loss by

radiation. (1A)

(b) Amount of energy absorbed by the water

= mcT (1M)

= 0.5 4200 (60 20)

= 84 000 J (1A)

(c) Add reflectors to the inside walls of the

cooker (1A)

to reflect additional sunlight into the

cooker. (1A)

13 (a) A dull surface is better absorber of

radiation than a shiny surface. (1A)

Therefore, the dull surface should face

outwards during baking because it can

absorb more energy and reduce the baking

time. (1A)



5

(b) No, the shinny surface should face

outwards in this case. (1A)

Since a shiny surface is a poor radiator of

heat, this arrangement can reduce energy

loss from the food through radiation. (1A)

(c) Evaporation takes away a large amount of

energy from the food. (1A)

The aluminium foil traps the steam

evaporated from the food. (1A)

This reduces the rate of evaporation and

thus the rate of heat transfer. (1A)

14 The steam of herbal tea rises. Glass plates can

reduce the energy loss due to evaporation(1A)

and convection. (1A)

Cold air from freezers sinks and the food in

freezers can be kept frozen even if they do not

have covers. (1A)

15 In winter, because of the large specific heat

capacity of water, the land loses energy more

quickly than the sea. (1A)

The warm air above the sea rises (1A)

and the cool air blows in from the land to

replace it. This results in winter monsoons.

(1A)

16 (a) Curve B (1A)

(b) Dull black objects are good radiators.(1A)

The beaker wrapped in dull black paper

loses energy through radiation more

quickly. (1A)

Therefore, curve B, which shows a more

rapid drop in temperature, corresponds to

this beaker. (1A)

(c) This is because the temperature difference

between the water and the room

temperature decreases. (1A)

17 (a) Black ‘fuel effect’ lumps burn and release

heat. Air around the lumps is heated. It

expands and rises. (1A)

Hot air leaves the fire at A. (1A)

Cold air flows in from C to replace the hot

air. (1A)

The air forms a convection current flowing

from C to A. (1A)

(b) (i) Radiation (1A)

(ii) The lumps should be dull black in

colour. (1A)

18 (a) (i) Since black bodies absorb radiation

better, (1A)

pipes painted in black heat water at a

higher rate. (1A)

(ii) Copper is a good conductor of heat,

but plastic is not. (1A)

(b) Advantage: It saves energy. (1A)

Disadvantage: It can only be used in sunny

days. (1A)

(c) Water at the top of the tank is warm (1A)

while water at the bottom is cold. (1A)

19 (a) (i)

(Air above the heater rises.) (1A)

(Rest of circulation) (1A)

(ii) Convection (1A)



6

(iii) The heater heats up the air around it.

The warm air expands (1A)

and becomes less dense. (1A)

Therefore, it rises. (1A)

The surrounding cold air moves in to

replace the warm air. (1A)

(b) Since metal foil is shiny, (1A)

it is a poor absorber of radiation which

reduces the energy loss from the heater to

the surroundings. (1A)

(c) Any two of the following or other

reasonable answers: (2A)

Use double-glazed window.

Paint the walls of the room with light

colours.

Add insulating materials to the walls.

Add insulating materials to the ceiling.

Put the heater away from the walls to the

outside.

20 (a) Their house acts as a control set-up. (1A)

(b) (i) The silver foil is poor radiator of

heat. (1A)

This reduces energy loss by

radiation. (1A)

(ii) The cotton wool is a poor conductor.

(1A)

This reduces energy loss by

conduction. (1A)

(iii) After the polythene is taken away,

the air can flow freely into and out

of the house. (1A)

This increases the energy loss by

convection. (1A)

Physics in articles (p. 132) 1 (a) Convection (1A)

and radiation (1A)

(b) We will feel hot first and then get burnt

after a long time. (1A)

This is because heat is transferred to our

hand by convection and radiation. (1A)

(c) Bamboo is a poor conductor of heat while

metal is a good conductor of heat. (1A)

Bamboo conducts less heat to our hand.

and so it is safer. (1A)


1 Heat and Gases Chapter 5 Gases


1

5 Gases Practice 5.1 (p. 150) 1 B

2 The reading on the syringe gives the volume

of air inside the syringe but does not include

that in the rubber tubing. The shorter the

rubber tubing, the smaller the error in

measuring the volume.

3 By Charles’ law,

1

1

T

V=

2

2

T

V

V2 = 21

1 TT

V

= 11727327273

1

= 1.3 m3

Its volume is 1.3 m3 at 117 C.

4 By general gas law,

1

11

T

Vp=

2

22

T

Vp

V2 =2

2

1

11

p

T

T

Vp

=5

35

100.1

273

20273

100.5109.0

= 4.19 10–3 m3

The volume of the balloon is 4.19 103 m3.

5 (a) Any two of the following:

Immerse the whole flask, including the

neck, in water.

Use a very short rubber tubing to connect

the Bourdon gauge and the flask.

Wait until the pressure and the

temperature have become steady before

taking the readings.

Do not allow the flask and the

thermometer to touch the bottom of the

beaker.

(b) By pressure law,

1

1

T

p=

2

2

T

p

p2 = 21

1 TT

p

= 8027340273

10120 3

= 135 000 Pa = 135 kPa

The pressure of the gas is 135 kPa.

6 Let p1 and p2 be the gas pressure in X before

and after the tap is opened respectively.

By the general gas law,

p1 =1

11

V

RTn=

1

15.1V

RT

The temperature of the gas does not change

after the tap is opened, and the pressure in X

and that in Y are the same after the tap is

opened. Therefore,

p2 =2

22

V

RTn=

1

1

2

4.25.1

V

TR=

1

195.1V

RT

Percentage change =1

12

p

pp 100%

=5.1

5.195.1 100% = 30%

Practice 5.2 (p. 164) 1 C

2 C

3 D

4 B

5 (a) The smoke particles move about

constantly along zigzag paths.

(b) They are bombarded by a large amount of

air molecules around them. The bombard-

ments come from all sides but not in equal

numbers. This results in a random motion



2

(c) Slower motion of smoke particles

6 (a) Increases

(b) Increases

7 The number of air particles inside the tyre

increases, so there are more frequent

bombardments on the tyre wall. As a result,

the pressure of the tyre increases.

8 Root-mean-square speed

=AmN

RT3

=

2326 1002.6106.5

2732531.83

= 469 m s1

9 Particles of the perfume vapour move at high

speeds and travel in all directions. This makes

the smell of the perfume spread.

10 Since total KE =2

3nRT,

increase in total KE

= 258031.81002.6

1028.7

2

323

24

= 8290 J

11 (a) The root-mean-square speed

becomes 3 times its original value.

(b) The root-mean-square speed triples.

(c) The root-mean-square speed remains

unchanged.

12 Average force exerted on surface W due to a

molecule with speed v

=change for the interval time

momentumin change

=

v

lmv2

2=

l

mv 2

Pressure exerted on surface W due to this

molecule

=area

force=

2

2

ll

mv

=3

2

l

mv

The pressure p due to N molecules

= 222

213

... Nvvvl

m = 2

3vN

l

m

where 2v is the mean value of v2 of all the

molecules.

Revision exercise 5 Multiple-choice (p. 168) 1 A

2 D

3 D

By crms =AmN

RT3,

the ratio of crms at 80 C to that at 20 C

=

A

A

mN

R

mN

R

)20273(3

)80273(3

= 1.10

4 B

Let v0 the volume of gas at X.

Let p0 the pressure of gas at Z.

By general gas law, T =nR

pV.

Temperature at X, Y, Z can be expresses as:

TX =nR

vp 005 =

nR

vp 005

TY =nR

vp 00 33 =

nR

vp 009

TZ =nR

vp 00 5=

nR

vp 005

ZXY TTT

5 A

By general gas law,

pV = nRT

V = nR 1

T

p



3

The volume of gas is inversely proportional to

T

p, which is the slope of the line connecting

the point to the origin as shown below.

The slope for stage X is greater than that for

stage Y, i.e.X

X

T

p>

Y

Y

T

p, so VY > VX.

6 C




10 (HKALE 2005 Paper II Q21)

Conventional (p. 169) 1 When the light bulb is switched on, the

temperature inside the bulb increases. Since

the volume of gas inside a light bulb is fixed,

the pressure is then increased. (1A)

The bulb will burst if the pressure is too high.

Filling it with a gas at low pressure can

prevent the bulb from bursting. (1A)

2 (a) By general gas law,

pV = nRT

Number of moles of gas

=RT

pV (1M)

= 2732731.8

510100 3

= 201 mol (1A)

(b) By general gas law,

1

11

T

Vp=

2

22

T

Vp (1M)

27327

510100 3

=2737

1080 23

V

V2 = 5.83 m3 (1A)

3 Root-mean-square speed

=AmN

RT3 (1M)

=

2327 1002.61035.3

2733131.83

= 1940 m s1 (1A)

4 While rising to the water surface, n and T of

the air inside his lungs remains constant.

By Boyle’s law,

p1V1 = p2V2 (1M)

2p2 V1 = p2V2

V2 = 2V1

Therefore, the volume of his lungs would

double. (1A)

5 By general gas law,

n =RT

pV

Percentage of air escape

=

1

21n

n 100% (1M)

=

1

11

2

22

1

RT

VpRT

Vp

100%

=

1

21p

p 100% (1M)

=

3

3

10220

101001 100%

= 54.5% (1A)

6 (a) (i) Decreases (1A)

(ii) Remains unchanged (1A)

(iii) Remains unchanged (1A)



4

(b) As the number of air molecules inside the

carton decreases, the number of

bombardments on the walls of the carton

decreases. (1A)

As a result, the pressure inside the carton

decreases (1A)

and the carton collapses due to the larger

pressure outside. (1A)

(c) Heat the sealed carton in a water bath.(1A)

The average kinetic energy of the air

molecules inside the carton increases; thus

the number of bombardments on the wall

of the carton increases. (1A)

As a result, the pressure inside the carton

increases and the carton gradually regain

its original shape due to the smaller

pressure outside. (1A)

7 (a) (i) (1) The piston floats lower. (1A)

(2) The piston floats higher. (1A)

(ii) The ball-bearings in the tube

represent the gas molecules; the

number of cardboard discs, the

voltage applied to the motor and the

height of the piston represent the

pressure, temperature and the volume

of the gas respectively. (1A)

When the voltage is gradually

increased, the piston floats higher.

(1A)

This shows that the gas volume

increases as the temperature increases.

This simulates Charles’ law. (1A)

(b) (i) Wait until the volume and

temperature have become steady

before taking the readings. (1A)

(ii) The open reservoir of oil keeps the

air inside the tube at atmospheric

pressure throughout the experiment.

(1A)

(iii) (1) The length of the air column

remains unchanged no matter

what temperature of the water

bath is. (1A)

(2) According to pressure law, at a

constant volume, the gas

pressure increases as the

temperature increases. (1A)

8 (a) By general gas law, pV = nRT. (1M)

Therefore,

nRT = 2

3

1cNm

Root-mean-square speed 2c

=Nm

nRT3 (1M)

=m

n

NRT

3

=AmN

RT3 (1A)

where m is the mass of a gas molecule and

T is the absolute temperature of the gas.

(b) Total kinetic energy of the molecules

= 2

2

1cmN (1M)

From (a), nRT = 2

3

1cNm

Rearrange the terms,

2

2

1cmN nRT

2

3 (1A)

where n is the number of moles and T is

the absolute temperature of the gas.



5

(c) By the equation in (b),

Total kinetic energy of the molecules

= nRT2

3 (1M)

= 2737031.85.32

3

= 15 000 J (1A)

9 (a) Gas particles gain more kinetic energy and

(1A)

bombard on the piston more frequently

and more violently. (1A)

Therefore, the gas pressure inside the

syringe becomes larger than that outside

(1A)

and so the piston moves outwards.

(b) (i) As the piston moves outwards, the

frequency of bombardment decreases,

(1A)

and so does the pressure inside. (1A)

When the pressure inside and outside

the syringe become the same, the

piston stops. (1A)

(ii) By Charles’ law,

1

1

T

V=

2

2

T

V (1M)

V2 = 21

1 TT

V

= 27310027320

50

= 63.7 cm3 (1A)

(c) (i) Before the syringe is turned vertical,

the pressure inside balances the

pressure outside. After turning

vertical, the weight of the piston is

not balanced. This becomes a net

force and so the piston drops. (1A)

When the piston drops, the volume of

the gas decreases, and the frequency

of bombardment increases, and so

does the pressure inside. (1A)

When the pressure inside balances

both the pressure outside and the

pressure provided by the weight of

the piston, the piston stops. (1A)

(ii) Pressure provided by the weight of

the piston

=A

F=

201.0π

1

= 3180 Pa (1M)

New pressure of carbon dioxide

= 100 103 + 3180

= 103 180 Pa (1A)

By Boyle’s law,

p1V1 = p2V2 (1M)

V2 =2

11

p

Vp

=180 103

7.6310100 3

= 61.7 cm3 (1A)

(iii) By Boyles law, the pressures of

carbon dioxide are 102.1 kPa and

101.1 kPa when its volume are

62.4 cm3 and 63 cm3 respectively.

(Correct labelled axes) (1A)



6

(A smooth curve with correct starting

point and ending point) (1A)

10 (a) If the collisions are not perfectly elastic,

energy will be lost during the collisions

and the speeds of the gas molecules will

become slower and slower. (1A)

We cannot get a definite value for the

change in momentum when a molecule

collides with a wall. (1A)

Therefore, we cannot find the pressure

exerted on the wall by the molecule and

can no longer make the derivation. (1A)

(b) If the gas molecules are not in random

motion, the mean values of the velocities

of the molecules in x, y and z directions

may not be the same. (1A)

Therefore, the average force, hence the

pressure, exerting on each wall may not be

the same. (1A)

In this case, a single value of pressure p is

not well defined. (1A)

(c) If the gas molecules have different mass,

the total pressure exerted on the wall in the

x-direction by all the molecules will

become:

p = 2222

211 ...

1xNNxx vmvmvm

V

= 2xmv

V

N

(instead of 2xvm

V

N) (1A)

Similarly, the total pressure exerted on the

walls in y- and z-directions will become

2ymv

V

N and 2

zmvV

N respectively.

(1A)

As a result we will derive the equation

pV = 2

3

1mcN instead of pV = 2

3

1cNm .

(1A)

11 (HKALE 2001 Paper I Q10)

12 (HKALE 2004 Paper II Q5)

13 (HKALE 2006 Paper I Q5)

Physics in articles (p. 173) (a) The gas pressure inside a ‘vacuum balloon’ is

zero. (1A)

The atmospheric pressure acts on the

‘balloon’ and so the ‘balloon’ collapses.(1A)

(b) The temperature of the air inside a hot air

balloon is higher than that outside. (1A)

Therefore, the kinetic energy of the air

molecules is higher. (1A)

These air molecules move faster and collide

with the wall of the balloon more frequently

and more violently. (1A)

As a result, the air inside the balloon can

provide the same pressure as the

surroundings with less air particles.

(c) By general gas law,

pV = nRT (1M)

90 103 10 = n 8.31 (70 + 273)

n = 316 mol (1A)

There are 316 mol of air particles inside the

balloon.

nss physics at work book 1 solution

Documents