Buku Teks Fizik Tingkatan 4/physics-form-4-chapter-4.pdf

Heat is closely related to human life.
Topics in this theme discuss concepts
and laws related to heat energy. We
will investigate the aspects of changes
in phases of matter, especially changes
in the properties of gas. Th ree gas
laws, Boyle’s Law, Charles’ Law and
Gay-Lussac’s (pressure) Law, will also
be introduced.

Heat

117117
Let’s StudyLet’s Study

4.14.1  Thermal Equilibrium  Thermal Equilibrium
4.24.2  Specifi c Heat Capacity  Specifi c Heat Capacity
4.34.3  Specifi c Latent Heat  Specifi c Latent Heat
4.44.4  Gas Laws  Gas Laws

Why is water suitable to be used Why is water suitable to be used
as a cooling agent?as a cooling agent?

What is the importance of specifi c What is the importance of specifi c
heat capacity of a substance?heat capacity of a substance?

What infl uences the behaviour of What infl uences the behaviour of
gas molecules?gas molecules?

HeatHeat4

118118
Kitchen is where a lot of the concepts related to heat
energy can be applied. When we heat up water in a kettle, the
rate of increase in water temperature depends on the quantity
of water heated. When the water boils, its temperature will
no longer increase. When the same quantity of oil and water
are heated separately at the same time, oil will be hotter fi rst.
All these examples involve the relationship and interaction
between physical properties of matter such as temperature,
pressure, volume and heat. Applications of the concept of heat
have greatly helped our daily life.

Video on application of physics
concepts in the kitchen

http://bt.sasbadi.com/p4119a

Learning Standards and Learning Standards and
List of FormulaeList of Formulae

PageInformation

119119
Thermal Equilibrium4.1

Observe Photograph 4.1. When a cold metal spoon
is put into a cup of hot coff ee, the spoon and the
coff ee are said to be in thermal contact because
heat energy can be transferred between the
two bodies. How can the metal spoon cool
down the hot coff ee? What is the fi nal
condition of the spoon and the coff ee?

Photograph 4.1 A cold metal spoon in a cup of
hot coffeeActivity 4.1

Aim: To show thermal equilibrium between two bodies in thermal contact

Apparatus: Two retort stands, two thermometers, 250 ml beaker labelled A, 50 ml beaker
labelled B, measuring cylinder and stopwatch

Materials: 50°C hot water, tap water and tissue paper

Instructions:

1. Wrap beaker A with tissue and fi ll it with 150 ml of tap water.

2. Fill 40 ml of 50°C hot water into beaker B.

3. Place beaker B into beaker A. Then, place thermometer A and thermometer B into

beaker A and beaker B respectively as shown in Figure 4.1.

Retort stand

Tissue paper

Hot water

Thermometer A Thermometer B

Tap water

Beaker A  Beaker B

Figure 4.1
4. Record the readings of thermometer A and thermometer B every 30 s until the readings

of both thermometers are the same. (This activity can normally be carried out in

fi ve minutes)

120120 4.1.14.1.1
Results:

Table 4.1

Time, t / s
Temperature of

thermometer A / °C
Temperature of

thermometer B / °C

0

30.0

60.0

Discussion:

1. Why is beaker A wrapped with tissue paper?
2. Describe the changes in temperature of the hot water and tap water.
3. What causes the changes in temperature?

Hot object, P is in thermal contact with cold
object, Q. Th e rate of heat transfer from P to
Q is higher than the rate of heat transfer from
Q to P.

Th ere is a net transfer of heat from P to Q.
Th erefore, temperature of P drops and
temperature of Q rises.

Temperature of P and temperature of Q become
the same. Th e rate of heat transfer from P to
Q is the same as the rate of heat transfer from
Q to P. Net heat transfer between P and Q
becomes zero. P and Q are in a state of thermal
equilibrium.

When two objects are in thermal contact, the temperature of the hot object will drop
while the temperature of the cold object will rise until the temperature of both objects become
the same. Net heat transfer between the two objects becomes zero. Both objects are said to be in
thermal equilibrium. Figure 4.2 explains the fl ow of heat between two objects in thermal contact
until thermal equilibrium is reached.

Figure 4.2 Flow of heat energy and thermal equilibrium

0

20

40

60

80

100

0

20

40

60

80

100

Heat

Heat
P Q

0

20

40

60

80

100

0

20

40

60

80

100

Heat

Heat
P Q

0

20

40

60

80

100

0

20

40

60

80

100

Heat

Heat
P Q

121121

Chapter 4 Heat

4.1.14.1.1
Heating ObjectHeating Object Cooling ObjectCooling Object

Thermal Equilibrium in Daily Life
Th ermal equilibrium causes two objects in thermal contact to reach the same temperature.
Figure 4.3 shows examples of thermal equilibrium in daily life.

Hot air in oven is in
thermal contact with
cake batter. Heat from
the hot air fl ows to the
cake batter. Th is causes
the cake batter to be
heated until it is baked.

Clinical thermometer is
placed under the tongue
of a patient. Heat from
the patient’s body fl ows to
the thermometer until the
temperature of the patient
and the thermometer
become the same. Th e
body temperature of the
patient can be determined
because thermal
equilibrium occurs.

When food is kept in the
refrigerator, heat from
the food fl ows to the air
in the refrigerator until
thermal equilibrium occurs.
Temperature of the food drops
and the food stays fresh for a
longer period.

Drinks are cooled
by adding several ice
cubes. Ice cubes absorb
heat from the drink
and melt. Melted ice
cubes absorb heat from
the drink until thermal
equilibrium is achieved.

Figure 4.3 Th ermal equilibrium in daily life

Activity 4.2
Aim: To discuss situations and applications of thermal equilibrium in daily life
Instructions:

1. Carry out this activity in groups.
2. Gather information on situations and other applications of thermal equilibrium in daily life.

The information can be obtained from reading resources or websites.
3. Discuss the fl ow of heat energy until thermal equilibrium is achieved.
4. Draw a mind map based on your fi ndings.

ISS ICS

122122 4.1.14.1.1
SCIENCE, TECHNOLOGY
SOCIETYand

ateway to

To Calibrate a Liquid-in-glass Thermometer Using Two Fixed Points

This thermometer does not have
a clear scale. We need another
thermometer.

This thermometer can still be
used. We just need to calibrate
the thermometer.

A thermometer that does not have a scale can be calibrated
using two fi xed temperature points. Two fi xed points used for
distilled water are melting point of ice, 0°C and boiling point of
water, 100°C.

Activity 4.3

Cooking thermometer is used
to measure the temperature
of food during and after food
preparation. Poor control of time
and temperature can cause food
poisoning. As such, periodic
calibration of the thermometer is
very important.

Aim: To calibrate a liquid-in-glass thermometer using boiling point of distilled water and
melting point of ice

Apparatus: Thermometer, ruler, 250 ml beaker, immersion heater, power supply and retort
stand

Materials: Ice, distilled water and masking tape

Instructions:

1.  Cover the scale of the thermometer with masking tape so that the scale cannot be seen.

2.  Prepare two beakers. Fill beaker A with ice and a small amount of distilled water. Fill
beaker B with distilled water and put in an immersion heater.

3.  Put a thermometer into beaker A. Wait until there is no more change in the level of liquid
column. Then, mark the level of liquid column on the stem of the thermometer. Label this
level as 0ºC (Figure 4.4).

4. Remove the thermometer from beaker A and switch on the immersion heater in beaker B.

5. When the distilled water in beaker B is boiling, put the thermometer into beaker B.
Wait until there is no more change in the level of liquid column. Then, mark the level
of liquid column on the stem of the thermometer. Label this level as 100°C (Figure 4.4).
Switch off the immersion heater.

F ile
The process of calibrating uses
the thermometric property of
liquids in glass. Thermometric
property means a physical
property which can be measured
(such as length of column of
liquid) which changes with
temperature.

123123

Chapter 4 Heat

4.1.24.1.2
Formative Practice 4.14.1

1. State what happens to two objects in thermal equilibrium.

2. Is our body in thermal equilibrium with the environment? Explain
your answer.

3. Aisyah uses an uncalibrated laboratory thermometer to determine the
temperature of a liquid, θ°C. She fi nds that the length of the liquid
column when the thermometer is put into the liquid is as shown in
Figure 4.6. Calculate the temperature of the liquid, θ°C.

Calibration is a process of making a scale of reading on a thermometer. 0°C is the fi xed
lower limit and 100°C is the fi xed upper limit. Th e length of liquid column between the fi xed
lower limit and the fi xed upper limit is divided into 100 equal divisions. Th e thermometer is
then calibrated and can be used to measure temperature between 0°C and 100°C.

Figure 4.4 Figure 4.5

6. Measure the length from the 0°C mark to the 100°C mark as L100 (Figure 4.5).
7. Prepare beaker C and fi ll it with tap water.
8. Put the calibrated thermometer into beaker C. Wait until there is no more change in the

level of the liquid column. Then, mark the level of the liquid column on the stem of the
thermometer.  Label this level as θ°C.

9. Measure the length from the 0°C mark to the θ°C mark as Lθ.

10. Calculate the temperature of tap water using the formula, θ  =
Lθ

L100

× 100°C

Discussion:
1. The bulb of the thermometer should not touch the base or side wall of the beaker while

taking measurement. Explain.
2. Why should you wait until there is no more change in the level of the liquid column

before making a mark on the stem of the thermometer?

Ice +
distilled water

Beaker A Beaker B

Thermometer  100°C

0°C

To power
supply

Immersion
heater
Boiling distilled
water

Retort
stand

100°C

4 c
m

12
 cm

θ °C

0°C

100°C

L100

0°C

Figure 4.6

124124124124 4.1.24.1.2
Chapter 4 Heat

Specific Heat Capacity4.2

Figure 4.7 Diff erence in temperature of sand and sea water at the beach

Figure 4.7 shows two diff erent situations. Th e sun heats up the sand and sea water at the same
period of time. However, sand gets hot quickly and sea water gets hot slowly.
 Th is can be explained based on the concept of heat capacity. Diff erent objects have diff erent
heat capacity. Sand has a low heat capacity and gets hot quickly while sea water has a high heat
capacity and gets hot slowly.

Heat capacity, C of an object is the quantity of heat needed to raise the temperature of
the object by 1°C.

C = Q
Δθ , that is Q = quantity of heat supplied

Δθ  = change in temperature

Unit for C = J °C–1

When 100 J of heat is supplied to objects X and Y, object X experiences a rise in temperature
of 1°C and object Y 2°C. What are the respective heat capacity of objects X and Y?

Heat capacity for object X, CX = 100 J
1°C

= 100 J °C–1

Heat capacity for object Y, CY = 100 J
2°C

= 50 J °C–1

Object X has a higher heat capacity than object Y.
Th erefore, the increase in temperature of object X is less than object Y.

Ouch, this sand
is hot!  This sea water is cold.

125125

Chapter 4 Heat

4.2.14.2.1
Figure 4.8 Daily situations which involve heat capacityFigure 4.8 Daily situations which involve heat capacity

Daily
Situations

Involving Heat
Capacity

Heat capacity of an object increases when the mass of the object increases. For example,
the water in a full kettle takes a longer time to boil compared to the water in a half-fi lled kettle.
Th is shows that water of bigger mass has a higher heat capacity compared to water of smaller mass.
Figure 4.8 shows several daily situations involving heat capacity.

At noon, there is a signifi cant diff erence in
temperature between cement court and grass.

Aft er being left  to cool for some time, the soup
in a large bowl is hotter compared to the same
soup in a small bowl.

Th e dashboard of a car has a lower
heat capacity compared to the cushion.
Absorption of heat energy from the Sun
causes the dashboard to experience a
higher rise in temperature compared to
the cushion.

Dashboard

Cushion

126126126126 4.2.14.2.1
Specifi c Heat Capacity of Substance
Figure 4.9 shows a material engineer tries to choose a suitable metal
as building material. He needs a material that does not heat up
easily. Since the heat capacity of a material diff ers with its mass, he
needs to make his choice based on specifi c heat capacity instead
– which means he has to choose the material based on the heat
capacity of every 1 kg of each material.

Figure 4.9 A material engineer is comparing specifi c heat capacity between diff erent metals

Specifi c heat capacity, c of a substance is the quantity of heat needed to raise
the temperature of 1 kg mass of the substance by 1°C.

c = Q
mΔθ , where  Q = quantity of heat supplied (J)

m = mass (kg)
  Δθ  = change of temperature (°C or K)

Unit for c  = J kg–1 °C–1 or J kg–1 K–1

Quantity of heat, Q that is absorbed or released by an object can be
determined using the formula Q = mcΔθ .

For example, the specifi c heat capacity of the metal
aluminium is 900 J kg–1 °C–1. Th is means 1 kg of aluminium
requires 900 J of heat to raise its temperature by 1°C.

Which metal gets
hot slowly?

127

INFOSMARTSMART

Heat capacity, C = Q
Δθ

Specifi c heat capacity, c = Q
mΔθ

Specifi c Heat Capacity

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127127

Chapter 4 Heat

4.2.24.2.2
Based on Table 4.2, the specifi c heat
capacity for water is higher compared
to metals such as aluminium.

How can these values
be determined?

Every substance has its own value of specifi c heat capacity. Table 4.2 shows examples of
substances and their specifi c heat capacity.

Table 4.2 Specifi c heat capacity of diff erent substances

Type of
substance Substance

Specifi c heat
capacity,

c / J kg–1 °C–1

Type of
substance Substance

Specifi c heat
capacity,

c / J kg–1 °C–1

Liquid Water 4 200 Metal Aluminium 900

Sea water 3 900 Iron 450

Ethanol 2 500 Copper 390

Paraffi  n 2 100 Gold 300

Cooking oil 1 850 Mercury 140

Olive oil 1 890 Lead 130

Gas Methane 2 200 Non-metal Polycarbonate 1 250

Steam
(at 100°C) 2 020

Wood 1 700

Concrete 850

Neon 1 030 Sand 800

Air 1 000 Glass 670

Water is a substance which has high specifi c heat capacity. Water needs to absorb a large
amount of heat to have a small rise in temperature. Th is makes water a good cooling agent. Metal
on the other hand has lower specifi c heat capacity compared to non-metal. Th erefore, objects
made from metal get hot quickly when supplied with an amount of heat.

128128 4.2.24.2.2
Aim:  To determine the specifi c heat capacity of water

Apparatus:  Power supply, immersion heater, beaker, stopwatch, thermometer, retort stand
and electronic balance

Materials: Water and tissue paper

Procedure:

1. Wrap a beaker with tissue paper.
2. Place the beaker on top of an electronic balance and reset the reading of the balance

to zero.
3. Fill the beaker with water until it is three-quarter full.

4. Record the reading of the mass of the water, m shown on the electronic balance.

5. Set up the apparatus as shown in Figure 4.10.

Figure 4.10

6. Record the initial temperature of the water, θ1.

7. Switch on the immersion heater and start the stopwatch
at the same time.

8. Observe the change in the thermometer reading.
9. After fi ve minutes, switch off the immersion heater.

Record the highest thermometer reading as the fi nal
water temperature, θ2.

Experiment 4.1

INFOSMARTSMART

Immersion heater converts
electrical energy to heat energy.
The heat energy supplied by the
immersion heater is
Q  =  Pt, where

P  =  power of heater and
t  =  period of time heater

is switched on

Change in water temperature,
Δθ = θ2 – θ1.
For this experiment, the equation
Q = mcΔθ is expressed as
Pt = mc (θ2 – θ1).

129129

Chapter 4 Heat

4.2.34.2.3
Experiment 4.2
Aim:  To determine the specifi c heat capacity of aluminium
Apparatus: Power supply, immersion heater, 1 kg aluminium block, stopwatch, thermometer

and retort stand
Material: Tissue paper

Procedure:

1. Set up the apparatus as shown in Figure 4.11.

Thermometer

Retort
stand

Immersion
heater Power supply

Aluminium
block

Tissue paper

Figure 4.11

Results:
Table 4.3

Power of immersion heater, P / W

Heating time, t / s

Mass of water, m / kg

Initial temperature of water, θ1 / °C

Final temperature of water, θ2 / °C

Analysis of data:
Calculate the specifi c heat capacity of water using the formula, c = Pt

m(θ2 – θ1)
.

Conclusion:

What conclusion can be made from this experiment?

Prepare a complete report on this experiment.

Discussion:
1. Why does the beaker need to be wrapped with tissue paper?
2. Why is the fi nal water temperature, θ2 not taken as soon as the fi ve-minute heating time
 ends?
3. Given specifi c heat capacity of water is 4 200 J kg–1 °C–1, compare the value of specifi c

heat capacity of water obtained from the experiment with the value given. Explain the
difference between the two values (if any).

4. Suggest methods to increase the accuracy of the result of this experiment.

130130 4.2.34.2.3
2. Record the initial temperature of the aluminium block, θ1.
3. Switch on the immersion heater and start the stopwatch at the same time.

4. After fi ve minutes, switch off the immersion heater. Record the highest thermometer
reading as the fi nal temperature of the aluminium block, θ2.

Results:

Table 4.4

Power of immersion heater, P / W

Heating time, t / s

Mass of aluminium, m / kg

Initial temperature of aluminium, θ1 / °C

Final temperature of aluminium, θ2 / °C

Analysis of data:
Calculate the specifi c heat capacity of aluminium using the formula, c = Pt

m(θ2 – θ1)
.

Conclusion:

What conclusion can be made from this experiment?

Prepare a complete report for this experiment.

Discussion:

1. What can be done to obtain a better thermal contact between the bulb of the
thermometer and the aluminium block?

2. Given specifi c heat capacity of aluminium is 900 J kg–1 °C–1, compare the value of
specifi c heat capacity of aluminium obtained from the experiment with the value given.
Explain the difference between the two values (if any).

131131131

Chapter 4 Heat

4.2.34.2.3
Metal wok
Clay pot

Cooking utensils

Woks are made of metal with low specifi c heat
capacity. As such, food can be fried at high
temperature in a short time. Clay pots on the
other hand are made of clay which has a high
specifi c heat capacity. As such, food can stay
hot for a long time.

Selection building materials of traditional houses in various climate zones

Car radiator system

Wood has a high specifi c heat
capacity and gets hot slowly. In
warm weather regions, traditional
houses are built from wood which
functions as an insulator of heat
from the scorching sun. In cold
weather regions, traditional houses
are also built from wood. Heat
from fi res lit in the wooden houses
cannot fl ow out because wood
functions as a good heat insulator.

Burning of fuel in car engines produces large
amounts of heat. Th is heat needs to be released
to avoid overheating the engine. Water has
a high specifi c heat capacity and is used as a
cooling agent. A pump will pump water into
the engine block. Water will fl ow through the
engine block to absorb heat produced. Hot
water fl ows to the radiator.  Cold air is sucked
in by fans so that heat in the hot water can be
released quickly through cooling fi ns.

Air flow

Fan

Radiator

Cover Cooling fin

Pump

Warm climate Cold climate

Knowledge on specifi c heat capacity is very important in daily life, material engineering and also
understanding several natural phenomena.

Video on car radiator system

http://bt.sasbadi.com/p4132

132132132132 4.2.44.2.4
Production of latest materials in the construction of green buildings

Cooking utensils

Th e Diamond Building, Energy Commission
is built with an insulating concrete roof, that is
a roof fi tted with insulators using styrofoam
boards. Styrofoam has a high specifi c heat
capacity and can reduce the absorption of heat
from the surroundings to reduce the temperature
inside the building.

Th e body of a pot is made of aluminium which has
a low specifi c heat capacity. Th is enables the pot to
heat up quickly. However, the handle of the pot is
made from plastic which has a high specifi c heat
capacity. Th is ensures that the handle of the pot gets
hot slowly and is safe to handle.

Handle of pot

Body of
pot

Outer layer of space capsule

Space capsule on its journey back to Earth
encounters air resistance when entering
the atmosphere. Th is friction increases the
temperature and causes the space capsule to
burn. Th erefore, the outer layer of a space
capsule is made from substance with a high
specifi c heat capacity and melting point.

The Diamond Building

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Chapter 4 Heat

Concrete
sheet

Cement
layer

Cement layer
Styrofoam
insulating
board

133133

Chapter 4 Heat

4.2.44.2.4
Land has a lower specifi c heat capacity than the sea.
Th erefore, temperature on land rises more quickly than
temperature in the sea during daytime. Th e air on land
becomes hot and rises upwards. Cold air from the sea
moves towards land as sea breeze.

Sea breeze

Land breeze

Sea has a higher specifi c heat capacity than land.
So, temperature in the sea drops more slowly than
temperature on land at night. Hot air above the sea
rises upwards. Cold air above the land moves towards
the sea as land breeze.

Activity 4.4

Aim: To search for information on applications of specifi c heat capacity

Instructions:

1. Carry out a Round Table activity.

2. Gather information on applications of specifi c heat capacity related to:
(a) Daily life
(b) Material engineering
(c) Natural phenomena

3. The information can be obtained from reading materials in the library or on the Internet.

4. One group member writes the information on a piece of paper. The paper is then
passed clockwise so other group members can add their information.

5. Present your group fi ndings in your class.

ISS ICS

134134 4.2.44.2.4
Example 1

Example 2

Solving Problems Involving Specifi c Heat Capacity

A 0.5 kg metal block is heated by a 50 W electric heater for 90 s. Th e temperature of the block
rises from 20°C to 45°C. Calculate the specifi c heat capacity of the metal.

Solution:

Step �
List the given information
in symbols.

Temperature rise, Δθ = 45 – 20
 = 25°C
Mass of block, m = 0.5 kg
Power of heater, P = 50 W
Heating time, t = 90 s

Step �
Identify and write down the
formula used.

c = Q
mΔθ

= Pt
mΔθ

Step �
Substitute numerical values into the
formula and perform the calculations.

c =  (50)(90)
(0.5)(25)

= 360 J kg–1 °C–1

Assumption: All heat supplied by the electric heater is absorbed by the metal block. No heat is lost to
the surroundings.

20 g of boiling water at 100°C is poured into a glass containing 200 g of water at 28°C. Calculate
the fi nal temperature of the mixture of water.

Solution:

Let y = fi nal temperature of mixture
For boiling water:
 Mass, m1 = 20 g
  = 0.02 kg
 Temperature change, Δθ1 = (100 – y)°C
For water at 28°C:
 Mass, m2 = 200 g
  = 0.20 kg
 Temperature change, Δθ2 = (y – 28)°C
Specifi c heat capacity of water, c = 4 200 J kg–1 °C–1

Q1 = Q2
 m1cΔθ1 = m2cΔθ2
 0.02 (4 200)(100 – y) = 0.20 (4 200)(y – 28)
 8 400 – 84y = 840y – 23 520
 924y = 31 920
 y = 34.55°C
Th erefore, the fi nal temperature of the mixture
of water is 34.55°C.

Heat is released
Q1

Heat is absorbed
Q2

200 g water, 28°CTemperature of mixture, y20 g water, 100°C

Assumption: No heat is absorbed or released to the surrounding. Heat transfer only occurs between
the boiling water and the water at 28°C. Therefore, heat released by the boiling water is the same as
the heat absorbed by the water at 28°C.

1
4

2
4

3
1

4
2

4
3

1
4

2
4

3

135135

Chapter 4 Heat

4.2.54.2.5
Activity 4.5

Aim:  To build a model of a cluster home which can overcome the problem of
extreme temperatures

Instructions:

1. Work in groups.
2. Read and understand the following information.

Cluster homes are homes which resemble
terrace houses. However, three walls of the house
are shared with the houses behind and beside it
(Figure 4.12).

Photograph 4.2 shows an example of a
cluster home which has only one door for exit
and entrance, while windows are only in the front
part of the home. Th e shape of the home can
minimize the use of land. However, when our
country experienced the El Nino phenomenon
with extreme rise in temperature, residents of
terrace cluster homes experienced extreme heat.

Figure 4.12 Plan of cluster home

Photograph 4.2 Example of cluster home

1 3

2

5 7 9 11

4 6 8 10 12

3. Based on the above information, analyse the situation by listing the facts and problems
related to the condition of extreme temperature in cluster homes.

4. Brainstorm several solutions to the problems. Sketch a model based on your solutions.

5. Build the model based on the sketch.

6. Display and present the model.

Formative Practice

1. What is the diff erence between heat capacity and specifi c heat capacity?

2. How much heat energy is needed to increase the temperature of a 0.2 kg mass of gold
by 10°C?

[Given the value of specifi c heat capacity of gold is 300 J kg–1 °C–1]

3. A container contains 200 g of water at initial temperature of 30°C.  An iron nail of mass 200 g
  at temperature of 50°C is immersed in the water. What is the fi nal water temperature? State

the assumptions you need to make in your calculations.
  [Given the value of specifi c heat capacity of water is 4 200 J kg–1 °C–1 and that of iron is

450 J kg–1 °C–1]

STEMCPS

136136

4.24.2

136136 4.2.54.2.5
Specific Latent Heat4.3

Latent Heat
Elements can exist in three states: solid, liquid and gas. Th e diff erence in the arrangement and
movement of molecules among the three states of matter shows that there are stronger molecular
bonds in solid than in liquid and gas. As gas molecules move freely at random, the bond between
gas molecules is the weakest.
 Figure 4.13 shows the changes in phase of matter. During the changes in the phase of matter
such as melting and boiling, the temperature remains constant even though heat is being supplied
continually. Heat that is absorbed during melting and boiling without change in temperature
is known as latent heat. During condensation and freezing, latent heat is released without
temperature change.

Latent heat
is released

Latent heat
is released

Latent heat
is absorbed

Latent heat
is absorbed

Figure 4.13 Changes in phases of matter

Specifi c Latent Heat
Th e quantity of heat needed to change the state of matter of
an object depends on the mass of the object and its material.
Specifi c latent heat, l of a substance is the quantity of heat, Q that
is absorbed or released during a change of phase of 1 kg of the
substance without any change in its temperature.

An object of mass, m absorbs a quantity of heat, Q during a change of
phase. Th erefore, specifi c latent heat of the substance of the object is

l = Q
m

S.I. unit for specifi c latent heat is J kg−1.

137

Latent Heat

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137137

Chapter 4 Heat

4.3.14.3.1 4.3.24.3.2
Specifi c latent heat of fusion, lf of a substance is the
quantity of heat, Q that is absorbed during melting or the
quantity of heat released during freezing of 1 kg of the substance
without any change in temperature.
 Specifi c latent heat of vaporisation, lv of a substance is
the quantity of heat, Q that is absorbed during boiling or the
quantity of heat released during condensation of 1 kg of the
substance without any change in temperature.
 Figure 4.14 shows the heating curve when an object
changes its state from solid to gas.

P

Q R

S T

U

Boiling point

Melting point
Solid

Solid and
liquid

0 Time

Liquid

Liquid and gas

Gas

PQ – Solid is heated until
melting point

Heat is absorbed
Temperature increases
Kinetic energy of

molecules increases

RS – Liquid is heated
until boiling point

Heat is absorbed
Temperature increases
Kinetic energy of

molecules increases

Temperature

TU – Gas is heated
Heat is absorbed
Temperature increases
Kinetic energy of

molecules increases

Melting
Solid Liquid
Q     –  Solid starts to melt
QR –  Solid is melting
R – The whole solid has melted

Latent heat is absorbed to weaken the bond
 between molecules

Molecules are released from their fixed position and
 move throughout the liquid

Kinetic energy of molecules does not increase
Constant temperature

Boiling
Liquid Gas
S    – Liquid starts to boil
ST  –  Liquid is boiling
T    –  The whole liquid has boiled

Latent heat is absorbed to break the bond
 between molecules

Molecules are separated far from one another
Kinetic energy of molecules does not increase
Constant temperature

Figure 4.14 Heating curve

138138

F ile
Based on the Kinetic Theory of
Matter, the higher the average
kinetic energy of a molecule, the
higher the temperature of the
object. Latent heat absorbed
during melting and boiling
does not increase the average
kinetic energy of the molecule.
Therefore, melting and boiling
occur at constant temperature.

138138 4.3.24.3.2
Figure 4.15 shows the cooling curve when an object changes its state from gas to solid.

P

Q R

S T

U

0 Time

Gas

Liquid

Liquid and solid

Solid

PQ – Gas cools to
boiling point

Heat is released
Temperature decreases
Kinetic energy of

molecules decreases
RS – Liquid cools to
freezing point

Heat is released
Temperature decreases
Kinetic energy of

molecules decreases TU – Solid cools
Heat is released
Temperature decreases
Kinetic energy of

molecules decreases

Temperature

Boiling point

Freezing point

Gas and
liquid

Condensation
Gas Liquid
Q      – Gas starts to condense
QR   – Gas is condensing
R      – The whole of the gas has condensed

bond is formed

decrease

Freezing
Liquid  Solid
S    – Liquid starts to freeze
ST  – Liquid is freezing
T – The whole liquid has frozen

between molecules is strengthened

64748

64748

64748

644744864748

Figure 4.15 Cooling curve

139139139139

Chapter 4 Heat

4.3.24.3.2
Activity 4.6

Aim: To compare and discuss:
 • specifi c latent heat of fusion of ice and wax
 • specifi c latent heat of vaporisation of water and oil

Instructions:

1. Carry out a Think-Pair-Share activity.

2. Study the information given in Table 4.5.

Table 4.5

Substance
 Phase at

room
temperature

Melting
point /

°C

Specifi c latent
heat of fusion,

lf / J kg−1

Boiling point /
°C

Specifi c latent heat
of vaporisation,

lv / J kg–1

Wax Solid 46 to 68 1.45 × 105 to
2.10 × 105

– –

Lead Solid 327 0.25 × 105 1 750 8.59 × 105

Copper Solid 1 083 2.07 × 105 2 566 47.3 × 105

Ice Solid 0 3.34 × 105 – –

Water Liquid – – 100 22.6 × 105

Petrol Liquid – – 35 to 200 3.49 × 105

Diesel Liquid – – 180 to 360 2.56 × 105

Olive oil Liquid 6 2.67 × 105 – –

Ethanol Liquid −114 1.04 × 105 78 8.55 × 105

Oxygen Gas −219 0.14 × 105 −183 2.13 × 105

Nitrogen Gas −210 0.26 × 105 −196 2.00 × 105

3. Based on the information above, discuss the following
questions:
(a) Compare the specifi c latent heat of fusion for ice and

wax. Then, state the difference between ice and wax
in terms of strength of bond between molecules.

(b) Compare the specifi c latent heat of vaporisation for
water and petrol. Then, state the difference between
water and petrol in terms of strength of bond between
molecules and distance of separation between
molecules in gaseous phase.

(c) For a specifi c substance, why is specifi c latent heat
 of vaporisation larger than specifi c latent heat of fusion?

4. Present the results of your discussion in a graphic form.

Note:
• Petrol and diesel are

diff erent types of
hydrocarbons and have

diff erent boiling points.

Logical Reasoning

140140 4.3.24.3.2
Experiment 4.3

Based on Activity 4.6, each substance has a diff erent specifi c latent heat. How is the value of
specifi c latent heat determined?

Aim: (i) To determine specifi c latent heat of fusion of ice, lf
 (ii) To determine specifi c latent heat of vaporisation of water, lv

A  Specifi c latent heat of fusion of ice, lf

Apparatus: Immersion heater, fi lter funnel, beaker, electronic balance, power supply,
stopwatch and retort stand

Material: Crushed ice

Procedure:

1. Place the beaker for the experiment set and the control set on the electronic balance
respectively. Reset the readings of both electronic balances to zero.

2. Set up the apparatus as shown in Figure 4.16. Initially, both beakers and electronic
balances are not below their respective fi lter funnels.

Immersion heater
Crushed ice
Filter funnel

Beaker
Water

Power supply

Electronic balance

Beaker
Water

Retort stand

Control set Experiment set

Beaker
Water

Power supply

Figure 4.16
3. Switch on the immersion heater for the experiment set only. When water is dripping out

of the fi lter funnel at a fi xed rate, place the beakers and electronic balances respectively
below the fi lter funnels. Start the stopwatch.

4. After time, t = 10 minutes, record the reading of the mass of water collected in the
beaker of experiment set, m1 and control set, m2.

5. Switch off the immersion heater and record the power of the heater, P.

Results:
Table 4.6

Mass of water collected in beaker of experiment set, m1 / kg
Mass of water collected in beaker of control set, m2 / kg
Power of heater, P / W
Heating time, t / s

141141

Chapter 4 Heat

4.3.34.3.3
Analysis of data:

Calculate the specifi c latent heat of fusion of ice using the formula, l = Pt
(m1 – m2)

.

Conclusion:
What conclusion can be made from this experiment?

B  Specifi c latent heat of vaporisation of water, lv

Apparatus:  High power immersion heater (500 W), power supply, beaker, electronic balance
and stopwatch

Materials: Water and tissue paper

Procedure:

1. Set up the apparatus as shown in Figure 4.17.

2. Switch on the immersion heater and wait
until the water boils.

3. When the water boils, start the stopwatch
and at the same time, record the reading
on the electronic balance, m1.

4. After time, t = 5 minutes, record the
reading on the electronic balance, m2.

5. Switch off the immersion heater and
record the power of the heater, P.

Results:

Table 4.7

Initial reading of electronic balance, m1 / kg

Final reading of electronic balance, m2 / kg

Time taken, t / s

Power of heater, P / W

Analysis of data:

Calculate the specifi c latent heat of vaporisation of water using the formula, l = Pt
(m1 – m2)

.

Conclusion:
What conclusion can be made from this experiment?

Prepare a complete report for this experiment.

Discussion:
1. Why does a control set need to be prepared for experiment A but not for experiment B?

2. Given specifi c latent heat of fusion of ice = 3.34 × 105 J kg–1, compare the value of
specifi c latent heat of fusion of ice obtained from experiment A with the value given.
Explain the difference between the two values (if any).

3. Given specifi c latent heat of vaporisation of water = 2.26 × 106 J kg–1, compare the value
of specifi c latent heat of vaporisation of water obtained from experiment B with the
value given. Explain the difference between the two values (if any).

4. Suggest ways to increase the accuracy of the results of this experiment.

17.
Immersion heater

Beaker

Tissue paper

Water

Electronic balance

To power supply

ON
OFF

Figure 4.17

142142 4.3.34.3.3
Activity 4.7

Observe Figure 4.18 which shows the changes of phase of water when latent heat is absorbed
and released.

Figure 4.18 Changes in phase of water

Absorption of latent heat during melting and evaporation can be used to give the eff ect of
cooling. Latent heat released during condensation however is used for the purpose of heating.

Aim: To show that evaporation causes cooling

Apparatus: 250 ml beaker, drinking straw and white tile

Materials: Alcohol and water

Instructions:

1. Set up the apparatus as shown in Figure 4.19.

2. Pour 100 ml alcohol into a beaker.

3. Touch the outside of the beaker and the water
around the base of the beaker. Record
your observations.

4. Blow air repeatedly into the alcohol.

5. Touch the outside of the beaker. Record
your observations.

Discussion:

1. What happens to the alcohol when air is blown into it?

2. Compare the level of coldness of the beaker before and after air is blown into the
alcohol. Explain your answer.

3. State the effect of evaporation.

Blow

Beaker

White tile

Alcohol

Water

Drinking
straw

Specifi c latent heat of vaporisation is required in the change of phase from liquid to gas. Th is
heat is absorbed from the surrounding. When a liquid evaporates, the liquid molecules absorb this
heat to break the bond between molecules. Th e surrounding loses heat. Th erefore, evaporation
causes cooling to the surrounding.

When ice melts, the ice
molecules absorb latent heat
of fusion causing ice to change
from solid to liquid.

When water boils, the water
molecules absorb latent heat
of vaporisation causing water
to change from liquid to gas.

When water vapour condenses,
the water vapour molecules
release latent heat of vaporisation
causing water vapour to change
from gas to liquid.

Figure 4.19

143143

Chapter 4 Heat

4.3.44.3.4
Activity 4.8

Figure 4.20 shows examples of phase change of matter that involve specifi c latent heat.

Figure 4.20 Examples involving specifi c latent heat

Aim: To discuss applications of specifi c latent heat in daily life

Instructions:

1. Work in groups.

2. Gather information on applications of specifi c latent heat in daily life:
(a) Evaporation of sweat
(b) Steaming food

3. Discuss how specifi c latent heat is applied in each situation.

4. Present the fi ndings in the form of a mind map.

Melting

Cooling drinks
and food with

ice cubes

Evaporation Boiling Condensation

Cooling body on
hot day

Cooling as liquid
nitrogen boils

Process of steaming
food

Cooling system
of refrigerator

Heating espresso
coff ee

Cooling system of
air conditioner

Specifi c Latent Heat

Video on cooling with liquid
nitrogen

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Video on heating espresso
coffee

http://bt.sasbadi.com/p4144b

ISS ICS

144144 4.3.44.3.4
Heat is
absorbed

Compressor

Evaporator

Expansion valve

Condenser

Heat is
released

� �

��

�

A refrigerator uses the cooling eff ect from evaporation. During circulation of the cooling agent,
heat is absorbed from inside the refrigerator and released outside.

Cooling system in refrigerator

Applications of Specifi c Latent Heat in Daily LifeApplications of Specifi c Latent Heat in Daily Life

C

Compressor
compresses cooling
agent (gas) to
increase pressure
and its temperature.

Cooling agent (gas)
fl ows out from
evaporator to
the compressor.

In the condenser,
cooling agent
(gas) releases
latent heat during
condensation.

Cooling agent
(liquid) fl ows
through expansion
valve.

In the evaporator,
cooling agent
(liquid) evaporates
and absorbs latent
heat from inside
the refrigerator.

We sweat on hot days or while doing
heavy work. When sweat evaporates,
heat is absorbed from the body
causing a cooling eff ect. Th e rate of
evaporation will increase when there
is air circulation.

Evaporation of sweat

Figure 4.21 Cooling system in a refrigerator

Wet your right hand. Put your
right hand which is wet and
your left hand which is dry
in front of a table fan. What
difference can you feel on your
right and left hands?

145145

Chapter 4 Heat

4.3.44.3.4
Example 1

Solving Problems Involving Latent Heat

Figure 4.22 shows a 480 W immersion heater
used to melt ice in a container. In 120 s, the
reading of the electronic balance decreases by
0.172 kg.
(a) What is the mass of ice that has melted

during the heating period?
(b) Calculate the specifi c latent heat of fusion of

ice, lf .

Solution:

(a) Make assumptions:
(i) Ice is melted by heat from the immersion heater only.
(ii) All water from the melting ice fl ows out of the container.

Relate the change in the readings of the electronic balance to the mass of ice which has
melted:

Mass of ice melted = Decrease in the reading of the electronic balance
 m = 0.172 kg
(b) Make assumptions:

(i) All heat supplied by the immersion heater is absorbed by the melting ice.
(ii) No transfer of heat from the surrounding into the apparatus.

Step �
List the given information
in symbols.

m = 0.172 kg
P = 480 W
t = 120 s

Step �
Identify and write down the
formula used.

Pt = mlf

Step �
Substitute numerical values into
the formula and perform the
calculations.

480 × 120 = 0.172 × lf
 lf = 480 × 120

0.172

= 3.35 × 105 J kg−1

1
4
2
4
3

1
2
3

1
4
2
4
3

146

Immersion heater

Electronic
balance

Crushed ice

To sink

ON
OFF

To power
supply

Figure 4.22

146146 4.3.54.3.5
Example 2

Formative Practice 4.34.3

What is the amount of heat supplied by a water heater to change 0.75 kg of water at 30°C to
steam at 100°C? State the assumptions you make in your calculations.
[Specifi c heat capacity of water, cwater = 4.20 × 103 J kg–1 °C−1,
specifi c latent heat of vaporisation of water, lv = 2.26 × 106 J kg−1]

Solution:

0.75 kg
water at 30°C

0.75 kg
water at 100°C

0.75 kg
steam at 100°C

Q1 = mcΔθ Q2 = ml

Make assumptions:
(i) All heat supplied by the heater is absorbed by the water.
(ii) No loss of heat to the surrounding during heating and change of phase.

Th ere are two stages of change:
(i) increase in water temperature from 30°C to its boiling point of 100°C
(ii) change of phase from water to steam without change in temperature.

Amount of heat supplied, Q = Q1 + Q2
   = mcΔθ + ml
   = [0.75 × 4.2 × 103 × (100 – 30)] + (0.75 × 2.26 × 106)
   = 1.92 × 106 J

1. Figure 4.23 shows an electric steamer. Explain how the fi sh
is heated.

2. What is the amount of heat released when 0.8 kg of water at
25°C cools until it becomes ice at –6°C? State the assumptions
you make in your calculations.

[Specifi c heat capacity of water, cwater = 4.2 × 103 J kg–1 °C–1,
  specifi c heat capacity of ice, cice = 2.0 × 103 J kg–1 °C–1 and
  specifi c latent heat of fusion of ice, lf = 3.34 × 105 J kg–1]

Fish

Contain
water

Figure 4.23

147147147147

Chapter 4 Heat

4.3.54.3.5
4.9Activity

4.4.14.4.1

Photograph 4.3 Air cushion
wrap being compressed

Gas Laws4.4

Pressure, Temperature and Volume of Gas
Photograph 4.3 shows an air cushion wrap used in packaging of goods.
When it is compressed, the air inside gives resistance. Th is observation
can be explained in terms of the behaviour of gas molecules based on
the Kinetic Th eory of Gas.

Aim: To observe the behaviour of gas molecules through computer simulation

Instructions:
1. Carry out a Think-Pair-Share activity.

2. Scan the QR code to see the simulation on the behaviour of gas molecules. Based on
the simulation, discuss:
(a)  movement of gas molecules
(b) space fi lled by gas molecules
(c) direction of motion of molecules
(d) collisions between gas molecules and the walls of

the container
(e)  effects of increasing and decreasing of pressure,

temperature and volume of gas on the behaviour of
the gas molecules

3. Present your fi ndings.

Table 4.8 explains pressure, temperature and volume of gas in a closed container based on
the Kinetic Th eory of Gas.

Table 4.8 Pressure, temperature and volume of gas based on the Kinetic Th eory of Gas

Characteristic of gas Description

Pressure • Gas molecules always move randomly.
• When gas molecules collide with the wall of the container and rebound, a

force is exerted on the wall of the container.
• Force per unit area is the pressure of the gas.

Temperature • Average kinetic energy of gas molecules increases with temperature.

Volume • Gas molecules move freely and fi ll the entire space of the container.
• Volume of gas is the same as the volume of its container.

Simulation of behaviour of
gas molecules

http://bt.sasbadi.com/p4148

ISS ICS

148148
Table 4.9 S.I. unit and other units for pressure, temperature and volume
of gas

Quantity S.I. unit Symbol for
S.I. unit Other units

Pressure, P pascal Pa cm Hg

Temperature, T kelvin K °C, °F

Volume, V (metre)3 m3 mm3, cm3, ml

Relationship between Pressure and Volume of Gas
Photograph 4.4 shows an exercise ball being compressed
when someone sits on it. What happens to the air pressure
inside the ball? Photograph 4.4 Exercise ball

being compressed

Experiment 4.4
Inference: Volume of gas infl uences pressure of gas

Hypothesis: The smaller the volume of gas, the higher the gas pressure

Aim:  To determine the relationship between volume and pressure of a fi xed mass of gas at
constant temperature

Variables:
(a) Manipulated variable: Volume, V
(b) Responding variable: Pressure, P
(c) Constant variable: Temperature and mass of air

Apparatus: 100 ml syringe, rubber tube, pressure gauge and retort stand

Procedure:

1. Set up the apparatus as shown in Figure 4.24.

Figure 4.24

Visit the following websites to
carry out virtual experiments on
Boyle’s Law

http://bt.sasbadi.com/p4149a

http://bt.sasbadi.com/p4149b

149149149

Chapter 4 Heat

4.4.14.4.1 4.4.24.4.2
2. Adjust the piston so that the volume of air in the syringe is 100 ml. Then, connect the
end of the syringe to a pressure gauge.

3. Take initial readings of the volume and pressure of the air in the syringe. Record the
readings in Table 4.10.

4. Push the piston slowly until the volume of air in the syringe becomes 90 ml. Take the
reading of the air pressure and record it in the table.

5. Repeat step 4 with volumes 80 ml, 70 ml and 60 ml.

6. Record all pressure, P in Table 4.10.

Results:

Table 4.10

Volume, V / ml Pressure, P / kPa 1
V

/ ml–1

60

70

80

90

100

Analysis of data:
Plot a graph of pressure, P against volume, V and a graph of P against 1

V
.

Conclusion:
What conclusion can be made from this experiment?

Prepare a complete report on this experiment.

Discussion:

1. Why is a syringe of larger volume used?
2. Why is the piston pushed slowly into the syringe?

Experiment 4.4 shows that gas pressure increases when the volume of gas decreases. What
is the relationship between pressure and volume of gas at constant temperature?

150150150 4.4.24.4.2
Figure 4.25 shows the relationship between pressure and volume of gas.

P

V0

P

1—V0

(a) Graph P against V (b) Graph P against 1
V

Figure 4.25 Relationship between pressure and volume of gas

Graph of P against V shows that pressure decreases with
volume. Graph of P against 1

V  however shows a straight line
passing through the origin. Th is shows that pressure is inversely
proportional to volume.

HISTORY
INTEGRATION OF

Robert Boyle (1627–1691) is a
scientist who emphasised the
use of scientifi c method when
carrying out investigations.
Through data from experiments,
he concluded that the volume of
gas is inversely proportional to
its pressure.

http://bt.sasbadi.com/p4151a

Boyle’s Law states that pressure is inversely proportional to
volume for a fi xed mass of gas at constant temperature.

P ∝ 1
V

P = k( 1
V )

Where k is a constant
  P = gas pressure (Pa)
  V = gas volume (m3)
 As such, PV = k

If a gas experiences a change in pressure and volume from
condition 1 to condition 2,

since PV = k, condition 1 of gas, P1V1 = k
 condition 2 of gas, P2V2 = k

therefore, P1V1 = P2V2

Boyle’s Law

http://bt.sasbadi.com/p4151b

151151

Chapter 4 Heat

4.4.24.4.2
Example 1

Figure 4.26 shows a fi xed mass of gas compressed at constant
temperature. When the volume of gas decreases, the same number
of molecules move in a smaller space. Th erefore, the number
of molecules per unit volume increases. Th is causes the rate of
collisions between molecules and the walls of the container to
increase. Force per unit area on the wall of the container also
increases. As such, gas pressure increases.

Figure 4.26 Fixed mass of a gas compressed at constant temperature

Air in a closed syringe has a volume of 60 cm3 and pressure of 108 kPa. Th e piston of the
syringe is pushed to compress the air to a volume of 48 cm3. Calculate the pressure of the
compressed air.

Solution:

Step �
List the given information
in symbols.

P1 = 108 kPa
P2 = compressed air pressure
V1 = 60 cm3

V2 = 48 cm3

Step �
Identify and write down the
formula used.

Temperature of gas does not change.
Boyle’s Law formula is used.
P1V1 = P2V2

Step �
Substitute numerical values into
the formula and perform the
calculations.

108 × 60 = P2 × 48

P2 = 108 × 60
48

= 135 kPa

1
4
2
4
3

1
4
2
4
3

INFOSMARTSMART

PVT triangle:

P V

T

For Boyle’s Law, temperature is
constant.

P V

PV = constant
P1V1 = P2V2

Volume
decreases

1
2
3

152152 4.4.24.4.2
Relationship between Volume and Temperature of Gas
Photograph 4.5 shows an empty plastic bottle fi lled with air in a refrigerator. What happened to
the volume of air in the bottle?

(a) Empty plastic bottle before being cooled (b) Empty plastic bottle aft er being cooled
Photograph 4.5 Condition of plastic bottle in refrigerator before and aft er being cooled

Experiment 4.5
Inference: Temperature of a gas infl uences the volume of gas

Hypothesis: The higher the temperature, the larger the volume of gas

Aim:  To determine the relationship between temperature and volume of a fi xed mass of gas
at constant pressure

Variables:
(a) Manipulated variable: Temperature, θ
(b) Responding variable: Volume, V represented by length of column of air, L in capillary tube
(c) Constant variable: Pressure and mass of air

Apparatus:  Capillary tube containing air trapped by a column of concentrated sulphuric
acid, 500 ml beaker, thermometer, ruler, Bunsen burner, tripod stand, wire
gauze, stirrer and retort stand

Materials: Water, ice and rubber band

Procedure:
1. Set up the apparatus as shown in Figure 4.27.

Thermometer
Ruler

Capillary tube

IceRubber band

Concentrated
sulphuric acid

Wire gauze

Tripod stand
Bunsen burner

Beaker

Stirrer

Retort stand

Water
L

Figure 4.27

153153

Chapter 4 Heat

4.4.34.4.3
2. Heat the water slowly and stir it continuously until the temperature of the water reaches
30°C.

3. Take the reading of the column of air, L in the capillary tube. Record the reading in
Table 4.11.

4. Repeat steps 2 and 3 for temperatures 40°C, 50°C, 60°C, 70°C and 80°C.

5. Record all lengths of column of air, L in Table 4.11.

Results:

Table 4.11

Temperature, θ / °C Length of column of air, L / cm

30

40

50

60

70

80

Analysis of data:

1. Plot a graph of length of column of air, L against temperature, θ. θ-axis has to cover the
range of 0°C to 100°C.

2. Extrapolate graph of L against θ until θ = 0°C.

3. Plot another graph of L against θ with θ-axis covering the range of –300°C to 100°C.

4. Extrapolate graph of L against θ until L = 0 cm.

Conclusion:
What conclusion can be made from this experiment?

Prepare a complete report on this experiment.

Discussion:

1. Why must the water be continuously stirred while being heated?

2. What assumption needs to be made so that the length of the column of air trapped in
the capillary tube can represent the volume of the trapped air?

[Key: Volume of column of air, V = length of column of air, L × cross sectional area of
capillary tube, A]

Volume of gas increases when the temperature of the gas rises. At 0°C, the air trapped in the
capillary tube still has a certain volume. Th is shows that at 0°C, gas molecules are still moving and
fi lling up the space in the container.

154154 4.4.34.4.3
Figure 4.28 shows the graph of V against θ extrapolated until V = 0 cm3.

V / cm3

/ °C
0 100–273

Figure 4.28 Extrapolation of graph V against θ

At temperature of –273°C, gas molecules no longer move and are unable to fi ll the space.
As such, volume of gas becomes zero. Th e temperature of –273°C is the lowest temperature
possible and is known as absolute zero. On the kelvin scale, absolute zero is given the value
0 kelvin or 0 K. Temperature that is stated in unit kelvin is absolute temperature.

Table 4.12 Temperature in units of degree Celsius, °C and kelvin, K
for three temperature points

Temperature point Temperature, θ / °C Temperature, T / K

Absolute zero –273 0

Melting ice 0 273

Steam 100 373

Conversion of units between degree Celsius, °C and kelvin, K
can be done using the following equation:

T = θ + 273
For θ °C and T K

Figure 4.29 shows the graph of V against T.

V / cm3

T /  K0

Figure 4.29 Graph of V against T for a gas

Th e graph of V against T for gas shows a straight line passing through the origin. Th is shows
that the volume of gas is directly proportional to absolute temperature.

HISTORY
INTEGRATION OF

Jacques Charles (1746–1823)
a French physicist and
chemist investigated how the
volume of gas depends on
the temperature of gas. He
built the fi rst hydrogen balloon
and succeeded in raising the
balloon to a height of 3.2 km.

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155155

Chapter 4 Heat

4.4.34.4.3
Charles’ Law states that volume is directly proportional
to absolute temperature for a fi xed mass of gas at
constant pressure.

V ∝ T
V = kT

where k is a constant
 T = absolute temperature (K)
 V = volume of gas (m3)

As such, V
T  = k

If a gas experiences a change in volume and temperature from
condition 1 to condition 2,

since V
T  = k, condition 1 of gas:

V1
T1

= k

condition 2 of gas:
V2
T2

= k

therefore,
V1
T1

=
V2
T2

Figure 4.30 shows a fi xed mass of gas being heated at constant pressure. When the
temperature of the gas increases, the average kinetic energy of its molecules increases, and the
molecules move with higher velocity. To keep a constant gas pressure, the volume of gas increases
so that the rate of collision of gas molecules with the walls of the container is unchanged.

Temperature increases

Key
Low velocity
High velocity

Figure 4.30 Fixed mass of gas heated at constant pressure

Charles’ Law

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INFOSMARTSMART

For Charles’ Law, pressure is
constant.

T

V

V
T  = constant
V1

T1
 =

V2

T2

156156156 4.4.34.4.3
Example 1

Relationship between Pressure and Temperature of Gas
Photograph 4.6 shows air pressure in the tyre of a car being measured on a hot day. Th e driver of
the car touched the tyre aft er the journey and found that the tyre is hotter than before the journey.
Photograph 4.7 shows the readings of the pressure gauge before and aft er the journey. What
happened to the air pressure inside the tyre?

An air bubble has a volume of 1.2 cm3 at a temperature of 27°C. What is the volume of the
air bubble if its temperature increases to 47°C?

Solution:

Step �
List the given information
in symbols.

V1 = 1.20 cm3
V2 = Final volume of air
T1 = (27 + 273) = 300 K
T2 = (47 + 273) = 320 K

Step �
Identify and write down the
formula used.

Gas pressure is constant.
Charles’ Law formula is used.

V1

T1
 = V2

T2

Step �
Substitute numerical values into
the formula and perform the
calculations.

1.2
300

=
V2

320

V2 = 1.2 × 320
300

= 1.28 cm3

157

Photograph 4.6Photograph 4.6 Air pressure of tyre being measured Air pressure of tyre being measured
Photograph 4.7Photograph 4.7 Readings of air pressure Readings of air pressure

(a) Before journey

(b) After journey

1
4
2
4
3

1
4
2
4
3

1
4
4
2
4
4
3

( ) B f j

(b) Aft j

157157

Chapter 4 Heat

4.4.34.4.3 4.4.44.4.4
Inference: Temperature of gas infl uences pressure of gas

Hypothesis: The higher the temperature, the higher the gas pressure

Aim:  To determine the relationship between temperature and pressure for a fi xed mass of
gas at constant volume

Variables:
(a) Manipulated variable: Temperature, θ
(b) Responding variable: Pressure, P
(c) Constant variable: Volume and mass of air

Apparatus:  Round-bottom fl ask, large beaker, thermometer, pressure gauge, Bunsen burner,
wire gauze, tripod stand, stirrer, retort stand and wooden block

Materials: Water and ice

Procedure:
1. Set up the apparatus as shown in Figure 4.31.

Thermometer

Stirrer

Beaker

Retort stand

Bunsen burner

Round-bottom
flask

Wire gauze

Tripod stand Wooden
block

Pressure gauge

Rubber tube

Figure 4.31

2. Heat the water slowly and stir it continuously until the temperature of the water
reaches 30°C.

3. Take the reading of the air pressure, P inside the fl ask. Record the reading in
Table 4.13.

4. Repeat steps 2 and 3 with temperatures 40°C, 50°C, 60°C, 70°C and 80°C.

5. Record all readings of air pressure, P in Table 4.13.

Experiment 4.6

158158 4.4.44.4.4
Experiment 4.6 shows that gas pressure increases when temperature of the gas rises.
Figure 4.32 shows the graph of P against θ extrapolated until P = 0 kPa.

P / kPa

/ °C
0 100–273

Figure 4.32 Extrapolation of graph P against θ

Results:

Table 4.13

Temperature, θ / °C Air pressure, P / kPa

30

40

50

60

70

80

Analysis of data:

1. Plot a graph of pressure, P against temperature, θ. θ-axis has to cover the range of
–300°C to 100°C.

2. Extrapolate the graph until P = 0 kPa. Determine the temperature when P = 0 kPa.

Conclusion:
What conclusion can be made from this experiment?

Prepare a complete report on this experiment.

Discussion:

1. What is the advantage of using a round-bottom fl ask to heat the air?
2. The thermometer is placed in the large beaker fi lled with water. What is the assumption

made so that the thermometer reading is the same as the temperature of the air in the
round-bottom fl ask?

159159

Chapter 4 Heat

4.4.44.4.4
Graph of P against θ shows that gas pressure increases
linearly when temperature of the gas rises. At 0°C, gas molecules
are still moving and the gas has pressure. At –273°C (absolute
zero), gas molecules no longer move and do not collide with
the walls of the container. Hence, gas pressure becomes zero.
Figure 4.33 shows the graph of P against T.

P / kPa

T /  K0

Figure 4.33 Graph of P against T

Graph of P against T of gas is a straight line through the
origin. Th is shows that gas pressure is directly proportional to
absolute temperature.

Gay-Lussac’s Law states that pressure is directly proportional
to absolute temperature of a fi xed mass of gas at constant
volume.

P ∝ T
 P = kT
 where k is a constant
  P = pressure (Pa)
  T = absolute temperature (K)
 As such, P

T
 = k

If a gas experiences change in pressure and temperature
from condition 1 to condition 2,

since P
T  = k, condition 1 of gas:

P1

T1
 = k

condition 2 of gas:
P2

T2
 = k

therefore,
P1

T1
 =

P2

T2

Gay-Lussac’s Law

http://bt.sasbadi.com/p4160b

HISTORY
INTEGRATION OF

Joseph Louis Gay-Lussac
(1778–1850) is a French
physicist and chemist who
made quantitative investigation
about the characteristics of
gas. He also investigated the
magnetic fi eld of the Earth and
composition of the atmosphere
at high altitudes. In addition, he
found two new elements, boron
and iodine.

http://bt.sasbadi.com/p4160a

INFOSMARTSMART

For Gay-Lussac’s Law, volume
is constant.

T

P

P
T  = constant
P1

T1
 =

P2

T2

160160 4.4.44.4.4
Figure 4.34 shows a fi xed mass of gas being heated at constant volume. When the temperature
of the gas increases, average kinetic energy of its molecules increases, and the molecules move
with higher velocity. As the volume of gas does not change, the rate of collision of gas molecules
with the walls of the container increases. Force per unit area on the wall of the container also
increases. As such, gas pressure increases.

Temperature increases

Key
Low velocity
High velocity

Figure 4.34 Fixed mass of gas heated at constant volume

161

Example 1

Gas in a closed steel cylinder has a pressure of 180 kPa at a temperature of 25°C. What is the
gas pressure when the cylinder is heated to a temperature of 52°C?

Solution:

Step �
List the given information in
symbols.

P1 = 180 kPa
P2 = Final pressure of air
T1 = (25 + 273) = 298 K
T2 = (52 + 273) = 325 K

Step �
Identify and write down the
formula used.

Volume of gas is constant.
Gay-Lussac’s Law formula is used.
P1

T1
 =

P2

T2

Step �
Substitute numerical values into
the formula and perform the
calculations.

180
298  =

P2

325

P2 = 180 × 325
298

= 196.3 kPa

1
4
2
4
3

1
4
2
4
3

1
4
2

4
3

161161

Chapter 4 Heat

4.4.44.4.4
162162

Example 1

Example 2

Photograph 4.8 shows a syringe with its nozzle closed. Air in the
syringe has an initial volume of  7.5 cm3 and pressure of 105 kPa.
Th e air is compressed to a volume of 2.5 cm3. What is the pressure
of the air?

Solution:

Step �
List the given information
in symbols.

P1 = 105 kPa
P2 = compressed air pressure
V1 = 7.5 cm3

V2 = 2.5 cm3

Step �
Identify and write down the
formula used.

P1V1 = P2V2

Step �
Substitute numerical values into
the formula and perform the
calculations.

105 × 7.5 = P2 × 2.5

P2 = 105 × 7.5
2.5

= 315 kPa

Air of volume 0.24 m3 in an expandable cylinder is heated from 27oC to 77oC at constant
pressure. What is the volume of the air at 77oC?

Solution:

V1 = 0.24 m3

V2 = Final volume of air
T1 = (27 + 273)

= 300 K
T2 = (77 + 273)

= 350 K

V1

T1
 =

V2

T2

0.24
300  =

V2
350

V2 = 0.24 × 350
300

= 0.28 m3

Photograph 4.8

Solving Problems Involving Pressure, Temperature and Volume of a Fixed Mass of
Gas Using Formulae from the Gas Laws

1
4
2
4
3

1
2
3

1
4
2
4
3

162162 4.4.54.4.54.4.5
Formative Practice 4.44.4

Thermometer

Ice

Retort
stand

Bunsen
burner

Water
Pressure gauge

Rubber tube

Round-
bottom
flask

Wire
gauze
Tripod
stand

Figure 4.35

163163

Initial pressure and temperature of air in the tyre of a car are 210 kPa and 25°C respectively.
Aft er a journey, the air pressure in the tyre is 240 kPa. Calculate the temperature of the air in
the tyre in °C.

Solution:

Assume the volume of the tyre does not change. Gay-Lussac’s Law is used.

P1 = 210 kPa
P2 = 240 kPa

T1 = 25°C + 273
 = 298 K
T2 = Final temperature of air

P1

T1
 =

P2

T2

210
298  = 240

T2

T2 = 240 × 298
210

= 340.6 K

Final temperature of air = 340.6 – 273
 = 67.6°C

1. State the physical quantities that are constant in Boyle’s Law, Charles’ Law and
Gay-Lussac’s Law.

2. A syringe contains  50 cm3 of air at a pressure of 110 kPa. Th e end of the syringe is closed
and its piston slowly pushed until the volume of air becomes 20 cm3. What is the pressure
of the compressed air?

3. An air bubble trapped under a leaf in a lake has a volume of 1.60 cm3 at a temperature of
38°C. Calculate the volume of the bubble if the temperature of the water in the lake drops
to 26°C.

4. Pressure in a gas cylinder is 175 kPa at a temperature of 27°C. Heat from a nearby furnace
causes the gas pressure to increase to 300 kPa. What is the temperature of the gas inside
the cylinder?

5. Figure 4.35 shows an apparatus set up to
study the relationship between pressure and
temperature for air inside a round-bottom fl ask.
(a) Identify four aspects in the apparatus set

up that can jeopardise the accuracy
of the results of the experiment.

(b) Suggest modifi cations to improve
the set up.

Example 3

163163

Chapter 4 Heat

4.4.54.4.5
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164164
SELF REFLECTIONSELF-REFLECTION

1. New things I learnt in this chapter on heat are ✎ .

2. Th e most interesting thing I learnt in this chapter on heat is ✎ .

3. Th ings I still do not fully understand or comprehend are ✎ .

4. My performance in this chapter ,

Poor  1 2 3 4 5   Excellent

5. I need to ✎  to improve my performance in
this chapter.

Download and print
Self-refl ection Chapter 4

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Photograph 1

Performance EvaluationPerformance Evaluation

1. Photograph 1 shows a steam injector machine which can
inject steam into water in a container.
(a) What is the meaning of latent heat?
(b) Explain how water in the container is heated by steam

injected into it.
(c) What is the advantage of heating water using the

injection of steam?

2. Tick (✓) for situations that show thermal equilibrium.

Situation Tick (✓)

(a) A hot object and a cold object placed side by side.

(b) An object is heated by a nearby source of fi re.

(c) Two objects at the same temperature and in contact so that
heat can be transferred between them but without net
heat transfer.

(d) Two objects at the same temperature but separated by a
heat barrier.

165165

Chapter 4 Heat
3. Block A has a high specifi c heat capacity and block B has a low specifi c heat capacity.
If both blocks have the same mass,
(a) which block needs more energy to raise its temperature by 10°C?
(b) which block heats up more quickly if supplied with the same amount of heat? Explain

your answer.

4. (a) Defi ne specifi c latent heat.
(b) Th e mass of a melting ice cube reduces by 0.68 kg. What is the amount of heat

absorbed from the surrounding by the ice cube?
 [Specifi c latent heat of fusion of ice = 3.34 × 105 J kg–1]

5. (a) What is meant by specifi c latent heat of vaporisation?
(b) Figure 1 shows the graph of mass of water, m against time, t when water in a beaker is

heated by a 1 800 W electric heater. At time t = 360 s, the water starts to boil.

Mass, m / g

Time, t / s0

200

360 710

480

Figure 1

Calculate:
(i) mass of water that becomes steam from t = 360 s until t = 710 s.
(ii) specifi c latent heat of vaporisation of water.

6. A gold ring of mass 5.5 g experiences a rise in temperature from 36°C to 39°C. How much
heat energy is absorbed by the ring?

[Specifi c heat capacity of gold = 300 J kg–1 °C–1]

7. Photograph 2 shows the power rating label of an electric kettle.
(a) What is the maximum power of this electric kettle?
(b) Calculate the time taken by this kettle to change

0.5 kg of boiling water at 100°C into steam at 100°C
when the kettle operates at maximum power.

[Specifi c latent heat of vaporisation of water
= 2.26 × 106 J kg–1]

(c) What assumptions are made in your calculations in 7(b)?
Photograph 2

166166
8. Air in the tyre of a car has a pressure of 220 kPa at initial temperature of 27°C. Aft er a
race, the temperature of the air increases to 87oC.
(a) Calculate the air pressure in the tyre aft er the race.

(b) What assumptions did you make in 8(a)?

9. An air bubble is trapped under a leaf fl oating on the water surface of a lake. Th e volume of
the air bubble is 3.6 cm3 when the temperature is 20°C.
(a) What is the volume of the air bubble when the water temperature rises to 38°C?
(b) State three assumptions that need to be made in your calculations in 9(a).

10. Figure 2 shows ice cubes being heated by a 500 W immersion heater for 80 seconds. Th e
melted ice cubes are collected in a beaker.
[Specifi c latent heat of fusion of ice = 3.34 × 105 J kg–1]

Immersion heater

Retort stand

Filter funnelIce
Power supply

Beaker

Water

Figure 2

(a)  What is meant by specifi c latent heat of fusion?
(b)  Why does the temperature remain unchanged when the ice cubes are changing into

liquid?
(c)  Calculate:

(i)  energy absorbed by the ice cubes.
(ii) mass of melted ice cubes.

(d) What assumptions are made in your calculations in 10(c)?

11. An electric kettle is fi lled with 500 g of water at 30°C. Th e power of the heating element of
the kettle is 0.8 kW. Assume that all heat from the heating element is transferred to
the water.

[Specifi c heat capacity of water = 4 200 J kg–1 °C–1]
(a) Calculate:

(i) heat energy needed to increase the temperature of the water to 100°C.
(ii) time taken by the kettle to heat water to a temperature of 100°C.

(b) Why is the handle of the kettle made of plastic?
(c) Why is the heating element of the kettle made of metal?
(d) Th e heating element of the kettle is located at the base of the kettle. Explain why.

167167

Chapter 4 Heat
14.  Khairi orders a cup of coff ee in a restaurant. He fi nds that the coff ee is too hot.
Photograph 4 shows two suggested ways to cool the coff ee.

Method A Method B

Photograph 4

(a) Discuss the suitability of methods A and B to cool the coff ee in the cup.
(b) State your choice. Give reasons for your choice.

IceIce

IceIce

12. A substance has a mass of 250 g. Th e substance loses 5 625 J of heat when cooled. Th ere is
a 25°C drop in temperature.
(a) Calculate the specifi c heat capacity of the substance. Identify the substance based on

Table 4.2 on page 128.
(b) Explain the use of the substance based on its specifi c heat capacity.

13. Photograph 3 shows a bamboo steamer. Amin receives an
order from a supermarket to supply 400 steamed buns per
day. Suggest and explain the design of the steamer needed.
Th e steamer should be durable, and able to prepare a large
quantity of steamed buns in a short time.

Photograph 3

Enrinchment Corner
Enrinchment Corner

168168
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