S54-Physics Chapter 4(HEAT)

Chapter 4: HEAT

Heat Temperature
Form of energy Degree of hotness of an object
Measure in Joule, J Measure in Kelvin, K
Temperature ∝ Average kinetic Energy

4.1 Thermal Equilibrium

Heat Transfer Heat is transferred from hotter object to colder object.
Thermal Contact Heat transfer occurs between two objects in thermal contact.
Thermal Equilibrium Net heat transfer between two objects becomes zero.
No net flow of heat between two objects.

Mechanism of Thermal Equilibrium

Hotter Object X Colder Object
HeaYt

 When two objects are in thermal contact, heat is transferred between them.
 Rate of heat transfer from X (Hotter) to Y (Colder) is higher.
 Temperature of X decreases and temperature of Y increases.
 Heat is transferred in the same rate and no net flow of heat between X and Y.

Thermal Equilibrium occurs.

Application of Thermal Equilibrium

Heating Object Cooling Object
Oven Refrigerator

 Hot air in the oven is in thermal contact with  Food is in thermal contact with the cold air in
cake batter. refrigerator.

 Heat is transferred from hot air to the cake  Heat is transferred from food to air until
batter. thermal equilibrium occurs.

 Cake batter is heated until it is baked.  Temperature of food drops and the food stays
fresh for longer period.
Thermometer
 Heat from patient’s body is in thermal contact Cooling Drinks
with thermometer.  Ice cubes are in thermal contact with the hot
 Heat is transferred from body to thermometer. drinks.
 The reading of thermometer can be  Heat is transferred from drinks to ice cubes.
determined when thermal equilibrium occurs.  Ice cubes absorb the heat and melt until
thermal equilibrium occurs.

1

Liquid-in-glass Thermometer

Working  Bulb of thermometer contains a fixed mass of mercury.
principle  Volume of mercury increases when heat is absorbed.
 Mercury expands and rises in the capillary tube.
Calibrating of  Length of mercury column indicates the temperature reading.
Thermometer

1. Temperature scale is obtained by using two fixed point.

Fixed point Position Value Symbol Procedure

Ice point Lower 0℃ 0 Place thermometer in melting ice.

Steam point Upper 100℃ 100 Place thermometer in boiling water.

2. Divide the length between two points into 100 equals division.
3. Each division is equal to 1℃.

Formula Reading of specific temperature, :

= − 0 × 100℃
100 − 0

Characteristic of 1. Good heat conductor
mercury as a 2. High boiling point (357℃)
liquid-in-glass 3. Expand uniformly when heated
thermometer 4. Opaque (not allow light to pass through)
5. Can be seen easily and not stick to the glass wall of capillary tube

Modification to Sensitivity of Thermometer
increase
sensitivity of Modification Reasons
thermometer
Small bulb Absorb heat in shorter time and respond faster towards

temperature change

Thinner wall Heat transfer faster

Narrower capillary tube Small change in volume produce larger change in length of
mercury column

2

4.2 Specific Heat Capacity

Physical property Heat Capacity Specific Heat Capacity

Symbol C c

S.I. Unit ℃−1/ −1 −1℃−1/ −1 −1

Definition Quantity of heat needed to raise the Quantity of heat needed to raise the
temperature of a substance by 1℃ temperature of 1kg mass of substance by
Formula involved 1℃

Physical quantity = ∆
involved = Quantity of heat supplied (J) = ∆
Calculation of ∆ = Change in temperature (℃)
quantity of heat = Quantity of heat supplied (J)
= ∆ = Mass of substance (kg)
∆ = Change in temperature (℃)

= ∆

Comparison between substances with high/low specific heat capacity

Value of Specific Heat Capacity High Low
Mass Equal
Heat supplied Equal Lower
Heat energy required to increase temperature by 1℃ Higher Higher
Increase in temperature Lower Shorter
Longer Higher
Time required to produce change in temperature Lower Lower
Rate of change in temperature Higher Shorter
Amount of heat stored Longer
Time required to cool down
Conversion of other forms of energy to heat energy

Kinetic Energy = 1 2 = ∆
2
= mass (kg)

= velocity ( −1)

Potential Energy = ℎ = ∆

= mass (kg) = ∆
= gravitational acceleration ( −2)

ℎ = height (m)

Electrical Energy = = ∆
= ower (W)
= time (s)

3

Calculation involve mixing of two substances

Hot Cold
Water Water

Hot Water Region Cold Water Region

Heat Loss Heat Gained

1 = 1 ( 1 − ) 2 = 2 ( − 2)

1 = 2

Phenomenon of Specific Heat Capacity

Phenomenon Sea Breeze Land Breeze

Occur during Day Time Night Time
Direction of breeze
Explanation From sea to land From land to sea
 During day time, land and sea  During night time, land and sea
receive same amount of heat release same amount of heat
energy. energy.

 Land becomes hot faster than sea  Sea becomes cool slower than land
water as sea has a higher specific as sea has a higher specific heat
heat capacity. capacity.

 Air above the land is hotter than air  Air above the sea is hotter than the
above the sea water. air above the land.

 Hot air above the land rises up to  Hot air above the sea rises up to the
the atmosphere whereas the cold atmosphere whereas the cold air
air above the sea moves towards above the land moves towards sea
land as sea breeze. as land breeze.

4

Application of Specific Heat Capacity

Building materials of Warm weather regions Wood has a high specific heat capacity and heat up slowly.
traditional houses in Cold weather regions  Wood can act as a heat insulator from the sun.
various climate zones
Wood act as a heat insulator.
 Heat from the fires lit in the wooden houses cannot
flow out.

Cooking utensils Woks Metal has low specific heat capacity.
 Food can be fried at high temperature in a short time.

Clay pot Clay has high specific heat capacity.
 Food can stay hot for a long time.

Pot Aluminium (body) has low specific heat capacity.
 Pot can heat up quickly.

Plastic (handle) has high specific heat capacity.
 Handle gets hot slowly and is safe to handle.

Car radiator system Engine Cooling Water has high specific heat capacity.
 Act as a cooling agent

-Water is pumped into engine block.
-Water absorbs heat produced from the burning of fuel.
-Hot water flow through radiator and is cooled by the air
drawn in by fan.
-Heat is released when water passes through fan.

Space capsule Outer layer -Space capsule encounters air resistance when entering
Thermal Radiator Cold weather country atmosphere.
-Friction increases temperature and cause capsule to burn.

 Outer layer of capsule made up of high specific
capacity and melting point material.

Water has high specific heat capacity.
 Act as a heating agent

-Hot water can store heat for a longer time with a little drop
in temperature.
-Hot water release warm air inside the house.

Rice Keeper/ Thermal Keep warm High specific heat capacity
Flask/Cooler Bag  Reduce heat loss
 Rate of temperature change lower

Ice cream container Ice cream box Low specific heat capacity
 Get cool faster

5

4.3 Specific Latent Heat

Latent heat Specific latent heat

Heat absorbed or released during change in phase of Quantity of heat required to change the phase of 1kg

matter at constant temperature. substance at constant temperature.

Symbol Reason of Constant Temperature:
Formula
During change of phase, transfer of heat does not affect average
Physical = kinetic energy of molecules.
quantities = Quantity of heat (J)  Melting:
involved = Mass (kg)
Heat absorbed is used to break the bonds between the solid
S.I. unit J kg−1 particles so that the particles are free and able to move
among each other.
Calculation of =  Boiling:
heat energy Heat absorbed is used to break the bonds completely
between the particles and do work against atmospheric
pressure when gas vapour expands into the atmosphere.

Change in phase of matter

Solid Melting Liquid Boiling Gas
Freezing Condensation
Type of specific latent heat
Process Change in phase Condition of latent heat Specific latent heat of fusion
Melting Solid  Liquid Absorbed Specific latent heat of vaporisation
Boiling Liquid  Gas Absorbed Specific latent heat of vaporisation
Condensation Gas  Liquid Released Specific latent heat of fusion
Freezing Liquid  Solid Released

Specific latent heat of fusion, Quantity of heat required to change the phase of 1kg substance from
solid to liquid phase at constant temperature.

Specific latent heat of vaporisation, Quantity of heat required to change the phase of 1kg substance from
liquid to gas phase at constant temperature.

Conversion of electrical energy to heat energy when changing phase

Electrical Energy = = =
= ower (W)
= time (s)

6

Heating Curve

Boiling Point D F
Melting Point C E

B

A

Phase Process Heat energy Kinetic Temperature State of Movement of
AB absorbed energy particles
Solid is heated to Specific heat Increase matter Vibrate in a
melting point. capacity fixed position
Increase Solid

BC B: Solid begins to Solid Molecular

melt. Solid bonds are
liquid
BC: Solid is Specific latent Constant Constant Liquid and weakened and

melting. heat of fusion Liquid molecules are

C: Solid has melted released from

completely. fixed position.

CD Liquid is heated to Specific heat Increase Increase Move freely,

boiling point. capacity randomly

throughout

liquid.

DE D: Liquid begins to Liquid Molecules are
Liquid and gas
boil. Specific latent Gas separated far

DE: Liquid is heat of Constant Constant away from

boiling. vaporisation each other.

E: Liquid has boiled

completely.

EF Gas is heated. Specific heat Increase Increase Gas Move rapidly
and randomly
capacity in all direction

DE (Boiling) will take longer time than BC (Melting). (DE>BC)

 During melting, latent heat is absorbed to weaken the bonds between molecules.
 During boiling, latent heat is absorbed to break the bonds between molecules and do work against the

existence of atmospheric pressure when gas vapours expand into the atmosphere.

7

Cooling curve

A

Boiling Point B C
D
Freezing Point E
F

Phase Process Heat energy Kinetic Temperature State of Movement of
Decrease matter
AB Gas is cooled released energy Gas particles
boiling point. Constant
to Specific heat Decrease Decrease Move rapidly

capacity Constant and randomly
Decrease
in all direction

BC B: Gas begins to Gas Molecular
Gas and liquid
condense. Specific Liquid bonds are
Liquid
BC: Gas is latent heat of Constant formed.

condensing. vaporisation

C: Gas has condensed

completely.

CD Liquid is cooled to Specific heat Decrease Move freely,

freezing point. capacity randomly

throughout

liquid

DE D: Liquid begins to Liquid Molecular

freeze. Specific Liquid bonds are
solid
DE: Liquid is freezing. latent heat of Constant Solid and strengthened.

fusion Solid

E: Liquid has frozen

completely.

EF Solid is cooled. Specific heat Decrease Vibrate in a
fixed position
capacity

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Application of Specific Latent Heat

Application Process Explanation
Cooling drink Melting Melting of ice absorbs a large amount of latent heat of fusion.

Maintaining Melting  Temperature of drink becomes lower.
freshness of fish Condensation Melting of ice absorbs a large amount of latent heat of fusion.
and meat
Steaming food  Food can be kept in low temperature for long period of time.

Sweating Evaporation Water has large specific latent heat of vaporisation.
 Heat from heater changes the water to the steam.
 Steam condenses on the food and releases latent heat directly into
the food enables food to be cooked in a faster rate.

Sweat absorbs large amount of latent heat of vaporisation when
evaporates.

 Body is cooled when heat is removed.

Cooling system in Condensation

refrigerator Evaporation

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

 Cooling agent (gas) releases latent heat of vaporisation through
condensation in condenser.

 Cooling agent (liquid) flows through expansion valve.
 Cooling agent (liquid) evaporates and absorbs latent heat of

vaporisation from inside the refrigerator in evaporator.
 Cooling agent (gas) flows out from evaporator to compressor.

Solving Problem involving Specific Latent Heat

Physical quantity Calculation Mass Temperature State of matter
Change Unchanged
Specific heat capacity = ∆ Fixed
Unchanged Changed
Specific latent heat = Fixed

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4.4 Gas Law

Kinetic Theory of Gas

Quantity Symbol S.I. unit Symbol of Other units Explanation
Pressure P pascal S.I. unit
Pa cm Hg, atm  Gas molecules move randomly.
 Force exerted on wall of container

when gas molecules collide with
the wall and rebound.
 Force per unit area is equal to
pressure of gas.

Temperature T kelvin K ℉, ℃  Average kinetic energy of gas
molecules increases with
temperature.

Volume V (metre)3 3 3 , 3 ,  Gas molecules move freely and
enter the entire space of container.

 Volume of gas same as the volume
of container.

Relationship between Pressure (P), Temperature (T) and Volume (V)
Boyle’s Law

Gay-Lussac’s Law PV Charles’ Law
T

Conversion between degree Celsius (℃) and kelvin (K)

℃ + = K

0℃ = 273K
-273℃ = 0K
27℃ = 300K

Absolute Zero Temperature (0K)

 Volume of gas equal to zero.
 Pressure of gas equal to zero.
 Kinetic energy of particles equal to zero.
 Particle at stationary.

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Gas Law Boyle’s Law Charles’ Law Gay-Lussac’s Law

Definition For a fixed mass of gas, For a fixed mass of gas, volume For a fixed mass of gas,
pressure of gas is inversely of gas is directly proportional to pressure of gas is directly
proportional to its volume its absolute temperature when proportional to its absolute
when its temperature is kept its pressure is kept constant. temperature when its
constant. volume is kept constant.

Relationship 1 ∝ ∝
∝

Constant T P V

Formula 1 1 = 2 2 1 = 2 1 = 2
1 2 1 2

Graph T must be in unit kelvin (K)

P

0 0 T/K 0 T/K
P

1 T/℃ -273℃ T/℃
0 -273℃

Explanation -Volume of container -When heat supplied, average -Temperature increases when
decreases. kinetic energy of molecules average kinetic energy of
increases. molecules increases.
-Number of molecules per
unit volume increases. -Temperature increases when -Speed of gas molecules
kinetic energy increases. increases.
-Frequency of collisions
increases. -Gas expands when temperature -Frequency of collisions
increases. increases.
-Force per unit area
increases. -Fast moving molecules move -Force per unit area increases.
in a bigger space.
-Pressure of gas increases. -Pressure of gas increases.
-Volume of gas increases.

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Formula Summary & Calculation

= − × ℃
−

The length of the liquid column is 0.5cm at 0℃ and 15.0 − 0.5
20.5cm at 100℃. Calculate the temperature shown by = 20.5 − 0.5 × 100℃

the thermometer when the length of liquid column is

15.0cm. = 72.5℃

= ∆

Calculate the heat capacity of object X when 2400J of
heat energy is supplied to raise the temperature of = ∆
object X from 20℃ to 35℃.
2400
= (35 − 20)

= 160 ℃−1

= ∆

Calculate the raise in temperature when 10kg of water
∆ =
is heated by 84000J of heat energy.
[Specific heat capacity of water= 4200 −1℃−1] 84000
= 10(4200)

= 2℃

= ∆


Calculate the specific heat capacity of a bullet when it 1 2 = ∆
hits a sand bag at − and the temperature rises 2

by 60℃. 2
= 2(∆ )

6002
= 2(60)

= 3000 −1℃−1

= ∆

Calculate the specific heat capacity of a coconut if it ℎ = ∆

falls from a height of 30m on the ground and cause ℎ
the temperature to rise by 2℃. = ∆
[g= 10 −2 ]
10(30)
=2
= 150 −1℃−1

12

= ∆

Calculate the power of heater required to heat 5kg of = ∆
∆
water for 5 minutes to increase its temperature by
=
20℃. 5(4200)(20)
[Specific heat capacity of water= 4200 −1℃−1]
= (5 × 60)

= 1400

=
( − . ) = ( . − )

A container contains 200g of water at initial 1 ( 1 − ) = 2 ( − 2 )
0.1(450)(60 − ) = 0.2(4200)( − 30)
temperature of 30 ℃ . An iron nail of 100g at
2700 − 45 = 840 − 25200
temperature of 60℃ is immersed in the water. What 885 = 27900
= 31.53℃
is the final temperature of water?
[Specific heat capacity of water = 4200 −1℃−1,
Specific heat capacity of iron = 450 −1℃−1]

=

Calculate the heat energy required to change 4kg of =
= 4(2.26 × 106)
water to steam at 100℃. = 9.04 × 106
[Specific latent heat of vaporisation of water =
2.26 × 106 −1]

=

50W electric heater is used to change 40g water to =
steam in 30 minutes at 100℃. Calculate the specific
latent heat of vaporisation of water.
=
50(30 × 60)

= 0.04
= 2.25 × 106 −1

= +
= ∆ +

What is the amount of heat needed to change 0.75kg = ∆ +
= 0.75(4200)(100 − 30) + 0.75(2.26 × 106)
of water at 30℃ to steam at 100℃? = 1.92 × 106
[Specific heat capacity of water = 4200 −1℃−1,

Specific latent heat of vaporisation of water = 2.26 ×
106 −1]

= Temperature of water increase (30℃ → 100℃)
= Phase change (Water → Steam) ( 100℃)

13

=

A balloon of volume has a pressure of . 1 1 = 2 2
3(28) = 2(21)
After being compressed, its volume changes to
21 . Find the final pressure of the balloon. 3(28)
2 = 21

= 4

=


A bubble in a water has a volume of . at a 1 = 38℃ = 311K
temperature of 38 ℃. Calculate the volume of the
bubble if the temperature of the water drops to 26℃. 2 = 26℃ = 299K

1 = 2
1 2

1.60 2
311 = 299

299(1.60)
2 = 311

= 1.54 3

=


Given pressure in a gas cylinder is 175 at 1 = 27℃ = 300K
temperature of 27 ℃ . When pressure increase to
, what is the temperature inside the cylinder? 1 = 2
1 2

175 300
300 = 2

300(300)
2 = 175

= 514.29K

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Objective Question

1. Hot milk is poured into a cold metal bowl. Which statement is true about heat flow?
(2019 MRSM TRIAL SPM) (CHAPTER 4.1)
A. The net heat flow is zero.
B. Heat flows only from metal bowl to hot milk.
C. Heat flows only from hot milk to metal bowl.
D. Net heat flow is from hot milk to metal bowl.

2. Diagram 1.1 shows a jar of hot ginger tea on the table is left to cool.

Diagram 1.1
Which statement below is correct about the tea?
(SPM 2012) (CHAPTER 4.1)
A. Heat lost hot tea is equal to heat lost by the surrounding.
B. Heat loss by hot tea is same as heat received by the surrounding
C. Heat loss by hot tea to surrounding is lower than heat received by the surrounding.
D. Heat loss by hot tea to surrounding is greater than heat received by the surrounding.

3. If a metal cube with an initial temperature 80oC is immersed in water at 40oC. What is the temperature
of water when thermal equilibrium is reached between the metal cube and water?
(2017 MRSM TRIAL SPM) (CHAPTER 4.1)
A. Higher than 80oC
B. Lower than 40oC
C. Same as room temperature
D. Between 40oC and 80oC

15

4. Diagram 1.2 shows 1 kg of liquid is heated using an electric heater for 10 minutes.

Diagram 1.2
Water temperature rises faster by
(SPM 2015)(CHAPTER 4.2)
A. replacing water with seawater
B. reducing heating time
C. reducing mass of water
D. remove the insulator

5. Diagram 1.3 shows a food is steamed.

Diagram 1.3
What concept is involved? (2020 MRSM SPM TRIAL) (CHAPTER 4.3)

A. Latent heat of vaporisation
B. Latent heat of fusion
C. Specific heat capacity
D. Thermal equilibrium

16

6. The specific heat capacity of water is 4200 Jkg-1oC-1.
This means that 4200J heat energy is required to
(TRIAL SPM 2020 KELANTAN) (CHAPTER 4.2)
A. Change the temperature of 1kg of water by 1oC
B. Change the temperature of 1kg of ice by 1oC
C. Change 1 kg of water to steam at 100oC
D. Change 1 kg of ice to water at 0℃

7. Calculate the time taken by an 800W immersion heater to change 1 kg of ice at 00C to water at 60oC.
(Specific heat capacity of water = 4200J, specific latent heat of fusion of ice = 3.4 x 105 Jkg-1)
(TRIAL SPM 2020 KELANTAN) (CHAPTER 4.3)
A. 315s
B. 425s
C. 740s
D. 1080s

8.

Diagram 1.4
Diagram 1.4 shows a temperature time graph of a liquid from its melting point. The liquid is heated by
using a 0.4 kW heater while its mass is 0.4kg.
What is the specific heat capacity of the liquid?
(TRIAL SPM 2019 PULAU PINANG) (CHAPTER 4.2)

A. 0.5 Jkg-1 0C-1
B. 50 Jkg-1 0C-1
C. 500Jkg-1 0C-1
D. 5000Jkg-1 0C-1

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9.

Diagram 1.5
Diagram 1.5 shows a graph of heating curve of a solid.
What is the time taken for all the solid to melt?
(TRIAL SPM 2019 SELANGOR) (CHAPTER 4.3)
A. 2s
B. 3s
C. 4s
D. 9s
10.

Diagram 1.6
Diagram 1.6 shows formation of sea breeze.
Which statement are true?
(2019 MRSM TRIAL SPM) (CHAPTER 4.2)
A. Speed of cool air is higher than the hot air.
B. The density of air above the land is higher than density of air above the sea water.
C. The temperature of land become lower than the temperature of sea water.
D. The specific heat capacity of sea water is higher than the specific heat capacity of land.

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11. Table 1.1 shows some pot with different specific heat capacity.
Which pot has the highest increase in temperature?
(TRIAL SPM 2015 KEDAH) (CHAPTER 4.2)

Cooking pot Specific heat capacity
Pot A 980
Pot B 500
Pot C 700
Pot D 120
Table 1.1
A. Pot A
B. Pot B
C. Pot C
D. Pot D

12. A piece of ice is added into 200g of water at a temperature of 30o C.After the ice has completely melted,the
temperature of water is 20oC.what is the mass of the ice?
(Specific heat capacity of water=4200Jkg-1 0C-1, specific latent heat of ice= 3.34 x 105 J kg-1)
(TRIAL SPM 2015 KEDAH) (CHAPTER 4.3)

A. 18.3g
B. 20.1g
C. 25.1g
D. 100.0g

13. At 27o C, a hot air balloon has a pressure of 100kPa. When the balloon reached certain height, the pressure
decreases to 95 kPa. What is the temperature at that height?
(TRIAL SPM 2016 SARAWAK) (CHAPTER 4.4)

A. -3oC
B. 3o C
C. -12o C
D. 12 o C

14. The air pressure in a tyre is 25kPa at 27o C. After a long journey, the air temperature in the tyre is 870 C.
What is the air pressure in the tyre now?
(TRIAL SPM 2014 SABAH) (CHAPTER 4.4)

A. 30 kPa
B. 23 kPa
C. 25 kPa
D. 15 kPa

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15.

Diagram 1.7 Diagram 1.8

Diagram 1.7 and Diagram 1.8 show a balloon in the air and a balloon submerged in water. If both
balloon are same, which physic laws can be used to explain the situation above?
(TRIAL SPM 2018 KEDAH) (CHAPTER 4.4)

A. Snell’s Law
B. Hooke’s Law
C. Boyle’s Law
D. Charles’ Law

16.

Diagram 1.9
Diagram 1.9 shows an experiment to investigate a gas law. Which gas law used the apparatus shown?
(2018 MRSM TRIAL SPM) (CHAPTER 4.4)
A. Boyle’s Law
B. Gay-Lussac’s Law
C. Charles’s Law

20

17. A gas of volume of 20 m3 at 27o C is heated until its temperature reached 77o C at constant pressure.What
is the change in the gas volume?
(TRIAL SPM 2016 SARAWAK) (CHAPTER 4.4)
A. Remain unchanged
B. Increase 3.3 m3
C. Decrease 3.3 m3
D. Increase 10 m3

18. Which of the following graphs shows Boyle’s Law?
(TRIAL SPM 2017 PAHANG) (CHAPTER 4.4)

19. A closed container contains gas at temperature 20o C, and a pressure of 100 kPa. If the gas is heated to
100 o C, what is the pressure of the gas?
(TRIAL SPM 2019 NEGERI SEMBILAN) (CHAPTER 4.4)
A. 227.3 kPa
B. 500.0 kPa
C. 112.5 kPa
D. 127.3 kPa

20. The volume of helium gas inside the balloon at room temperature 40o C is 350 cm3. If the volume of the
helium gas increase to 400 cm3, what is the temperature?
(TRIAL SPM 2019 PAHANG) (CHAPTER 4.4)
A. 34.6o C
B. 87.7o C
C. 33.4o C
D. 67.2o C

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SECTION A

1. Diagram 1.0 shows an uncalibrated mercury thermometer placed in a beaker containing ice cubes and
water at 0℃. The length of mercury column, = 6 .
Diagram 1.1 shows the length of mercury column, = 20 when the ice cubes and water in the beaker
are boiled until 100℃.
Diagram 1.2 shows the length of mercury column, = 16 when the thermometer is placed in a hot
liquid with temperature, ℃.
(TRIAL SPM 2018 JOHOR)

Diagram 1.0 Diagram 1.1

Diagram 1.2

(a) Complete the following sentence by ticking the correct box.
Working principle of thermometer is
( ) Thermal equilibrium
( ) Force in equilibrium

(b) (i) Based on the Diagram 1.0 and Diagram 1.1, determine the difference in length of mercury
columns.

(1 mark)
(ii) Based on Diagram 1.2 and answer in (b)(i), calculate the temperature, of the hot liquid.

(2 marks)

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(iii) What will happen to the boiling point of the water if it is boiled on the top of the mountain?
____________________________________________________________________________
____________________________________________________________________________
(2 marks)

(c) Diagram 1.3 shows the design of a thermometer which can be used to determine the body
temperature of a patient.

Diagram 1.3
Table 1.1 shows several characteristics of different thermometers.

Table 1.1
Based on the Table 1.1,
(i) Choose the suitable range of thermometer.
__________________________________________________________________________ (1 mark)
(ii) Give 1 reason for answer in (c)(i).
__________________________________________________________________________ (1 mark)
(iii) Choose the suitable diameter of the capillary tube of the thermometer.
__________________________________________________________________________ (1 mark)
(iv) Give one reason for the answer in (c)(iii).
__________________________________________________________________________ (1 mark)

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(v) Choose the suitable accuracy of the thermometer.
__________________________________________________________________________ (1 mark)
(vi) Give one reason for the answer in (c)(v).
__________________________________________________________________________ (1 mark)
(vii) Based on the answer in (c)(iii) and (c)(v),choose the most suitable thermometer which can be used
to determine the body thermometer of a patient.
__________________________________________________________________________ (1 mark)

2. Diagram 2.1 and Diagram 2.2 show two sets of apparatus used to determine the specific heat capacity of
the aluminium block.
(TRIAL SPM 2017 PAHANG)

Diagram 2.1 Diagram 2.2

(a) What is the meaning of specific heat capacity?

_____________________________________________________________________________________(2 marks)

(b) Based on Diagram 2.1 and Diagram 2.2, state the suitable characteristics of the arrangement of the
apparatus to increase the accuracy of determining the specific heat capacity of the aluminium block. Give
reason for the suitability of the characteristics
(i) The type of plate to be used as the base. State reason.

_____________________________________________________________________________________

______________________________________________________________________________ (2 marks)

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(ii) The type of liquid poured in the hole. State reason.

______________________________________________________________________________

______________________________________________________________________________ (2 marks)

(iii) Material used to wrap the aluminium block. State reason.
_____________________________________________________________________________________
______________________________________________________________________________ (2 marks)

(c) The aluminium blocks in both diagrams have 1 kg mass and being heated by using the electric heater of power
200 W for 4 minutes. The increasing of temperature in Diagram 2.4 is 30℃ whereas in Diagram 2.5 is 50℃.
Calculate the specific heat capacity of the aluminium blocks in:
(i) Diagram 2.1

(2 marks)

(ii) Diagram 2.2

(2 marks)
(d) Based on your answer in (c)(i) and (c)(ii), determine the most suitable apparatus that can give an accurate

result to determine the specific heat capacity of the aluminium block.
______________________________________________________________________________
______________________________________________________________________________ (1 mark)

25

3. Diagram 3.1 shows the graph of temperature against time for water when heated.
(TRIAL SPM MRSM 2019)

Diagram 3.1
(a)(i) State the boiling point.
______________________________________________________________________ (1 mark)

(ii) Lots of salt is added to the water. Draw the new heating curve on Diagram 2.6.

(2 marks)
(iii) Explain your answer.
____________________________________________________________________
____________________________________________________________________ (2 marks)
Diagram 2.7 shows a basic refrigeration cycle of the air conditioner in a house.

Diagram 3.2

26

The basic regeneration cycle of this air conditioner contains compressor, condenser and evaporator for
the vaporisation and condensation process.
(b) What is meant by condensation?

______________________________________________________________________
______________________________________________________________________ (1 mark)
(c) K, L and M are three refrigerants that can be used in the air conditioner above.
Table 2.2 shows the characteristics of those refrigerants.

Table 1.2
(i) Calculate the specific latent heat of vaporisation for each of the refrigerants.

(4 marks)
(ii) State the suitable boiling point of the refrigerant.
_________________________________________________________________ (1 mark)
Reason
_________________________________________________________________ (1 mark)
(iii) Based on the answer in (c)(i) and (c)(ii), choose the most suitable refrigerant.
_________________________________________________________________ (1 mark)

27

4. Diagram 4.1 and Diagram 4.2 show an experiment is carried out to investigate a law of gas. A column of
air is trapped in the capillary tube and is heated for a period of time.
(TRIAL SPM 2020 MELAKA)

Diagram 2.8 Diagram 2.9

(a) What is meant by heat?

_____________________________________________________________________

_____________________________________________________________________ (1 mark)

(b) Based on Diagram 2.8 and Diagram 2.9, compare
(i) the length of the air trapped in the capillary tube

_____________________________________________________________________ (1 mark)

(ii) the temperature of the air trapped in the capillary tube

_____________________________________________________________________ (1 mark)

(iii) Name the physical quantity represented by the length of the air trapped.

_____________________________________________________________________ (1 mark)

(c) Using your answer in (b), state the relationship between the physical quantity in (b) (iii) and the
temperature.

_____________________________________________________________________ (1 mark)

(d) Name physical law involved in (c).

_____________________________________________________________________ (1 mark)

28

(e) Diagram 4.3 shows a graph obtained from the experiment. The temperature, is -273℃ when it is
extrapolated and intersect x-axis.

Diagram 4.3
(i) What is the name given to the temperature, ?
________________________________________________________________________ (1 mark)
(ii) State one characteristic of the gas at temperature, ?
________________________________________________________________________ (1 mark)
5. Diagram 5.1 and Diagram 5.2 shows the arrangement of apparatus used in experiment regarding a
physic law.
(TRIAL SPM 2013 PAHANG)

(a) State the function of Bourdon Gauge.

_________________________________________________________________________ (1 mark)

(b)Based on Diagram 5.1 and Diagram 5.2, compare
(i) The reading of Bourdon Gauge

_________________________________________________________________________ (1 mark)

(ii) The volume of the air in the airtight container.

______________________________________________________________________ (1 mark)

29

(iii)The temperature of the air in the airtight container.
______________________________________________________________________ (1 mark)
(c) Based on the answers in (b)(i) and (b)(ii) ,state the relationship between the reading of Bourdon
Gauge and the volume of trapped air.
______________________________________________________________________
______________________________________________________________________ (1 mark)
(d)Name the physic law involved in (c).
______________________________________________________________________ (1 mark)
(e)Explain why the air bubbles formed by a fish expand as they float towards the surface of water.
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________ (2 marks)

30

Section B

6. Diagram 6.1 shows two identical metal balls heated for a few minutes in boiling water.
The two metal balls are then transferred into two beakers containing liquid X and liquid Y.
Diagram 6.2 shows the initial and final readings of thermometers in liquid X and liquid Y.
(SBP TRIAL SPM 2018)

Diagram 6.1

Diagram 6.2
Table below shows the mass and specific heat capacity of the liquid X and liquid Y.

Table 6.1

(a) (i) What is the meaning of specific heat capacity?
[1 mark]

(ii) Explain in term of specific heat capacity, why the final temperature of liquid X is higher than liquid Y.

[4 marks]
(b) (i) Based on the information given in the Diagram 6.2 and Table 6.1, calculate the amount of heat energy
involved for liquid X and liquid Y.

[4 marks]

(ii) State one assumption that you have made in 7(b)(i).

[1 mark]

31

(c) You are required to determine the most suitable specifications for a design of a pressure cooker. Explain the
suitability of each specification. Choose the most suitable pressure cooker and give reasons for your choice.
(10 marks)

Table 6.2

32

Section C

7. Diagram 7.1 shows two test tubes filled with aluminium dust and copper dust of the same mass. Both
test tubes are heated with the same rate of heat energy.
(SBP TRIAL SPM 2018)

Diagram 7.1

Table 6.1 shows the temperature before and after heating, and the specific heat capacity of aluminium
dust and copper dust.

Temperature before Temperature after Specific heat capacity/Jkg−1°C−1

heating /°C heating/°C

Aluminium dust 27 53 900
77 380
Copper dust 27 Table 7.1

(a)(i) What is the meaning of specific heat capacity?

[1 mark]

(ii) Based on the informations for Diagram 7.1 and in Table 7.1, compare the amount of heat supplied, the
specific heat capacity, the final temperature and change of temperature of aluminium dust and copper dust.
Relate the change of temperature and the specific heat capacity.

[5 marks]

(b) Explain why the body of a cooking pot is made of good heat conductor whereas the handle of the pot is
made of poor heat conductor.

[4 marks]

33

(c) Diagram 7.2 shows an aluminium pail filled with some ice cubes is used to cool the canned drinks.

Diagram 7.2
Suggest and explain how to produce a portable container that is able to cool canned drinks in a shorter time and
to ensure the canned drinks remain cold for a longer period, based on the following aspects:
(i) quantity of ice
(ii) characteristics of the container
(iii) additional features needed

[10 marks]

34

ANSWER
OBJECTIVE

1D 2B 3D 4C 5A
6A 7C 8C 9B 10 D
11 D 12 B 13 D 14 A 15 C
16 C 17 B 18 C 19 D 20 B

SECTION A

1. (a) Thermal equilibrium 1 1
(b) (i) 20 − 6 = 14 1 1
(ii) 16 − 6 1
20 − 6 × 100℃ 2
1
= 71.43℃ 1
1
(iii) Less than 100℃ 1 1
(c) (i) 30 − 45//small 1 1
1 1
(ii) Normal body temperature is 37℃ 1 1
(iii) Small 1 1
(iv) Not too big// increase sensitivity 1 1
(v) 0.1//small 1 12
(vi) Increase sensitivity/accuracy 1
(vii) Q

Total

2. (a) The quantity of heat required to increase the temperature of 1 kg 22
of substance by 1℃.
1
(b) (i) Polystyrene 12
1
Avoid heat loss to surrounding 12
1
(ii) Oil 12

To produce good thermal contact between aluminium block and 1
12
thermometer
35
(iii) Felt cloth

Avoid heat loss to surrounding

(c) (i) = ∆


= ∆
200(240)

= 1(30)

= 1600 −1℃−1

(ii) 1 2
(d) = ∆
1 1
200(240) 1 13
= 1(50) Total
= 960 −1℃−1

Apparatus used in Diagram 2.2

3. (a) (i) 100℃ 11
(ii) 22
Temperature/℃

100

Time/s
0

(iii) More energy needed to break the bond between particles. 1 1
1
(b) The change of the physical state of matter from gas to liquid phase. 1
(c) (i) 4
1
= 1 2
54 1 1
12
= 0.25
= 216 −1 1
1
= 328 −1 1
1
= 800 −1 1 1
(ii) Low boiling point 1 1
1 1
Easy to change to vapour state 1 1
8
(iii) Refrigerant K
36
Total

4. (a) Heat is a form of energy. 1

(b) (i) 5.2 > 5.1 1

(ii) 5.2 > 5.1 1

(iii) Volume 1

(c) Directly proportional 1
(d) Charles’ Law 1

(e) (i) Absolute Zero 1

(ii) Volume of air is zero//Air molecules are at rest//Kinetic energy is zero 1

Total

5. (a) To measure pressure. 1 1
1
(b) (i) 5.2 > 5.1 1 1
1
(ii) 5.1 > 5.2 1 1
1
(iii) Temperature of air is same. 1 2
8
(c) The higher the reading of Bourdon gauge, the lower the volume of air. 1
1
(d) Boyle’s Law 1
4
(e) -When the air bubbles move upward to water surface, the depth decreases. 1
4
-Pressure decreases as the depth of water decreases causing the volume of 1

air in the air bubbles increases. 1

Total

SECTION B

6. (a) (i) The quantity of heat required to increase the temperature by 1 °C for 1kg 1
(ii) of material.
-Liquid X and liquid Y have the same initial temperature and receive a 1
metal ball of the same temperature. 1
-Liquid X has a smaller specific heat capacity than liquid Y. 1
-The temperature in liquid X increases faster than liquid Y. 1
-The final temperature of liquid X is higher than liquid Y.

(b) (i) Total energy involved in liquid X: 1
= ∆ 1
(ii)
(c) = 0.3 × 2100 −1℃−1 × (40 − 25)℃ 1
= 9450 1
1
Total energy involved in liquid Y:
= ∆ 2

= 0.3 × 4200 −1℃−1 × (30 − 25)℃
= 6300

No heat is released to the environment.

Characteristic Description/ Reason

i. Steel pot Not easily corroded

ii. Has a release valve Release high gas pressure in the pot 2

iii. Has sealer ring Prevent leakage due to high pressure in the 2
iv. Multilayer wall pot 2

To withstand high pressure in the pot

Q model because of the steel pot, has a release valve, has a sealer ring 2 10
and multilayer wall.
20
Total 37

SECTION C 1 1
5
7. (a) (i) Heat energy needed to change the temperature of 1 kg object by 1°C. 1 4
1
(ii) -Amount of heat supplied is same 1 Max.10
-Specific heat capacity of aluminium dust > copper dust 1 20
-Final temperature of aluminium dust < copper dust 1
-Change of temperature of aluminium dust < copper dust
-Change of temperature increases, specific heat capacity decreases

(b) -Good heat conductor has a smaller specific heat capacity. 1

-The body of the cooking pot can be heated up very fast/easy to hot. 1

-Poor heat conductor has a larger specific heat capacity. 1

-The handle may absorb a great amount of heat without a high increase in

temperature//will not become too hot. 1

(c) Suggestion Reason

Add more ice Large mass of ice can absorb more 2

heat from the drinks.

Container made of heat To reduce heat absorbed from the 2

insulator/ polystyrene surrounding.

Container made of substance Heat up slower 2

with high specific heat capacity

Use white container Absorb less heat 2

Low density of container Lighter 2

Cover the container Reduce heat absorbed through 2

radiation from the surrounding.

Total

38