ENGINEERING SCIENCE

Example 15
In a hydraulic lift, the radius of the piston are 2.5 cm and 10 cm. A car weighing 10 kN is to
be lifted by the force of the large piston.
a) What force, F1 must be applied to the small piston?
b) When the small piston is pushed by 10 cm, how far is the car lifted?

Wcar = 10 kN

h1=10 cm F1

? h2 ?
F2
r1 =r2 2=.52.c5mcm
r2 = 10 cm

Solution :

Given: (input): Radius, r 1 = 2.5 = 0.025 m, Downwards distance, h 1 = 10 = 0.1 m
100
100

10
(output); Radius, r 2 = 100 = 0.1 m, Output force, F 2 = 10 × 1000 = 10 000 N

a) , ?

P 1 = P2

= F 2

A1 A2

= 10 000 Note that,
π (0.025)2 π (0.1)2 , = π r2

10 000
0.00196 = 0.0314

= 318 471.34
0.00196

= .

Chapter 5- Solid and Fluid / Page 194

b) , ?

V1 = V2

A1h1 = A2

0.00196 (0.1) = 0.0314

0.00196 (0.1) =

0.0314

= . @ .

Example 16

A barber ‘s chair rest on a hydraulic piston 10 cm in diameter. The
input side has a piston with a cross-sectional area of 10 cm2, which
is pumped on using a foot pedal. If the chair and the client together
have a mass of 160 kg, what force must be applied to the input
piston ?

Solution :

Given: (output): diameter, d 2 = 10 cm , r2 = 5 = 0.05 m
100

(input): Contact area, A 1 = 10 = 0.001 m2 , mass on input piston, m 1 = 160 kg
1002

, ?

P 1 = P2

F1 = F2

A1 A2

= 160 (9.81) Note that,
0.001 π (0.05)2 , =

= 1569.6 , = π r2
0.001 0.00785

= 199 949.04
0.001

= .

Chapter 5- Solid and Fluid / Page 195

QUICK CHECK

The Figure shows a simple hydraulics system. Assuming there is no friction in
the system ,
a) What is the pressure at A ?
b) What is the pressure at B ?
c) What is the upward force produced?
d) What is the effect on the hydraulics system if the cross sectional area of

the large piston is increased?

Answer: a) P = 200 Pa b) P = 200 Pa c) F = 100 N d) Upward force is increased

Chapter 5- Solid and Fluid / Page 196

5.3 ARCHIMEDES' PRINCIPLE

1. Figure 5.7 shows an object attached to a spring balance. The weight of the object in
air, w1 is greater (10 N) than its weight when immersed in water, w2. The difference
in weight is due to an upward buoyant force acting on the object by the water.

spring balance w2 = 8 N When an object being wholly or
partially immersed in a fluid, the
w1 =10 N object lighter due to Buoyant
Force which is push upward to
the object.

object

fluid

FB = 2N

Figure 5.7: An object immersed in fluid

Therefore,

, = , − , w2

FB = Bouyant Force (Newton, N)

w1 = actual weight of an object (Newton, N)

w2 = apparent weight of an object or weight og
an object in fluid (Newton, N)

Video description: Archimedes'
Principle

Chapter 5- Solid and Fluid / Page 197

2. Archimedes’ principle states that for a body immersed wholly or partially in a
fluid, the upward Buoyant Force acting on the body is equal to the weight of
the fluid it displaces.

Object Fb Liquid displaced
Liquid by the object

Figure 5.8: Liquid displaced by the object

Therefore,

, = ,

= or FB = Bouyant Force (Newton, N)
= m = Mass of liquid displaced (kilogram, kg)

g = gravity acceleration (9.18 ms-2 )

ρ = density of liquid displaced (kg ms-3 )

V = volume of liquid displaced (meter cube, m3)

Why a ship remains upright?

The weight of a ship acts through the ship's centre of
gravity (G). It is counteracted by buoyancy—the force
of displaced water—which acts upward through a
centre of buoyancy (B). When a ship is upright (left),
the forces are in direct opposition. When the ship heels
(right), B shifts to the low side. Buoyancy then acts
through the metacentre (M), a point on the ship's
centreline above G. Encyclopædia Britannica, Inc.

Chapter 5- Solid and Fluid / Page 198

3. Figure 5.8 shows four situations of objects in a liquid .

Buoyant Force, Fb Rising Weight of object is smaller than the Buoyant
Weight of Object, Wo Force : wO < Fb , There is a net upward force
acting on the object. Thus the object rises up.

For totally submerged object, volume of object
is equal to volume of liquid displaces, Vo = Vf

The density of object is less than the density of
the liquid, ρ < ρ . The object will rise up.

Buoyant Force, Fb Sinkin Weight of object is bigger than the Buoyant
Weight of Object, Wo g Force : wO > Fb , There is a net downward force
acting on the object. Thus the object sinks.

For totally submerged object, volume of object
is equal to volume of liquid displaces, Vo = Vf

The density of object is more than the density of
the liquid, ρ > ρ . The object will sink.

Buoyant Force, Fb Floating Weight of object is equal to Buoyant Force :
Weight of Object, Wo wO = Fb , The net force acting on the object is
zero. Thus the object floats.

For partially submerged object, volume of object
is bigger than the volume of liquid displaces,
Vo > Vf

The density of object is less than the density of
the liquid, ρ < ρ

Buoyant Force, Fb Floating Weight of object is equal to Buoyant Force :
wO = Fb , The net force acting on the object is
zero. Thus the object floats.

Volume of object is equal to volume of liquid
displaces, Vo = Vf

Weight of Object, Wo The density of object is equal to density of the
liquid, ρ = ρ

Figure 5.9: Four situations of objects in a liquid

Chapter 5- Solid and Fluid / Page 199

Example 17

A stone weighed in the air and then in the water. What is the bouyant force, FB acting on the
stone?

Spring balance

0.8 N 0.6 N

air water
FB
stone

Solution :
Given: actual weight, w = 0.8 N , apparent weight = 0.6 N , Buoyant Force, ?
= Actual weight − apparent weight
= 0.8 − 0.6
= .

Example 18

Figure shows an object which has a weight of 0.25 N fully immersed in water. Its apparent

weight is 0.22 N. ( Density of water = 1000 kg m-3 )

(Physics SPM Trial Paper 2 2009 MRSM) Water surface

a) What is the buoyant force on the object? Buoyant
b) Determine the volume of the object Force

Solution :

Given: (Object ) : actual weight, w = 0.25 N , apparent weight = 0.22 N

(Water displaced) ; density, ρ = 1000 kg⁄m3

Chapter 5- Solid and Fluid / Page 200

a) Buoyant force, ?
= Actual weight − apparent weight
= 0.25 − 0.22
= .

b) V ? ∗Note that object is submerged, the volume of object is equal to volume of
fluid displaced,

FB = ρ g (fluid displaced)

0.03 = 1000 (9.81)

0.03 = 9810

= 3.058 × 10−6m3 (fluid displaced)

∴ = . × − @ .

Example 19

A stone weights 2.5 N. When it is fully submerged in a
solution, its apparent weight is 2.2 N. Calculate the density
of the solution if its volume displaced by the stone is 25
cm3.
[g = 9.81 ms-2]

Solution :

Given: (Stone) ; actual weight, wo = 0.25 N apparent weight = 0.22 N

(Solution displaced) ; Volume, V = 25 m3 = 25 = 2.5 × 10−5 m3
1003

, ?

FB = g V (solution displaced) FB = Actual weight − apparent weight
0.3 = (9.81 × 2.5 × 10−5) FB = 2.5 − 2.2
0.3 = 2.45 × 10−4 FB = 0.3 N
= . /

Chapter 5- Solid and Fluid / Page 201

Example 20

A rock with a weight of 52 N is found to have an apparent weight of 36 N when completely
submerged in water. The density of water is 1000 kg m-3.
a) Calculate the buoyant force acting on the rock when it is submerged in water.
b) What is weight of the water displaced by the rock?
c) Calculate the volume of the rock in cm3.

Solution :
Given: (Rock) ; actual weight, wo = 52 N , apparent weight = 36 N , volume, v ?

(Water displaced) ; ρ = 1000 kg m −3

a) Buoyant force, ?
= Actual weight − apparent weight
= 52 − 36
=

b) weight of the water displaced = =

c) ? ∗Note that object is submerged, the volume of object is equal to volume of fluid
displaced,

FB = ρg (fluid displaced)

16 = (1000 × 9.81)

16 = 9810
= 1.63 × 10−3 m3
= 1.63 × 10−3 × 1003
= 1630 cm3 (fluid displaced)
= 1630 cm3

Chapter 5- Solid and Fluid / Page 202

QUICK CHECK

A student carried out an experiment as shown in figure below.
a) What was the buoyant force of the water on the object?
b) Find the volume of water displaced.
c) Name the principle used.
(Density of water = 1000 kg m-3)

Weight of water
displaced = 2 N

Answer : a) Fb = 2 N b) v = 2 x10-4 m3 c) Archimedes’s Principle

Chapter 5- Solid and Fluid / Page 203

ACTIVITY CHAPTER 5

1) Compare solids, liquids and gases and relate them to their properties to their
structures.

Answer : Refer notes

2) The average radius of the earth is 6.4 x 106 m. Its mass is 6.0 x 1024

kg. What is its average density? (Notes : V sphere = 4  r3)
3

Answer : = 5 464.15 kg m-3

3) What is the mass of a steel block with a density of 7800 kg/m3? The block is the cube
with sides of 3 cm.

Answer : m = 0.2106 kg

Chapter 5- Solid and Fluid / Page 204

4) A scaffolding pole has an external diameter of 40 mm and internal diameter is 32 mm.
It is made of steel. What is the mass of a 5 meter length of pole?
(notes: Vcylinder = π r2h and ρ steel = 7860 kg m-3)

Answer : m = 17.76 kg

5) The figure shows a container with the mass of 2 kg filled with 0.04 m3 of oil. The
density of the oil is 800 kg/m3.
a) What is the mass of the oil?
b) What is the total weight of the container of oil?
c) Calculate the pressure exerted by the container of oil on the floor.

OIL

CAP ROS

5 cm

8 cm

Answer : a) m = 32 kg b) w = 313.92 N c) P = 784.8 Pa

6) If the diameter of the tip of the thumbtack is 0.2 cm, find the pressure exerted on the tip when

a force of 20 N is applied. Chapter 5- Solid and Fluid / Page 205

6) A book has a mass of 1.3 kg is placed on the table as shown in Figure
below. The area of the book in contact with the table is 0.04 m2.
Calculate the pressure exerted by the book on the table.

table

Answer : P = 318.83 Nm-2 @ 318.83 Pa

7) If the diameter of the tip of the thumbtack is 0.2 cm, find the pressure exerted on the
tip when a force of 20 N is applied.

8) A cylindrical container with a base diameter of 40 cm in Answer : 6.37 x 106 Pa
contact with the floor exerted a pressure with the floor of
8730 Pa on the floor. What is the mass of the container? 40 cm

8730 Pa

Answer : m = 111.83 kg

Chapter 5- Solid and Fluid / Page 206

9) State the definition of pressure, the formula and SI unit of pressure.

Answer : Refer notes

10) A man who has a mass of 60 kg lies on a bed which has 750 nails. The end of each
nail has an area of 8.0 x 10-5 m2. Calculate the pressure acting on each nail.

Answer : P = 9810 Pa

11) The weight of a tractor and its driver is 8000 N. The area of one of the front tyres in
contact with the ground is 1 x 10-2 m2. The area of one of the back tyres in contact
with the ground is 4 x 10-2 m2. If the weight is equally distributed among the tyres,
find the pressure exerted by the front and back tyres.

Answer : Pressure front tyre = 2.0 x 105 Pa , Pressure back tyre = 5.0 x 104 Pa

Chapter 5- Solid and Fluid / Page 207

12) What is the pressure due to sea water at the depth of 100 m below the sea level if
the density of the sea water is 1020 kg m-3 ?

Answer : P = 1 000 620 Pa @ 1.00 x 106 Pa

13) A fish is swimming at a depth of 120 m below the surface of a sea. [The density
of the sea water is 1020 kg m-3; g = 9.81 N kg-1)
a) What is the pressure exerted by the water on the fish?
b) Explain why the fish can withstand this pressure without being crushed?

Answer : a) P = 1.2 x 106 Pa @ 1 200 744 Pa
b) The pressure in the body of the fish is equal to the external pressure acting on the fish. The net force
acting on the fish is zero

Chapter 5- Solid and Fluid / Page 208

Answer : a) 1.2 x 106 Pa @ 1 200 744 Pa
b) The pressure in the body of the fish is equal to the external pressure acting on the fish. The net force

14) What is Pascal Principle ?

Answer : Refer notes

15) A force of 40 N pushes down on the movable piston of closed cyclinder containing
a gas. The piston’s area is 0.5 m2. What pressure does this produced in the gas?
40 N

Piston area = 0.5 m2

Pressur

Answer : P = 80 Pa

16) The pressure of a gas contained in a cylinder with a movable piston is 450 Pa. The
area of the piston is 0.2 m2. What is the magnitude of the force (F) exerted on the
piston by the gas?
F
Piston area = 0.2 m2

450 Pa

Answer : F = 90 N

Chapter 5- Solid and Fluid / Page 209

17) A hydraulic car lift has a pump piston with radius r1 = 0.0120 m. The resultant
piston has a radius of r2 = 0.150 m. The total weight of the car is 2500 N. What
input force is required to stabilize the car?

Input force ?

r1 = 0.0120 m r2 = 0.150 m

Answer : F = 16 N

18) The figure shows a hydraulic lift used to raise 2000 kg car. 250 cm2
a) Name the principle used
b) What minimum force must be applied to the small piston?

4 cm2

Answer : a) Pascal Principle b) F = 313.92 N

Chapter 5- Solid and Fluid / Page 210

19) A force of 10 N is applied to a circular piston with an area of 2 cm2 in a hydraulic
jack. The output piston for the jack has an area of 100 cm2.
a) What is the pressure in the fluid?
b) What is the force exerted on the output piston by the fluid?

Answer : a) P = 50000 Pa b) F = 500 N

20) The face of the piston on the left has an area of 21.2 cm2, and the one on the right
has an area of 63.6 cm2. If the mass on the left is 26.2 kg, how much mass should
be placed on the right to balance the system?

26.2 kg

21.2 cm2 63.6 cm2

Answer : m = 8.73 kg

Chapter 5- Solid and Fluid / Page 211

21) The figure shows an apparatus to study Pascal’s Principle.

F1 = 12 N F2

9 cm2

oil 108 cm2
a) What is Pascal’s Principle?
b) What is the pressure exerted by the small piston on the oil in N cm-2 ?
c) What is the pressure exerted by the large piston?
d) Calculate the force, F2 as the result of the application of the force, F1

Answer : a) Refer notes b) P = 1.33 N cm-2 c) P = 1.33 N cm-2 d) F = 143.64 N

Chapter 5- Solid and Fluid / Page 212

22) The diagram shows a hydraulic system used to raise a load. A force of 50 N is
applied on piston A of cross-sectional area 2 cm2. A load is placed on piston B of
cross-sectional area 15 cm2. (Physics SPM Trial Paper 2 2009 MRSM)

50 N Load

Piston A moves Piston A Piston B
21 cm downwards (2 cm2) (15 cm2)

Liquid

a) Calculate the pressure acting on piston A
b) Calculate the pressure acting on piston B
c) Calculate the weight of the load
d) Calculate the distance moved by piston B if the distance moved by piston A is

21 cm.

Answers : a) P = 250 000 Pa b) P = 250 000 Pa c) w = 375 N d) h = 2.8 cm

Chapter 5- Solid and Fluid / Page 213

23) Refer to the figure , what is the buoyant force acting on the nut when immersed in
water?

Answer : Fb = 0.05 N

24) A ring weight 6.327 x 10-3 N when measured in a air and 6.033 x 10-3 N when
submerged in water.
a) Calculate the buoyant force acting on the ring
b) What is its volume ?

Answer : a) Fb = 2.94 x 10-4 N b) v = 2.99 x 10-8 m3

Chapter 5- Solid and Fluid / Page 214

25) A piece of signboard weighing 28.2 N and 480 kg m-3 is tied by a string to bottom
of a fish tank as shown in the figure.
(Density of water = 1000 kg m-3, g = 9.81 N kg-1)
a) Calculate the volume of the signboard
b) Calculate the buoyant force acting on the signboard
c) What is the tension in the string ?

Answer : a) v = 5.99 x 10-3 m3 b) Fb = 58.76 N c) T = 30.56 N

Chapter 5- Solid and Fluid / Page 215

26) A glass block of volume 500 cm3 is submerged in water. The density of the glass
is 2500 kg m-3 and the density of the water is 1000 kg m-3. Calculate:
a) The actual weight of the glass.
b) The apparent weight when it is submerged.

Answer : a) 12.26 N b) 7.35 N

27) A glass cube of density 2500 kgm-3 and mass 1.25 kg is immersed in a water of
density 1000 kgm-3 . Calculate :
a) The volume of glass cube
b) The mass of water displaced
c) The buoyant force
Gravity g = 9.81 ms-2

Answer : a) v = 5 x 10-4 m3 b) m = 0.5 kg c) Fb = 4.905 N

Chapter 5- Solid and Fluid / Page 216

CONCEPT MAP

An externally States that in
applied enclosed

PRESSURE FLUID PASCAL’S
PRINCIPLE

Is defined Is transmitted uniformly States that an object whether
as the in all completely or partially
immersed in a
Direction

FORCE ARCHIMEDES’
PRINCIPLE
Acting
perpendicular on a Is acted
unit on by a

Area Buoyant force

Which is equal
to the weight of
the

of Displaced fluid
Surface

Chapter 5- Solid and Fluid / Page 217

6 Temperature
and Heat

6.1 The concept of temperature and heat
6.2 Concept of heat energy

Many cooking utensils such as pans and pots are made
of material with high thermal conductivity. This allows
heat to be conducted to the food quickly, enabling
shorter cooking time. Conversely, the handles of
cooking utensils are made of material with low thermal
conductivity which prevents heat from conducting
easily. As such, the handle will not become too hot to
handle.

Chapter 6- Temperature and Heat/ Page 218

CHAPTER Temperature
and Heat
6

Learning Outcomes

A student should be able to:
❖ Define temperature and heat and its unit.
❖ Describe the process of heat transfer.
❖ Define heat quantity, specific heat capacity, latent heat and its unit
❖ Calculate heat energy quantity due to temperature change, = Δ
❖ Calculate heat energy quantity due to phase change, =
❖ Apply the concept and formula in solving problems on specific heat capacity and latent

heat
❖ Calculate heat energy transferred between two objects at different temperature
❖ Calculate temperature at thermal equilibrium

Chapter 6- Temperature and Heat/ Page 219

MIND MAP

TEMPERATURE
AND HEAT

Latent Heat Specific Heat Heat Transfer Thermal
Capacity Equilibrium

= = Conduction Convection



Latent Heat of Fusion Radiation
Latent Heat of Vaporisation

KEYWORDS Latent Heat – Haba pendam pelakuran
Radiation – Sinaran
Absorbed – menyerap
Conduction – konduksi Released - melepaskan
Convection – perolakan
Copper – tembaga Specific heat capacity- muatan haba
Equilibrium temperature – suhu tentu
keseimbangan Substance – bahan
Gained – diperolehi,dapat Temperature – suhu
Heat – haba Thermal equilibrium – keseimbangan
Heat capacity – muatan haba terma
Heat transfer – penghantaran haba
Increase – meningkat

Chapter 6- Temperature and Heat/ Page 220

6.1 TEMPERATURE AND HEAT

1. The measurement of temperature is part of everyday life. We measure the temperature of
the air outdoor to decide how to dress when going outside, a thermostat measures the air
temperature indoors to control heating and cooling systems to keep our homes and office
comfortable.

2. Heat also important to our lives. Heat of oven is important in baking.
3. Table 6.1 shows the difference between heat and temperature.

Table 6.1: The difference between heat and temperature

TEMPERATURE HEAT

Definition The amount of thermal energy that can
The degree of hotness of a body be transferred from one object to
Type of
Quantity another
Unit
Base quantity Derived quantity
Apparatus
Kelvin (K) or degrees Celsius Joule (J)
(0C) No specific measuring equipment
Can be measured using a
thermometer

Video description: Difference
between heat and temperature

Chapter 6- Temperature and Heat/ Page 221

Thermal Equilibrium

‘ Two objects at thermal equilibrium have the same temperature and there is no net
transfer of heat energy between the objects ’

1. For example : A cup of cold water is poured into the cup of hot coffee. The cold water
warms up while the hot water cools down. After some time, they are at a state known as
thermal equilibrium. The mixture of water are at the same degree of hotness.

2. The mechanism of thermal equilibrium is shown schematically in Figure 6.2 :

A B Before thermal equilibrium At thermal equilibrium

Hot Cold AB AB
object object

Initial state Net transfer of heat No net transfer of heat

Figure 6.2 : The mechanism of thermal equilibrium

BEFORE THERMAL EQUILIBRIUM AT THERMAL EQUILIBRIUM

Object A and B are at different temperature. Object A and B have equal temperature.
The rates of energy transfer are equal.
The rate of energy transfer are different:
a) the heat flow from A to B at a

faster rate
b) the heat flow from B to A at a

slower rate

The hotter object ( Object A) loses There is no net gain or net loss of energy
energy while the colder object ( Object by either object.
B) gains energy.

The hotter object ( Object A) cools down Both of the objects remain at the
while the colder object ( Object B) heats same temperature.
up.

Energy is transferred between the two objects.

Chapter 6- Temperature and Heat/ Page 222

3. Two objects at different temperatures in thermal contact will eventually come to a state
of thermal equilibrium regardless of the:
a) Mass
b) Size
c) Shape
d) Type of material
of the two objects.

Process Of Heat Transfer

There are three methods of heat transfer :

1. When one end of a rod is heated, the atoms ( in the rod above the flame ) vibrate faster,

bump into the neighboring atoms and start them vibrating. In this way, the atoms

conduct heat from the hot end to the cool end. But during the process, the atoms

themselves do not move from one end of the rod to the other.

2. Similarly, heat can be transferred between 2 bodies, e.g. from a hot drink into and

along a metal spoon.

3. Conduction is the transfer of heat as a result of the direct contact of rapidly

moving atoms through a medium or from one medium to another, without

movement of the media.

4. Materials that allow heat to travel through them in this way are called conductors.

Metals are good conductors of heat. Non-metals such as plastic, clay, and wood are

poor conductors.

Chapter 6- Temperature and Heat/ Page 223

5. Convection occurs when a liquid or gas is in contact with
a solid body at a different temperature and is always
accompanied by the motion of the liquid or gas. In the
kettle, hot water rises and cold water descends until all
the water is at the same temperature. In the atmosphere,
convection results in the movement of hot and cold air-
winds.

6. Convection is the transfer of heat by the physical movement of the heated
medium itself. Convection occurs in liquids and gasses but not in solids.

7. The sun cannot transfer heat to earth by conduction
(because there is no physical contact with the earth) or
convection (because there is no liquid or gas between
them). The sun heats our earth by radiation which does
not require contact or presence of any matter between
them.

8. Radiation is the transfer of heat in the form of waves through space (vacuum). Dull
block surfaces are better than white shining ones at absorbing radiated heat.

Video description: Heat transfer

Chapter 6- Temperature and Heat/ Page 224

QUICK CHECK

State the methods of heat transfer for each statement:
a) Energy is transferred by direct contact
b) Energy is transferred by mass motion of the molecules
c) Heat is transferred in the form of waves through space (vacuum)
d) Heat transfer between two bodies need conductors (solid or liquid)

6.2 HEAT ENERGY

Heat Quantity, Q
1. The Heat Quantity is the amount of energy received or given off by a system in a heat

transfer process.

=

2. The unit of heat quantity is Joule
3. When two objects are in equilibrium state, the heat released by hotter object is equal to the

heat absorbed by colder object.

=

Heat Capacity, C
1. The Heat Capacity of a substance is defined as the quantity of heat required to increase

its temperature by 1 0C.


=

Chapter 6- Temperature and Heat/ Page 225

Specific Heat Capacity, c
1. The specific heat capacity of a substance is the quantity of heat needed to increase the

temperature of mass of 1 kg by 10C or 1 K
2. The unit of specific heat capacity is J kg-1 0C-1 or Jg-1 0C-1 or Jg-1 K-1

3. The specific heat capacity of iron is 500 J kg-1 0C-1 . This means that 500 J of heat is required
to raise the temperaure of a 1 kg block of iron through 1OC.

4. Table 6.3 lists the specific heat capacities of some common substances.

Table 6.3: Specific Heat Capacity

SUBSTANCES SPECIFIC HEAT CAPACITY, c
( J kg-1º C-1 )

Ice 2100

Aluminium 900

Concrete 800

Glass 700

Iron 500

Copper 400

Water 4200

Alcohol 2500

Paraffin 2100

Mercury 140

QUICK CHECK

What is meant by “ water has a specific heat capacity of 4200 J kg-1 OC-1” ?

Chapter 6- Temperature and Heat/ Page 226

Example 1

An iron spoon of mass 500 g is heated from 200C to 1000C. How much heat is absorbed by
the iron spoon? (Given that ciron = 452 J kg-1 0C-1)

Solution:

Given: , = 500 = 500 = 0.5

1000

ℎ , = 100 − 20 = 80℃

ℎ , = 452 −1 ℃−1

, ? (ℎ )

=

= 0.5 × 452 × 80

=

Example 2

A thermometer containing 0.10 g of mercury is cooled from 15 oC to 8.5 oC. How much
energy was lost by the mercury in this process? ( CHg = 140 J kg-1 ºC-1)

Solution:
Given: , = 0.10 = 0.1 = 1 × 10−4

1000

ℎ , = 15 ℃ − 8.5 ℃ = 6.5 ℃
ℎ , = 140 −1 ℃−1
, ? (ℎ )
=
= (1 × 10−4) × 140 × 6.5
= .

Chapter 6- Temperature and Heat/ Page 227

Example 3
The amount of heat needed to increase the temperature of piece of marble from 270C to 370C is
2.64 kJ. The mass of the marble is 0.30 kg. What is the specific heat capacity of marble?

Solution:
Given: , = 0.30
ℎ , = 37 ℃ − 27 ℃ = 10 ℃
, = 2.64 = 2.64 × 1000 = = 2 640
, ?

=
2 640 = 0.30 × × 10
2 640 = 3

= − ℃−

Example 4
A copper is given 8190 J of energy and the temperature rises by 15 K. What is the mass of the
copper? The specific heat capacity of copper is 390 J kg-1 K-1

Solution:
Given: , = 8 190 (ℎ )
ℎ , = 15
ℎ , = 390 −1 −1
, ?

=
8 190 = × 390 × 15
8 190 = 5 850
= .

Chapter 6- Temperature and Heat/ Page 228

Example 5

When a mass of 3 kg of water heated from 150C to 1000C in an 800 g aluminum kettle, how
much heat does the kettle and water absorbed?

(Given that cwater = 4.2 x 103 J kg-1 0C-1 and cal = 0.88 x 103 J kg-1 0C-1)

Solution:

WATER HEAT ABSORBED

m = 3 kg ALUMINUM KETTLE
c = 4.2 × 103 J kg−1 ℃−1 m = 800 g = 800 = 0.8 kg
= 100 − 15 = 85 ℃
1000

c = 0.88 × 103 J kg−1 ℃−1
= 100 − 15 = 85 ℃

= ( ) + ( )

= [3 × ( 4.2 × 103) × 85] + [0.8 × (0.88 × 103) × 85 ]
= (1.07 × 106 ) + (5.98 × 104 )
= . ×

QUICK CHECK

1. How much is heat is absorbed by 60 g of copper when its temperature is raised from
200C to 800C ? (Ccooper = 400 J kg-1 0C-1)

Answer: Q = 1440 J

Chapter 6- Temperature and Heat/ Page 229

2. A 4 kJ of energy are given out when 2 kg of metal block is cooled from 50 0C to 40
0C. What is the Specific Heat Capacity of the metal?

Answer: c = 200 J kg-1 0C-1

Example 6

A piece of metal of mass 220 g at 100 0C is immersed in 440 g of water at 15 0C. When thermal
equilibrium is reached, the temperature of the water is 23 0C. What is the specific heat capacity
of the metal? (Given that cwater = 4.2 x 103 J kg-1 0C-1)

Solution:

HEAT RELEASED, METAL HEAT ABSORBED, WATER

= 220 = 220 = 0.22 = 440 = 440 = 0.44

1000 1000

? = 4.2 × 103 −1 ℃−1
= 100 − 23 = 77 ℃ = 23 − 15 = 8 ℃

=
=

0.22 × × 77 = 0.44 × ( 4.2 × 103) × 8
16.94 = 14 784
= . − ℃−

Chapter 6- Temperature and Heat/ Page 230

QUICK CHECK

A silver spoon of mass 50 g is at temperature of 20 0C. This spoon is used to stir
coffee which is at a temperature of 90 0C. After stirring the final temperature
reached by the spoon and coffee are 89 0C. The mass of the coffee is 200 g.
Given that specific heat capacity of the spoon = 230 J / kg 0C
a) How much heat is absorbed by the spoon?
b) Calculate the specific heat capacity of the coffee.

Answer: a) Q = 793.5CJhapbt)erc6=- T3e9m67p.e5raJt/ukrge and Heat/ Page 231

0C

Example 7

A mass of 50 g of aluminum at a temperature of 98 0C is immersed in 80 g of water at 12 0C.
Assuming that no heat is lost to the surrounding, what is the temperature of mixture?
Given that cwater = 4200 J / kg 0C and cal = 880 J / kg 0C )

Solution: HEAT ABSORBED, WATER

HEAT RELEASED, ALUMINUM = 80 = 80 = 0.08

= 50 = 50 = 0.05 1000

1000 = 4200 ⁄ ℃
= − 12
= 880 ⁄ ℃
= 98 −

Qreleased = Qabsorbed
0.05 × 880 × (98 − ) = 0.08 × 4200 × ( − 12)

44 (98 − ) = 336 ( − 12)
4312 − 44 = 336 − 4032
4312 + 4032 = 336 + 44

8344 = 380
= . ℃

Chapter 6- Temperature and Heat/ Page 232

QUICK CHECK

A mechanic dropped a steel nut of mass 0.02 kg and temperature of 90 0C into 0.25
kg of water at 24 0C in a polystyrene cup. What is the temperature when the steel
nut and water have come to thermal equilibrium?
( C water = 4200 J kg-1 0C-1 and C steel = 450 J kg-1 0C-1 )

Answer: =24.56 0C

Chapter 6- Temperature and Heat/ Page 233

Phase Change and Latent Heat

1. Specific latent heat, L is defined as the amount of thermal energy (heat, Q) that is

absorbed or released when a body undergoes a constant-temperature process. The

equation for specific latent heat is:

= where:



• l is the specific latent heat
• Q is the heat absorbed or released
• m is the mass of a substance
2. The most common types of constant-temperature processes are phase changes, such as
melting, freezing, vaporization, or condensation. The energy is considered to be "latent"
because it is essentially hidden within the molecules until the phase change occurs. It is
"specific" because it is expressed in terms of energy per unit mass.
3. The most common units of specific latent heat are J/g and kJ/kg.
4. Specific latent heat is an intensive property of matter. Its value does not depend on sample
size or where within a substance the sample is taken.
5. There are 3 types of latent heat transfer :
• Latent Heat of Fusion: Latent heat of fusion is the heat absorbed or released when matter
melts, changing phase from solid to liquid form at constant temperature.
• Latent Heat of Vaporization: Latent heat of vaporization is the heat absorbed or released
when matter vaporizes, changing phase from liquid to gas phase at constant temperature.
• Sensible Heat: Although sensible heat is often called latent heat, it isn't a constant-
temperature situation, nor is a phase change involved. Sensible heat reflects heat transfer
between matter and its surroundings. It is the heat that can be "sensed" as a change in an
object's temperature
6. Latent heat and sensible heat are two types of heat transfer between an object and its
environment. Tables 6.4 are compiled for latent heat of fusion and latent heat of
vaporization. Sensible heat, in turn, depends on the composition of a body.

Chapter 6- Temperature and Heat/ Page 234

Table 6.4: Specific Latent Heat Values

Material Melting Boiling Specific Latent Specific Latent Heat of
Point (°C) Point (°C) Heat of Fusion Vaporization
Ammonia (kJ/kg)
Carbon Dioxide −77.74 −33.34 (kJ/kg)
Ethyl Alcohol −78 −57 1369
Hydrogen −114 78.3 332.17 574
Lead −259 −253 184 855
Nitrogen 327.5 1750 108 455
Oxygen −210 −196 58 871
Refrigerant134A −219 −183 23.0 200
Toluene −26.6 25.7 213
Water −101 110.6 13.9
−93 100 — 215.9
72.1 351
0 334 2264.705

7. Daily life is filled with examples of latent heat and Sensible Heat:
• Boiling water on a stove occurs when thermal energy from the heating element is
transferred to the pot and in turn to the water. When enough energy is supplied, liquid
water expands to form water vapor and the water boils. An enormous amount of energy
is released when water boils. Because water has such a high heat of vaporization, it's
easy to get burned by steam.
• Similarly, considerable energy must be absorbed to convert liquid water to ice in a
freezer. The freezer removes thermal energy, allowing the phase transition to occur.
Water has a high latent heat of fusion, so turning water into ice requires removal
of more energy than freezing liquid oxygen into solid oxygen, per unit gram.
• Latent heat causes hurricanes to intensify. Air heats as it crosses warm water and
picks up water vapor. As the vapor condenses to form clouds, latent heat is
released into the atmosphere. This added heat warms the air, producing instability

Chapter 6- Temperature and Heat/ Page 235

and helping clouds to rise and the storm to intensify.
• Sensible heat is released when soil absorbs energy from sunlight and gets warmer.
• Cooling via perspiration is affected by latent and sensible heat. When there is a

breeze, evaporative cooling is highly effective. Heat is dissipated away from the
body due to the high latent heat of vaporization of water. However, it's much harder to
cool down in a sunny location than in a shady one because sensible heat from absorbed
sunlight competes with the effect from evaporation.

Example 8
Calculate the amount of heat required to melt 0.3 kg of ice at 0oC
(Specific latent heat of fusion of ice = 3.36 x 105 J kg-1)
Solution :

Given : , = 0.3
ecif ℎ , = 3.36 × 105 −1

, ?

= (phase change from ice to water)

= 0.3 × 3.36 × 105
=

Example 9

Calculate the amount of heat required to boil away 0.25 kg of water at 100oC
(Specific latent heat of vaporisation of water = 2.26 x 106 J kg-1 )

Solution :

Given: , = 0.25
ecif ℎ , = 2.26 × 106 −1

, ?

= (phase change from water to steam)

= 0.25 × 2.26 × 106

Chapter 6- Temperature and Heat/ Page 236

=
Example 10

In a boiler, water boils at 120oC. How much heat must be supplied to 4.0 kg of water at
30oC to change it to steam at the same temperature?
(Specific heat capacity of water, c = 4200 −1 ℃−1
Specific latent heat of vaporisation of water, L = 2.26 x 106 J kg-1)

Solution:
Given : , = 4.0

ecif ℎ , = 2.26 × 106 −1
ecif ℎ , = 4200 −1 ℃−1
ℎ , = 120 − 30 = 90℃
, ?

= mL (phase change from water to steam)
= 4.0 × 2.26 × 106
= 9 040 000 J

= (temperature change from 30oC to 120oC)
= 4.0 × 4200 × 90
= 1 512 000 J

Qtotal = 9 040 000 + 1 512 000
Qtotal = or . x

Chapter 6- Temperature and Heat/ Page 237

QUICK CHECK

How much heat transfer is required to thaw a 0.450 kg package of frozen
vegetables originally at 0oC if their heat of fusion is the same as that of water.
(Specific latent heat of fusion of water = 3.36 x 105 J kg-1)

Answer: Q = 1.51 x 105 J or 151 200 J

Video description: Latent Heat

Chapter 6- Temperature and Heat/ Page 238

ACTIVITY CHAPTER 6

1) Fill in the blank with suitable answer : Kelvin Heat
Temperature Specific Heat Capacity J/kg0C Unit

Quantities Definition

The quantity of heat needed to
increase the temperature of mass of 1
kg by 10C

The degree of hotness of a body

The amount of thermal energy that Joule
can be transferred from one object to
another object

2) A 0.40 kg aluminum teakettle contains 2.0 kg of water at 15.0 0C. How much heat
is required to raise the water to 100 0C? (Cwater = 4.2 x 103 J kg-1 OC-1 )

Answer: Q = 714 000 J

3) Heat is the __________________ transferred from one object to another due to the
_____________ between them. The heat transfer will stop when both objects will
have attained _________________. At this instant, both objects will have the same
_______________________

Answer: Refer notes

Chapter 6- Temperature and Heat/ Page 239

4) State the method of heat transfer in the following box :

Answer: Refer notes

5) State the method of heat transfer in the following box :

Answer: Refer notes

Chapter 6- Temperature and Heat/ Page 240

6) How much heat energy is required to raise the temperature of a 3 kg sheet of glass
from 24 0C to 36 0C? (Specific heat capacity of glass = 840 J kg-1 0C-1)

Answer: Q = 30 240 J

7) A 0.40 kg aluminum teakettle contains 2.0 kg of water at 15.0 0C. How much heat is
required to raise the water to 100oC?
( C water = 4.2 x 103 J kg-1 0C-1 )

Answer: Q = 714 000 J

8) How much heat is required to raise the temperature of 20 g of water from 10oC to
20oC if the Specific Heat Capacity of water is 4.2 x 103 J / kg 0C

Answer: Q = 840 J
Chapter 6- Temperature and Heat/ Page 241

9) A 0.4 kg block of iron is heated from 295 K to 325 K. How much heat had
to be transferred to the iron?
(Specific Heat Capacity of iron = 450 J/kg K)

Answer: Q = 5400 J
10) 2 kg of oil is heated from 30oC to 40oC . What is the amount of heat absorbed by

the oil? (Given : Coil = 8 kJ kg-1 oC-1)

Answer: Q = 160 000 J
11) It takes 880 Joule heat to raise the temperature of 350 g of lead from 0oC to 20oC.

What is the specific heat of lead?

Answer: c = 125.71 J kg-1 ºC-1
Chapter 6- Temperature and Heat/ Page 242

12) 0.5 kJ of heat is needed to raise the temperature of 20 g of silver spoon by 30oC.
What is the specific heat capacity of silver?

Answer : c = 833.33 J/ kg ºC
13) Water has specific heat capacity of 4200 J kg-1 oC-1. 21 kJ of heat is supplied to

the water of mass 1 kg. What is the rise in temperature of the water?

Answer : = 5 oC

14) If 600 J of heat are added to 50 g of water initially at 20 oC, what is the final
temperature of the water? (Specific Heat Capacity of water = 4200 J/kg 0C)

Answer : Tf = 22.86 oC

Chapter 6- Temperature and Heat/ Page 243