Properties of Matter
Heat and Thermodynamics
Dr. / Mohamed Abdel-Rahman El-Sayed
Head of Nuclear Engineering department
CONTENTS
Part (1) Properties of Matter
 Chapter (1) : Units and Dimensions
 Chapter (2) : Elastic Properties of Matter
 Chapter (3) : Fluid mechanics
Part (2) Heat and Thermodynamics
 Chapter (4): Heat
 Chapter (5): Thermodynamics
Chapter 1
Units and Dimensions
1 Units of Physical quantities
• The study of properties of matter or any other branch in physics starts
by defining a number of fundamental physical quantities as length,
math and time and the unit of measuring each of them.
• Many other quantities called derived quantities can be expressed in
terms of these fundamental quantities, as velocity, pressure and
force.
Systems of units:
The units of the fundamental quantities mass, length and time in the
three systems of units are:
French System
: gm – cm – second.
International System (SI): kg – meter – second.
British System
: Pound – foot – second.
In this book we use only the international system of units of the three
fundamental quantities.
Remarks
It was found that it is necessary to introduce other fundamental
quantities as:
• Absolute temperature in Kelvin
• Electric current intensity in Ampere
• Quantity of matter in mole
• Luminous intensity in Candela
Relation between different systems is important in some engineering
applications to change the unit of a quantity into another unit.
Transformation of units
1 m = 100 cm
1 cm = 10-2 m
1 Angstrom = 1Χ10-10 m
1 m2 = 100Χ100 cm2
1 cm2 = 10-4 m2
1 m3 = 1000 liter
1 Km = 1000 m
1 cm3 = 10-6 m2
1 Gal 3.78 liter
1 mile = 1.609 Km
1 Km = 0.62 mile
1 yard = 0.914 meter
1 yard = 3 ft
1 Kg = 2.2 1b
1 1b = 0.45 Kg
1 Calorie = 4.186 joule 1 inch = 2.54 cm
1 Ton = 1000 Kg
1 KJ = 1000 J
Multiples of 10
1015
peta
P
10-15
femto
f
1012
tera
T
10-12
peco
p
109
giga
G
10-9
nano
n
106
mega
M
10-6
micro
μ
103
kilo
k
10-3
mili
mm
Dimensions:
Dimensional formula is a method to express a derived quantity in terms of
the fundamental quantities whose dimensions are:
Mass : M
Length : L
Time : T
The dimension of a derived quantity can be in general written in the form:
MxLyTz
The power of M, L and T depends on the definition of the physical
quantity.
Example:
What is the dimensional formula of the following derived quantities:
area – density – velocity – acceleration – force – momentum – work
- kinetic energy - potential energy - power and pressure
Quantity
Definition
Units
Dimensions
Mass
-
Kg
M
Length
-
Meter (m)
L
Time
-
Second (sec)
T
Area
A=LχW
m2
L2
Density
ρ =mass/volume
Kg/m3
ML-3
Velocity
v=s/t
m/s
LT-1
Acceleration
a = Δv / Δt
m/s2
LT-2
Force
F=ma
Kg. m/s2 = N
MLT-2
Momentum
P = mass χ velocity
Kg. m/s
MLT-1
Work
W = Force χ
Displacement
N.m = Joule
ML2T-2
K.E
K.E = ½ m v2
Kg.m2/s2 =Joule
ML2T-2
P.E
P.E = m g h
Joule
ML2T-2
Power
P = Energy / time
J/s = Watt
ML2T-3
Pressure
P = Force / Area
N/m2
ML-1T-2
Use of the dimensional formula
1- To check the equation
The units and hence the dimensions of each term of the two sides of the
equation must be the same. Otherwise the equation is not correct.
Example:
Show that the equation describing the change of the velocity of transverse
wave v in a string stretched by a tension force T is given by:
v
T

Where μ is the mass of the string per unit length
The dimensions of the LHS:
LT-1
The dimensions of the RHS:
(MLT-2/ML-1 )1/2 = (L2T-2)1/2 = LT-1
The equation is correct since the two sides have the same dimensions.
Use of the dimensional formula
2- To find new relation containing 3 variables:
The dimensions of any physical quantity can be written in the form of :
MxLyTz
The powers x, y and z can be obtained by equating the powers of M, L
and T of the quantity in the left side and those in the right side , then
solving the three equations.
Example:
The time T of one vibration of the simple pendulum depends on :
(a) the mass of sphere attached to its end (m)
(b) the length of the string (L)
(c) the acceleration of the gravity (g)
Find the equation relating the above quantities if the constant is 2π
T  m x Ly g z
[T ]  M  [L] y [LT -2 ] z
x
 [ M 0 L0T 1 ]  M  [L] y [LT -2 ] z
x
 [ M 0 L0T 1 ]  M  [L] y  z [T] -2z
x
M :  0x
L:  0 y z
T :  1  -2 z
1
2
1
y
2
 z-
 T  2
L
g
It is seen that:
• The mass of the sphere doesn't affect the time of one vibration.
• The period is directly proportional to the square root of the length of the
pendulum.
• The acceleration of gravity on the earth's surface is nearly constant. At the
moon's surface the period is longer than on earth since the gravity is only 1/6
that on earth.
Chapter 2
Elastic Properties of Matter
1- Introduction
• An object is deformed when a contact force is applied on it. A
deformation is a change in the size or shape of the object.
• When the contact forces are removed, an elastic object returns to its
original shape and size. Many objects are elastic as long as the
deforming forces are not too large.
• Any object may be permanently deformed or even broken if the acting
forces are too large.
• Elasticity is the ability of the body to maintain its original shape and
dimensions after removing the external force acting on it.
2- Hooks law for Tensile and compressive forces:
For an object of initial length (L) subjected to a tensile or compressive force
of magnitude “F”, the length changes by an amount “ΔL”.
Hooks law states that “the deformation is proportional to the deforming
forces as long as they are not too large”
F  K L
(1)
where K is a constant, called the spring constant.
3- Tensile Stress and Tensile Strain
• (a) Tensile Stress: It is the force per unit area.
F
Tensile Stress 
A
N
Units : 2
Dimensions : [ML-1T -2 ]
m
• (b) Tensile Strain: It is the ratio between the
extension and the original length.
L
Tensile Strain 
L
No units
Dimensionl ess
• (c) Young modulus of elasticity Y: It is the ratio between Tensile Stress
and Tensile Strain.
F
L
F/A
Y
Y 
A
L
L/L
N
Units : 2
Dimensions : [ML-1T -2 ]
m
Young’s modulus measures the resistance of a solid to a change in its length.
Example:
A man whose weight is 0.8 KN is standing upright. By how much is his
femur shortened compared to when he is lying ?
Assume that the compressive force on each femur is half his weight. The
average cross-sectional area of the femur is 8 cm2 and its length
when lying down is 43 cm. Young modulus for a femur is 9.4×109 Pa.
L  43 cm  0.43 m
A  8 cm 2  8  10 -4 m 2
F  0.4 KN  400 N
F
L
FL
Y
 L 
A
L
YA
400  0.43
 L 
 0.0023 cm
9
-4
9.4  10  8  10
4- Stress – Strain Curve
(I) Ductile Material:
When a load is attached to one end of a thin long
wire its length increases according to the
following graph:
(1) Proportional limit: For small loads, the
stress is directly proportional to the strain tell
the proportional limit.
(2) Elastic limit: The solid returns to its original length when the stress is
removed as long as the stress doesn't exceed the elastic limit. If the stress
exceeds the elastic limit, the material is permanently deformed.
(3) Ultimate strength: Maximum stress that can be withstood without
breaking.
(4) Breaking point: Examples of ductile materials are soft metals such as
gold, silver and copper.
(II) Brittle Materials:
The ultimate strength and the breaking point are close to each other.
Examples: Bone, glass and ceramics.
5- Shear and volume deformations
(I) Shear deformation:
Shear deformation is the result of a pair of equal and opposite forces that act
parallel to two opposite surfaces.
Shear Stress 
Tangential Force
F

Surface Area
A
Shear Strain 
displaceme nt of surfaces
x

 tan 
separation of surfaces
L
The Shear Strain is proportional to the Shear Stress as long as the stress is not
too large. The constant of proportionality is the shear modulus S.
F
x
 S
A
L
(II) Volume deformation – Bulk modulus:
A solid immersed in a fluid (gas or liquid) is
subjected to compression. The fluid pressure P is
the force per unit area which is the volume stress.
F
volume Stress  P 
A
The volume of the solid “V” is compressed and
decreases by an amount ΔV (final volume < initial
volume) so ΔV is negative.
V
V
The volume strain is proportional to the volume stress (as long as the stress is
not too large) and the constant of proportionality is called the bulk
modulus (B).
V
P  - B
V
where ΔP… change in pressure from the atmospheric pressure.
volume Strain 
The negative sign allows “B” to be positive, since ΔV is negative.
(II) Volume deformation – Bulk modulus:
Bulk modulus: it is the ratio between the change in pressure and the
relative change in volume.
- P
B
V/V
Bulk modulus measures the resistance of solids or liquids to changes in
their volume. The negative sign means that the volume decreases as the
pressure increases.
Units : pascal 
N
m2
Dimensions : [ML-1T -2 ]
In solids the value of bulk modulus is very large, which means that the
change in volume requires very high pressure.
In liquids the bulk modulus is also so large such that the liquid is said to be
incompressible.
In gases the value of B is relatively small. It is easy to change the volume of
gas, since the intermolecular force between gas molecules is weak and
the intermolecular spaces are wide.
Chapter 3
Fluid mechanics
1- states of matter
• Ordinary matter is classified into three states (phases): solids, liquids and
gases.
• We mean by fluid, liquid or gas, the substance that flows and has no
definite shape. Fluid takes the shape of the container.
• The streamline flow of fluids of no viscosity in a tube from one section
to another satisfies two equations: Continuity equation and Bernoulli's
equation.
2- Pressure:
F
A
where F…magnitude of the force acting perpendicularly to a surface (N)
A…. Surface area (m2)
Average pressure
Units of P
Pav 
(N/m2) or Pascal (Pa)
1 atm = 1.013×105 Pa
Example:
A woman weighting 534 N. find the average pressure due to its weight on
the floor when
(i) she wears tennis shoes of area 60 cm2.
(ii) she wears spike-heeled shoes of area 1 cm2.
(i ) F  534 N
A  60 cm 2  60  10 -4 m 2
F
534
4
Pav  
 8.9  10 Pa
-4
A 60  10
(ii )
A  1 cm  1  10
2
-4
m
2
F
534
6
Pav  

5.34

10
Pa
-4
A 1  10
1- Pascal's Principle
“ A change in pressure at any point in a confined fluid is transmitted
everywhere throughout the fluid ”.
Hydraulic Lift
The pressure increase ΔP due to the force F1
on the small piston of area A1 is :
P 
F1
A1
This increase in pressure is transmitted
everywhere in the liquid, then
F1
F
 2
A1 A 2
 F2  F1
A2
A1
So, if A2=10 A1 (10 times larger), then F2=10 F1
Since the volume of liquid displaced by each piston is the same, so
A1 d1  A2 d 2
 F2  F1

A2
d
 1
A1 d 2
A2
d
 F2  F1 1
A1
d2
 F1  d 1  F2  d 2
The work done by moving the small piston = the work done by the large piston.
Example:
In a hydraulic lift, if the radius of the small piston is 2 cm, and the radius of
the large piston is 20 cm.
(a) What weight can the large piston support when a force of 250 N is
applied to the smaller piston ?
(b) If the small piston moves 50 cm. what is the distance moved by the large
piston ?
F
F
A
(a ) 1  2  F2  F1 2
A1 A 2
A1


  r    20  10 
  20  10 

 100
  2  10 
A 1   r1    2  10
2
A2
A2
A1
-2 2
-2 2
2
2
-2 2
-2 2
 F2  100 F1  100  250  25000 N
(b) F1  d 1  F2  d 2
 d2 
F1  d 1 250  50

 0.5 cm
F2
25000
4- The effect of gravity on fluid pressure
 The pressure variation in a static fluid is :
P2  P1   g d
where point 2 is a depth “d” below point 1 in a liquid at rest in a
vessel.
 The pressure “ P ” at a depth “ d ” below the surface of a liquid open to
the atmosphere is :
P  Patm   g d
where Patm ….. is the atmospheric pressure.
ρ….. Density of the fluid (ρ = mass/volume)
Example:
A diver swims to a depth of 3.2 m in water ( density = 1000 Kg/m3 ).
What is the increase in the force pushing in on her eardrum compared to
what it was at the lake surface ?
Area of the eardrum is 0.6 cm2.
P2  P1   g d
P  P2  P1   g d
P  1000  9.8  3.2  31.4  10 3 Pa
F  P . A



 F  31.4  10 3 0.6  10  4  1.9 N
5- Archimedes Principle
 When an object is immersed in a fluid, the pressure on
the lower surface of the object is higher than the
pressure on the upper surface. This difference in
pressure leads to an upward force acting on the object,
called the Buoyant force.
Archimedes Principle:
 “ A fluid exerts an upward buoyant force on a submerged object equal in
magnitude to the weight of the volume of fluid displaced by the object.”
FB   g Vsub
where Vsub ….. volume of the submerged part of the object.
ρ….. density of the fluid (ρ = mass/volume)
Example:
What percentage of a floating icebergs volume is above the water ?
The density of ice is 917 Kg/m3
while the density of sea water is 1025 Kg/m3
 At equilibrium, the weight of the iceberg equals the buoyant force.
FB   water g Vsubmerged ..............(1)
weight  m ice  g
 (  ice  Vice )  g ..........(2)
then equating (1) and (2) leads to
Vsubmerged
Vice
 ice
917


 0.895
 water 1025
So, 89.5 % of the ice is below the surface of water and 10.5 % is above
the surface.
6- The Continuity Equation
 The volume of liquid entering the tube equals the
volume of liquid leaving it. The rate of flow is the
volume passing per second.
volume
m3
Q
......
time
s
Q
AL
 Q  Av
t
 The quantity of liquid flowing per second equals the
product of area of the section and the velocity at this
section. According to the continuity equation: for
incompressible fluid,
Q1  Q2
 A1 v1  A 2 v 2
The velocity of liquid is inversely proportional to the area.
where Av is the volume flow rate (m3/s). So, the volume flow rate is
constant (the volume of liquid in = the volume of liquid out)
7- Bernoulli's Equation
 It is one form of the conservation of energy
which in general states that : “ Energy is
neither created nor destroyed but can be
transformed into other forms “.
 The Bernoulli equation, which applied for
non-viscous and incompressible streamline
flow, states that : “ The sum of the pressure,
Kinetic and Potential energy per unit
volume is constant “.
1
1
2
2
P1  gy1  v1  P2  gy 2  v 2
2
2
P ……. pressure (Pa)
ρ ……. fluid density (Kg/m3)
y ……. height above a reference level (m)
v ……. fluid speed (m/sec)
8- Applications of Bernoulli's Equation
(A)- Horizontal Pipe:
y1  y 2
 gy1  gy 2
1
1
2
2
 P1  v1  P2  v 2
2
2
(B)- Pressure at depth y below the surface:
at poit (1)
at point (2)
y1  y
y2  0
P1  P0
P2  P
v1  0
v2  0
 by applying Bernoulli equation
 P  P0  gy
1
y
reference
2
(C)- Torricelli's theorem:
Applying the Bernoulli's equation at point “1” and “2” .
1
1
2
2
P1  gy1  v1  P2  gy 2  v 2
2
2
1
y
2
v2
(D)- The Venturi Meter:
It measures fluid speed in a pipe. A
construction of cross-sectional area A2 is
put in a pipe of normal cross-sectional
area A1.
P1  gy1 
1
1
v1 2  P2  gy 2  v 2 2
2
2
Chapter 4
Heat
Chapter 5
Thermodynamics
Thank you for your
attention!