PHYSICS 1401
SEMESTER EXAM REVIEW
1.1 Measurements
Micrometer
Vernier caliper
Photogate (millisec)
% error 
theoretica l value - experiment al value
theoretica l value
(100%)
1.2 Resultant and Equilibrant
2.4 Motion Graphs
2.4 Equations of Kinematics for Constant Acceleration
Equations of Kinematics for Constant Acceleration
POSITION, VELOCITY & ACCELERATION
v  vo  at
x
1
2
vo  v  t
v  v  2ax
2
2
o
x  vot  at
1
2
2
3.2 Equations of Kinematics in Two Dimensions
vx  vox  axt
x  voxt  a x t
1
2
x
2
1
2
vox  vx  t
v  v  2a x x
2
x
2
ox
3.2 Equations of Kinematics in Two Dimensions
v y  voy  a y t
y  voyt  a yt
1
2
y
1
2
v
oy
2
 vy  t
v  v  2a y y
2
y
2
oy
3.3 Projectile Motion
Under the influence of gravity alone, an object near the
surface of the Earth will accelerate downwards at 9.80 m/s2.
a y  9.80 m s
2
ax  0
vx  vox  constant
3.3 Projectile Motion
Objects falling in a vacuum will experience the same speed.
Galileo started experimenting to
test the theories of other
scientists such as Aristotle.
3.3 Projectile Motion
Properties of Projectile Motion
1. Horizontal velocity stays constant.
2. No vertical velocity when object is thrown horizontally
from the top of hill.
3. When object is launched from the ground, velocity
has horizontal and vertical components.
4. At the top of the trajectory, no vertical velocity, but
there is acceleration due to gravity.
5. The time for a projectile to reach the top is equal to
the time for it to go back to the ground.
6. The initial launching velocity is equal to the final
lvelocity just before it hits the ground.
3.3 Projectile Motion
y
ay
-1050 m -9.80 m/s2
vy
voy
t
?
0 m/s
14.6 s
3.3 Projectile Motion
y
ay
vy
voy
?
-9.80 m/s2
0
14 m/s
t
3.3 Projectile Motion
Example 7 The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
3.3 Projectile Motion
y
ay
0
-9.80 m/s2
vy
voy
t
14 m/s
?
3.3 Projectile Motion
y
ay
vy
0
-9.80 m/s2
voy
t
14 m/s
?
y  voyt  a yt
1
2
0  14 m s t 
1
2
2
 9.80 m s t
2

2

0  214 m s    9.80 m s t
t  2.9 s
2
3.3 Projectile Motion
Example 8 The Range of a Kickoff
Calculate the range R of the projectile.
x  voxt  axt  voxt
1
2
2
 17 m s 2.9 s   49 m
4.2 Newton’s First Law of Motion
An object continues in a state of rest
or in a state of motion at a constant
speed along a straight line, unless
compelled to change that state by a
net force.
The net force is the vector sum of all
of the forces acting on an object.
If the vector sum is equal to zero,
then the system is in equilibrium.
4.2 Newton’s First Law of Motion
Inertia is the natural tendency of an
object to remain at rest in motion at
a constant speed along a straight line.
The mass of an object is a quantitative
measure of inertia.
SI Unit of Mass: kilogram (kg)
4.3 Newton’s Second Law of Motion
Newton’s Second Law
When a net external force acts on an object
of mass m, the acceleration that results is
directly proportional to the net force and has
a magnitude that is inversely proportional to
the mass. The direction of the acceleration is
the same as the direction of the net force.

a


F
m



F  ma
4.3 Newton’s Second Law of Motion
SI Unit for Force
 m  kg  m
kg   2   2
s
s 
This combination of units is called a newton (N).
4.4 The Vector Nature of Newton’s Second Law
The direction of force and acceleration vectors
can be taken into account by using x and y
components.



F  ma
is equivalent to
F
y
 ma y
F
x
 max
4.5 Newton’s Third Law of Motion
Newton’s Third Law of Motion
Whenever one body exerts a force on a
second body, the second body exerts an
oppositely directed force of equal
magnitude on the first body.
It involves TWO objects to form an
action-reaction pair.
4.6 Types of Forces: An Overview
In nature there are two general types of forces,
fundamental and nonfundamental.
Fundamental Forces
1. Gravitational force
2. Strong Nuclear force
3. Electroweak force
4.6 Types of Forces: An Overview
Examples of nonfundamental forces:
friction
tension in a rope
normal or support forces
4.7 The Gravitational Force
Newton’s Law of Universal Gravitation
Every particle in the universe exerts an attractive force on every
other particle.
He said gravity is universal.
The force that each exerts on the other is directed along the line
joining the particles.
4.7 The Gravitational Force
For two particles that have masses m1 and m2 and are
separated by a distance r, the force has a magnitude
given by
m1m2
F G 2
r
G  6.673 1011 N  m 2 kg 2
4.7 The Gravitational Force
m1m2
F G 2
r

 6.67 10
8
11
 1.4 10 N
N  m kg
2
2

12 kg 25 kg 

2
1.2 m 
4.7 The Gravitational Force
4.9 Static and Kinetic Frictional Forces
When the two surfaces are
not sliding across one another
the friction is called
static friction.
4.9 Static and Kinetic Frictional Forces
The magnitude of the static frictional force can have any value
from zero up to a maximum value.
fs  f
f
MAX
s
0  s  1
MAX
s
  s FN
is called the coefficient of static friction.
4.9 Static and Kinetic Frictional Forces
Note that the magnitude of the frictional force does
NOT depend on the contact area of the surfaces.
4.9 Static and Kinetic Frictional Forces
Static friction opposes the impending relative motion between
two objects.
Kinetic friction opposes the relative sliding motion motions that
actually does occur.
f k   k FN
0  k  1
is called the coefficient of kinetic friction.
4.9 Static and Kinetic Frictional Forces
4.10 The Tension Force
Cables and ropes transmit
forces through tension.
4.11 Equilibrium Application of Newton’s Laws of Motion
Definition of Equilibrium
An object is in equilibrium when it has zero acceleration.

Fx  0

Fy  0
4.12 Nonequilibrium Application of Newton’s Laws of Motion
When an object is accelerating, it is not in equilibrium.
F
x

 max
Fy  may
5.1 Uniform Circular Motion
Let T be the time it takes for the object to
travel once around the circle.
r
2 r
v
T
5.2 Centripetal Acceleration
The direction of the centripetal acceleration is towards the
center of the circle; in the same direction as the change in
velocity.
2
v
ac 
r
5.3 Centripetal Force
Recall Newton’s Second Law
When a net external force acts on an object
of mass m, the acceleration that results is
directly proportional to the net force and has
a magnitude that is inversely proportional to
the mass. The direction of the acceleration is
the same as the direction of the net force.

a


F
m



F  ma
5.3 Centripetal Force
Thus, in uniform circular motion there must be a net
force to produce the centripetal acceleration.
The centripetal force is the name given to the net force
required to keep an object moving on a circular path.
The direction of the centripetal force always points toward
the center of the circle and continually changes direction
as the object moves.
2
v
Fc  mac  m
r
5.7 Vertical Circular Motion
2
1
v
FN 1  mg  m
r
FN 2
FN 4
2
2
v
m
r
2
4
v
m
r
2
3
v
FN 3  mg  m
r
6.1 Work Done by a Constant Force
W  Fs
1 N  m  1 joule J 
6.1 Work Done by a Constant Force
W  F cos  s
cos 0  1
cos 90  0
cos 180  1
6.1 Work Done by a Constant Force
W  F cos 0s  Fs
W  F cos180s  Fs
6.2 The Work-Energy Theorem and Kinetic Energy
THE WORK-ENERGY THEOREM
When a net external force does work on and object, the kinetic
energy of the object changes according to
W  KEf  KEo  mv  mv
1
2
2
f
1
2
2
o
6.3 Gravitational Potential Energy
W  F cos  s
Wgravity  mg ho  h f 
6.3 Gravitational Potential Energy
Wgravity  mg ho  h f 
6.3 Gravitational Potential Energy
Wgravity  mgho  mgh f
DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY
The gravitational potential energy PE is the energy that an
object of mass m has by virtue of its position relative to the
surface of the earth. That position is measured by the height
h of the object relative to an arbitrary zero level:
PE  mgh
1 N  m  1 joule J 
6.3 Gravitational Potential Energy
W  12 mvf2  12 mvo2
mgho  h f    12 mvo2
Wgravity  mg ho  h f 
vo   2 g ho  h f 


vo   2 9.80 m s 2 1.20 m  4.80 m   8.40 m s
6.4 Conservative Versus Nonconservative Forces
Version 1 A force is conservative when the work it does
on a moving object is independent of the path between the
object’s initial and final positions.
Wgravity  mg ho  h f 
6.4 Conservative Versus Nonconservative Forces
Version 2 A force is conservative when it does no work
on an object moving around a closed path, starting and
finishing at the same point.
Wgravity  mg ho  h f 
ho  h f
6.5 The Conservation of Mechanical Energy
THE PRINCIPLE OF CONSERVATION OF
MECHANICAL ENERGY
The total mechanical energy (E = KE + PE) of an object
remains constant as the object moves, provided that the net
work done by external nonconservative forces is zero.
6.5 The Conservation of Mechanical Energy
6.5 The Conservation of Mechanical Energy
Ef  Eo
mghf  mv  mgho  mv
1
2
2
f
1
2
gh f  12 v 2f  gho  12 vo2
2
o
6.7 Power
DEFINITION OF AVERAGE POWER
Average power is the rate at which work is done, and it
is obtained by dividing the work by the time required to
perform the work.
Work W
P

Time
t
joule s  watt (W)
6.7 Power
Change in energy
P
Time
1 horsepower  550 foot  pounds second  745.7 watts
P  Fv
6.8 Other Forms of Energy and the Conservation of Energy
THE PRINCIPLE OF CONSERVATION OF ENERGY
Energy can neither be created nor destroyed, but can
only be converted from one form to another.
6.9 Work Done by a Variable Force
Constant Force
W  F cos  s
Variable Force
W  F cos  1 s1  F cos  2 s2  
7.1 The Impulse-Momentum Theorem
There are many situations when the
force on an object is not constant.
7.1 The Impulse-Momentum Theorem
DEFINITION OF IMPULSE
The impulse of a force is the product of the average
force and the time interval during which the force acts:
 
J  F t
Impulse is a vector quantity and has the same direction
as the average force.
newton  seconds (N  s)
7.1 The Impulse-Momentum Theorem
 
J  F t
7.1 The Impulse-Momentum Theorem
DEFINITION OF LINEAR MOMENTUM
The linear momentum of an object is the product
of the object’s mass times its velocity:


p  mv
Linear momentum is a vector quantity and has the same
direction as the velocity.
kilogram  meter/seco nd (kg  m/s)
7.1 The Impulse-Momentum Theorem
 
 vf  vo
a
t


 F  ma
 mv f  mv o
 F  t
 



 F t  mvf  mvo
7.1 The Impulse-Momentum Theorem
IMPULSE-MOMENTUM THEOREM
When a net force acts on an object, the impulse of
this force is equal to the change in the momentum
of the object
impulse
 



 F t  mvf  mvo
final momentum
initial momentum
7.2 The Principle of Conservation of Linear Momentum
 
sum of average external forces t  Pf  Po
If the sum of the external forces is zero, then
 
0  Pf  Po
 
Pf  Po
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
The total linear momentum of an isolated system is constant
(conserved). An isolated system is one for which the sum of
the average external forces acting on the system is zero.
7.3 Collisions in One Dimension
The total linear momentum is conserved when two objects
collide, provided they constitute an isolated system.
Elastic collision -- One in which the total kinetic
energy of the system after the collision is equal to
the total kinetic energy before the collision.
Momentum and KE are constant.
Inelastic collision -- One in which the total kinetic
energy of the system after the collision is not equal
to the total kinetic energy before the collision; if the
objects stick together after colliding, the collision is
said to be completely inelastic.
Momentum is constant but not KE.
7.2 The Principle of Conservation of Linear Momentum
 
Pf  Po
m1v f 1  m2 v f 2  0
vf 2  
vf 2
m1v f 1
m2

54 kg  2.5 m s 

 1.5 m s
88 kg
7.3 Perfectly Inelastic Collision
Momentum is conserved. Kinetic energy is NOT conserved.
7.5 Center of Mass
The center of mass is a point that represents the average location for
the total mass of a system.
xcm
m1 x1  m2 x2

m1  m2
7.5 Center of Mass
xcm
m1x1  m2 x2

m1  m2
vcm
m1v1  m2 v2

m1  m2
7.5 Center of Mass
vcm
m1v1  m2 v2

m1  m2
In an isolated system, the total linear momentum does not change,
therefore the velocity of the center of mass does not change.
7.5 Center of Mass
BEFORE
vcm
m1v1  m2 v2

0
m1  m2
AFTER
vcm

88 kg  1.5 m s   54 kg  2.5 m s 

 0.002  0
88 kg  54 kg
8.1 Rotational Motion and Angular Displacement
Arc length s
 (in radians) 

Radius
r
For a full revolution:
2 r

 2 rad
r
2 rad  360
8.1 Rotational Motion and Angular Displacement
Arc length s
 (in radians) 

Radius
r
 2 rad 
  0.0349 rad
2.00 deg
 360 deg 


s  r  4.23 107 m 0.0349 rad 
 1.48 106 m (920 miles)
8.2 Angular Velocity and Angular Acceleration
DEFINITION OF AVERAGE ANGULAR VELOCITY
Angular displaceme nt
Average angular ve locity 
Elapsed time
  o



t  to
t
SI Unit of Angular Velocity: radian per second (rad/s)
8.2 Angular Velocity and Angular Acceleration
Example 3 Gymnast on a High Bar
A gymnast on a high bar swings through
two revolutions in a time of 1.90 s.
Find the average angular velocity
of the gymnast.
8.2 Angular Velocity and Angular Acceleration
 2 rad 
  2.00 rev
  12.6 rad
 1 rev 
 12.6 rad

 6.63 rad s
1.90 s
8.2 Angular Velocity and Angular Acceleration
Changing angular velocity means that an angular
acceleration is occurring.
DEFINITION OF AVERAGE ANGULAR ACCELERATION
Change in angular ve locity
Average angular accelerati on 
Elapsed time
 
  o
t  to


t
SI Unit of Angular acceleration: radian per second squared (rad/s2)
8.3 The Equations of Rotational Kinematics
8.3 The Equations of Rotational Kinematics
Example 5 Blending with a Blender
The blades are whirling with an
angular velocity of +375 rad/s when
the “puree” button is pushed in.
When the “blend” button is pushed,
the blades accelerate and reach a
greater angular velocity after the
blades have rotated through an
angular displacement of +44.0 rad.
The angular acceleration has a
constant value of +1740 rad/s2.
Find the final angular velocity of the blades.
8.3 The Equations of Rotational Kinematics
θ
α
ω
ωo
+44.0 rad
+1740 rad/s2
?
+375 rad/s
t
 2  o2  2
  o2  2

375 rad s 
2

 2 1740 rad s
2
44.0rad   542 rad s
8.4 Angular Variables and Tangential Variables

vT  tangentia l velocity
vT  tangentia l speed
8.4 Angular Variables and Tangential Variables
s r
 
vT  
 r 
t
t
t
vT  r ( in rad/s)


t
8.4 Angular Variables and Tangential Variables
Total acceleration is the vector sum of centripetal acceleration
and tangential acceleration.

 
atotal  ac  aT
8.4 Angular Variables and Tangential Variables

vT  vTo r   ro     0 
aT 

 r

t
t
 t 
aT  r
( in rad/s )
2
  o
t
8.4 Angular Variables and Tangential Variables
Example 6 A Helicopter Blade
A helicopter blade has an angular speed of 6.50 rev/s and an
angular acceleration of 1.30 rev/s2.
For point 1 on the blade, find
the magnitude of (a) the
tangential speed and (b) the
tangential acceleration.
8.4 Angular Variables and Tangential Variables


   6.50
rev  2 rad 

  40.8 rad s
s  1 rev 
vT  r  3.00 m40.8 rad s  122 m s
8.4 Angular Variables and Tangential Variables


  1.30
rev  2 rad 
2

8
.
17
rad
s



s 2  1 rev 


aT  r  3.00 m 8.17 rad s 2  24.5 m s 2
9.1 The Action of Forces and Torques on Rigid Objects
In pure translational motion, all points on an
object travel on parallel paths.
The most general motion is a combination of
translation and rotation.
9.1 The Action of Forces and Torques on Rigid Objects
According to Newton’s second law, a net force causes an
object to have an acceleration.
What causes an object to have an angular acceleration?
TORQUE
9.1 The Action of Forces and Torques on Rigid Objects
DEFINITION OF TORQUE
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
  F
Direction: The torque is positive when the force tends to produce a
counterclockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m)
9.1 The Action of Forces and Torques on Rigid Objects
  F

cos 55 
3.6 10  2 m

790 N
  720 N 3.6 10 2 m cos 55
 15 N  m
9.2 Rigid Objects in Equilibrium
EQUILIBRIUM OF A RIGID BODY
A rigid body is in equilibrium if it has zero translational
acceleration and zero angular acceleration. In equilibrium,
the sum of the externally applied forces is zero, and the
sum of the externally applied torques is zero.
F
x
0
F
y
0
  0
9.2 Rigid Objects in Equilibrium
  F 
2 2
 W W  0
W W
F2 
2
F2

530 N 3.90 m 

 1480 N
1.40 m
9.2 Rigid Objects in Equilibrium
F
y
 F1  F2  W  0
 F1  1480 N  530 N  0
F1  950 N
9.2 Rigid Objects in Equilibrium
Example 5 Bodybuilding
The arm is horizontal and weighs 31.0 N. The deltoid muscle can supply
1840 N of force. What is the weight of the heaviest dumbbell he can hold?
9.2 Rigid Objects in Equilibrium
  W 
a a
 Wd  d  M M  0
 M  0.150 msin 13.0
9.2 Rigid Objects in Equilibrium
 Wa  a  M M
Wd 
d
 31.0 N 0.280 m   1840 N 0.150 m sin 13.0

 86.1 N
0.620 m
9.3 Center of Gravity
When an object has a symmetrical shape and its weight is distributed
uniformly, the center of gravity lies at its geometrical center.
9.3 Center of Gravity
W1 x1  W2 x2  
xcg 
W1  W2  
9.6 Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L of a body rotating about a
fixed axis is the product of the body’s moment of
inertia and its angular velocity with respect to that
axis:
L  I
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Angular Momentum: kg·m2/s
9.6 Angular Momentum
PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM
The angular momentum of a system remains constant (is
conserved) if the net external torque acting on the system
is zero.
9.6 Angular Momentum
Conceptual Example 14 A Spinning Skater
An ice skater is spinning with both
arms and a leg outstretched. She
pulls her arms and leg inward and
her spinning motion changes
dramatically.
Use the principle of conservation
of angular momentum to explain
how and why her spinning motion
changes.
10.1 The Ideal Spring and Simple Harmonic Motion
Applied
x
F
 kx
spring constant
Units: N/m
10.1 The Ideal Spring and Simple Harmonic Motion
HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING
The restoring force on an ideal spring is
Fx  k x
10.2 Simple Harmonic Motion and the Reference Circle
amplitude A: the maximum displacement
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
1
f 
T
2
  2 f 
T
10.3 Energy and Simple Harmonic Motion
DEFINITION OF ELASTIC POTENTIAL ENERGY
The elastic potential energy is the energy that a spring
has by virtue of being stretched or compressed. For an
ideal spring, the elastic potential energy is
PEelastic  12 kx2
SI Unit of Elastic Potential Energy: joule (J)
10.3 Energy and Simple Harmonic Motion
Example 8 Changing the Mass of a Simple Harmonic Oscillator
A 0.20-kg ball is attached to a vertical spring. The spring constant
is 28 N/m. When released from rest, how far does the ball fall
before being brought to a momentary stop by the spring?
10.3 Energy and Simple Harmonic Motion
E f  Eo
1
2
mv2f  mghf  12 ky2f  12 mvo2  mgho  12 kyo2
1
2
kho2  mgho
2mg
ho 
k
20.20 kg  9.8 m s 2

 0.14 m
28 N m


10.4 The Pendulum
Example 10 Keeping Time
Determine the length of a simple pendulum that will
swing back and forth in simple harmonic motion with
a period of 1.00 s.
2
  2 f 

T
g
L
T 2g
L
4 2
T 2 g 1.00 s  9.80 m s 2 
L

 0.248 m
2
2
4
4
2
Period of simple pendulum is T 
2

 2
L
g
11.1 Mass Density
DEFINITION OF MASS DENSITY
The mass density of a substance is CONSTANT and
is the mass of a substance divided by its volume:
m

V
SI Unit of Mass Density: kg/m3
11.1 Mass Density
11.2 Pressure
F
P
A
SI Unit of Pressure: 1 N/m2 = 1Pa
Pascal
11.2 Pressure
Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere
11.3 Pressure and Depth in a Static Fluid
V  Ah
P2 A  P1 A   Vg
P2 A  P1 A   Ahg
P2  P1   hg
11.3 Pressure and Depth in a Static Fluid
P2  P1   gh
atmospheric pressure




P2  1.01105 Pa  1.00 103 kg m3 9.80 m s 2 5.50 m 

 1.55 105 Pa
 


11.4 Pressure Gauges
P2  PB  PA
PA  P1   gh
absolute pressure
P2  Patm   gh



gauge pressure
11.5 Pascal’s Principle
PASCAL’S PRINCIPLE
Any change in the pressure applied
to a completely enclosed fluid is transmitted
undiminished to all parts of the fluid and
enclosing walls.
11.5 Pascal’s Principle
P2  P1   g 0 m 
F2 F1

A2 A1
 A2 
F2  F1  
 A1 
11.6 Archimedes’ Principle
P2  P1   gh
FB  P2 A  P1 A  P2  P1 A
V  hA
FB   ghA
FB  
V g
mass of
displaced
fluid
11.6 Archimedes’ Principle
ARCHIMEDES’ PRINCIPLE
Any fluid applies a buoyant force to an object that is partially
or completely immersed in it; the magnitude of the buoyant
force equals the weight of the fluid that the object displaces:
F
B
Magnitudeof
buoyantforce
 Wfluid

Weight of
displacedfluid
11.6 Archimedes’ Principle
If the object is floating then the
magnitude of the buoyant force
is equal to the magnitude of its
weight.
11.8 The Equation of Continuity
Incompressible fluid:
Volume flow rate Q:
A1v1  A2v2
Q  Av
11.9 Bernoulli’s Equation
The fluid accelerates toward the
lower pressure regions.
According to the pressure-depth
relationship, the pressure is lower
at higher levels, provided the area
of the pipe does not change.
11.9 Bernoulli’s Equation
W   F s  F s  PAs  P2  P1 V
Wnc 

1
2
 
mv12  mgy1 
1
2
mv22  mgy2

11.9 Bernoulli’s Equation
P2  P1 V  12 mv12  mgy1   12 mv22  mgy2 
P2  P1   12 v12  gy1   12 v22  gy2 
BERNOULLI’S EQUATION
In steady flow of a nonviscous, incompressible fluid, the pressure, the
fluid speed, and the elevation at two points are related by:
P1  12 v12  gy1  P2  12 v22  gy2
11.10 Applications of Bernoulli’s Equation
Conceptual Example 14 Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the
tarpaulin lies flat, but it bulges outward
when the truck is speeding down
the highway.
Account for this behavior.
11.10 Applications of Bernoulli’s Equation
11.10 Applications of Bernoulli’s Equation
Example 16 Efflux Speed
The tank is open to the atmosphere at
the top. Find an expression for the speed
of the liquid leaving the pipe at
the bottom.
11.10 Applications of Bernoulli’s Equation
P1  P2  Patm
P1  12 v12  gy1  P2  12 v22  gy2
y2  y1  h
1
2
v12  gh
v1  2 gh
v2  0
12.1 Common Temperature Scales
AT SEA LEVEL
Temperatures are reported in degrees
Celsius or degrees Fahrenheit.
Temperatures changed, on the
other hand, are reported in Celsius
degrees or Fahrenheit degrees:
1 C 
9
F
5
12.2 The Kelvin Temperature Scale
AT SEA LEVEL
Kelvin temperature
T  Tc  273.15
12.2 The Kelvin Temperature Scale
absolute zero point = -273.15oC
12.4 Linear Thermal Expansion
LINEAR THERMAL EXPANSION OF A SOLID
The length of an object changes when its temperature changes:
L   Lo T
coefficient of
linear expansion
Common Unit for the Coefficient of Linear Expansion:
 
1


C
C
1
12.4 Linear Thermal Expansion
12.4 Linear Thermal Expansion
THE BIMETALLIC STRIP
12.4 Linear Thermal Expansion
12.5 Volume Thermal Expansion
VOLUME THERMAL EXPANSION
The volume of an object changes when its temperature changes:
V   Vo T
coefficient of
volume expansion
 
1


C
Common Unit for the Coefficient of Volume Expansion:
C
1
12.5 Volume Thermal Expansion
Expansion of water.
The physics of bursting
water pipes.
12.6 Heat and Internal Energy
DEFINITION OF HEAT
Heat is energy that flows from a highertemperature object to a lower-temperature
object because of a difference in temperatures.
SI Unit of Heat: joule (J)
12.6 Heat and Internal Energy
The heat that flows from hot to cold
originates in the internal energy of
the hot substance.
It is not correct to say that a substance
contains heat.
13.1 Convection
CONVECTION
Convection is the process in which heat is carried from one place
to another by the bulk movement of a fluid.
convection
currents
13.2 Conduction
CONDUCTION
Conduction is the process whereby heat is transferred directly through
a material, with any bulk motion of the material playing no role in the
transfer.
One mechanism for conduction occurs when the atoms or molecules
in a hotter part of the material vibrate or move with greater energy than
those in a cooler part.
By means of collisions, the more energetic molecules pass on some of
their energy to their less energetic neighbors.
Materials that conduct heat well are called thermal conductors, and those
that conduct heat poorly are called thermal insulators.
13.2 Conduction
The amount of heat Q that is conducted through the bar depends on
a number of factors:
1.
2.
3.
4.
The time during which conduction takes place.
The temperature difference between the ends of the bar.
The cross sectional area of the bar.
The length of the bar.
13.3 Radiation
RADIATION
Radiation is the process in which
energy is transferred by means of
electromagnetic waves.
A material that is a good absorber
is also a good emitter.
A material that absorbs completely
is called a perfect blackbody.
13.4 Applications
A thermos bottle minimizes heat
transfer via conduction, convection,
and radiation.
The space between the inner
glass walls minimizes heat transfer by
conduction and convection.
The silvered surfaces reflect radiated
heat back to the inside.
13.4 Applications
GREENHOUSE EFFECT
-Depletion of the ozone layer is harmful to Earth
-Harmful effects of technology and urbanization
-Most heat transfer is by radiation.
14.1 Molecular Mass, the Mole, and Avogadro’s Number
One mole of a substance contains as many
particles as there are atoms in 12 grams of
the isotope cabron-12.
The number of atoms per mole is known as
Avogadro’s number, NA.
N A  6.022 1023 mol 1
N
n
NA
number of
moles
number of
atoms
14.2 The Ideal Gas Law
An ideal gas is an idealized model for real gases
that have sufficiently low densities.
The condition of low density means that the
molecules are so far apart that they do not
interact except during collisions, which are
effectively ELASTIC.
At constant volume, the pressure is
directly proportional to the temperature.
P T
14.2 The Ideal Gas Law
At constant temperature, the pressure is
inversely proportional to the volume.
P 1 V
The pressure is also proportional
to the amount of gas.
Pn
14.2 The Ideal Gas Law
THE IDEAL GAS LAW
The absolute pressure of an ideal gas is directly proportional to the Kelvin
temperature and the number of moles of the gas and is inversely proportional
to the volume of the gas.
nRT
P
V
PV  nRT
R  8.31J mol  K 
14.2 The Ideal Gas Law
Consider a sample of an ideal gas that is taken from an initial to a final
state, with the amount of the gas remaining constant.
PV  nRT
PV
 nR  constant
T
Pf V f
Tf
PiVi

Ti
14.2 The Ideal Gas Law
Pf V f
Tf
Constant T, constant n:
Constant P, constant n:
Constant V, constant n:
PiVi

Ti
Pf V f  PiVi
Vf
Vi

T f Ti
Pf
Pi

T f Ti
Boyle’s law
Charles’ law
Gay Lussac’s law
14.3 Kinetic Theory of Gases
The particles are in constant, random
motion, colliding with each other
and with the walls of the container.
Each collision changes the
particle’s speed.
As a result, the atoms and
molecules have different
speeds.
14.3 Kinetic Theory of Gases
THE INTERNAL ENERGY OF A MONATOMIC IDEAL GAS
2
KE  12 mvrms
 32 kBT
U  N 32 k BT  32 nRT
15.1 Thermodynamic Systems and Their Surroundings
Thermodynamics is the branch of physics that is built
upon the fundamental laws that heat and work obey.
The collection of objects on which attention is being
focused is called the system, while everything else
in the environment is called the surroundings.
Walls that permit heat flow are called diathermal walls,
while walls that do not permit heat flow are called
adiabatic walls.
To understand thermodynamics, it is necessary to
describe the state of a system.
15.2 The Zeroth Law of Thermodynamics
Two systems are said to be in thermal
equilibrium if there is no heat flow
between then when they are brought
into contact.
Temperature is the indicator of thermal
equilibrium in the sense that there is no
net flow of heat between two systems
in thermal contact that have the same
temperature.
15.2 The Zeroth Law of Thermodynamics
THE ZEROTH LAW OF THERMODYNAMICS
Two systems individually in thermal equilibrium
with a third system are in thermal equilibrium
with each other.
15.3 The First Law of Thermodynamics
Suppose that a system gains heat Q and that is the only effect occurring.
Consistent with the law of conservation of energy, the internal energy
of the system changes:
U  U f  U i  Q
Heat is positive when the system gains heat and negative when the system
loses heat.
15.3 The First Law of Thermodynamics
Thermodynamics is a conservation law; i.e. heat added to a system is used
by the system to increase its internal energy or to do work in expanding.
An increase in internal energy due to heat added to the system (positive) or
work done on the system (positive).
U  Q  W
Work done on a system, according to this convention, would result in a
decrease in volume:
W   P(V )
15.3 The First Law of Thermodynamics
THE FIRST LAW OF THERMODYNAMICS
Process
Definition
Result
Isothermal
T  0  U  0
Q  W
Adiabatic
Q0
U  W
V  0  W  0
U  Q
Isochoric or
Isovolumetric
15.3 The First Law of Thermodynamics
Example 1 Positive and Negative Work
In part a, the system gains 1500J of heat
and 2200J of work is done BY the system on its
surroundings.
In part b, the system also gains 1500J of heat, but
2200J of work is done ON the system.
In each case, determine the change in internal energy
of the system.
15.4 Thermal Processes
An isobaric process is one that occurs at
constant pressure.
W  Fs  P As  P(V )
At constant pressure, if volume decreases,
ΔV is negative, and work done is positive.
Isobaric process:

W  P(V )  P V f  Vi

15.4 Thermal Processes
Example 3 Isobaric Expansion of Water
One gram of water is placed in the cylinder and
the pressure is maintained at 2.0x105Pa. The
temperature of the water is raised by 31oC. The
water is in the liquid phase and expands by the
small amount of 1.0x10-8m3.
Find the work done and the change in internal
energy.
15.4 Thermal Processes
W   PV



  2.0 105 Pa 1.0 108 m3  0.0020J
U  Q  W  130 J  0.0020 J  130 J




Q  mcT  0.0010 kg  4186 J kg  C 31 C  130 J
15.4 Thermal Processes

W  P(V )  P V f  Vi

15.4 Thermal Processes
isochoric: constant volume
U  Q  W  Q
W 0
15.4 Thermal Processes
Example 4 Work and the Area Under a
Pressure-Volume Graph
Determine the work for the process in
which the pressure, volume, and temperature of a gas are changed along the
straight line in the figure.
The area under a pressure-volume graph is
the work for any kind of process.
15.4 Thermal Processes
Since the volume increases, the work
is negative.
Estimate that there are 8.9 colored
squares in the drawing.
W  8.92.0 105 Pa 1.0 10 4 m 3 
 180 J
15.5 Thermal Processes Using an Ideal Gas
ISOTHERMAL EXPANSION OR COMPRESSION
Isothermal
expansion or
compression of
an ideal gas
Vf
W  nRT ln 
 Vi



15.5 Thermal Processes Using an Ideal Gas
Example 5 Isothermal Expansion of an Ideal Gas
Two moles of the monatomic gas argon expand isothermally at 298K
from and initial volume of 0.025m3 to a final volume of 0.050m3. Assuming
that argon is an ideal gas, find (a) the work done by the gas, (b) the
change in internal energy of the gas, and (c) the heat supplied to the
gas. ??
15.5 Thermal Processes Using an Ideal Gas
(a)
Vf
W  nRT ln 
 Vi



 0.050 m 3 
  3400 J
 2.0 mol 8.31 J mol  K 298 K  ln 
3 
 0.025 m 
(b)
(c)
U  32 nRT f  32 nRTi  0
U  Q  W
Q  W  3400 J
15.3 The First Law of Thermodynamics
(a)
U  Q  W
  1500 J    2200 J   700 J
(b)
U  Q  W
  1500 J    2200 J   3700 J
15.3 The First Law of Thermodynamics
Example 2 An Ideal Gas
The temperature of three moles of a monatomic ideal gas is reduced
from 540K to 350K as 5500J of heat flows into the gas.
Find (a) the change in internal energy and (b) the work done by the
gas. ???
U  U f  U i  Q  W
U  32 nRT
15.3 The First Law of Thermodynamics
(a)
U  32 nRT f  32 nRTi

(b)
3
2
3.0 mol 8.31 J mol  K 350 K  540 K   7100 J
W  Q  U  5500 J   7100 J   12600 J
15.4 Thermal Processes
A quasi-static process is one that occurs slowly enough that a uniform
temperature and pressure exist throughout all regions of the system at all
times.
isobaric: constant pressure
isochoric: constant volume
isothermal: constant temperature
adiabatic: no transfer of heat
15.7 The Second Law of Thermodynamics
The second law is a statement about the natural tendency of heat to
flow from hot to cold, whereas the first law deals with energy conservation
and focuses on both heat and work.
THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENT
Heat flows spontaneously from a substance at a higher temperature to a substance
at a lower temperature and does not flow spontaneously in the reverse direction.
15.8 Heat Engines
A heat engine is any device that uses heat to
perform work. It has three essential features.
1. Heat is supplied to the engine at a relatively
high temperature from a place called the hot
reservoir.
2. Part of the input heat is used to perform
work by the working substance of the engine.
3. The remainder of the input heat is rejected
to a place called the cold reservoir.
QH  magnitude of input heat
QC  magnitude of rejected heat
W  magnitude of the work done
15.8 Heat Engines
The efficiency of a heat engine is defined as
the ratio of the work done to the input heat:
e
W
QH
If there are no other losses, then
QH  W  QC
e  1
QC
QH
15.8 Heat Engines
Example 6 An Automobile Engine
An automobile engine has an efficiency of 22.0% and produces
2510 J of work. How much heat is rejected by the engine?
e
W
QH  W  QC
QH
QH 
W
e
15.8 Heat Engines
QH  W  QC
QH 
W
QC  QH  W
1 
QC 
 W  W   1
e
e 
W
 1

 2510 J 
 1  8900 J
 0.220 
e
15.9 Carnot’s Principle and the Carnot Engine
A reversible process is one in which both the system and the
environment can be returned to exactly the states they were in
before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND
LAW OF THERMODYNAMICS
No irreversible engine operating between two reservoirs at constant temperatures
can have a greater efficiency than a reversible engine operating between the same
temperatures. Furthermore, all reversible engines operating between the same
temperatures have the same efficiency.
15.9 Carnot’s Principle and the Carnot Engine
The Carnot engine is useful as an idealized
model.
All of the heat input originates from a single
temperature, and all the rejected heat goes
into a cold reservoir at a single temperature.
Since the efficiency can only depend on
the reservoir temperatures, the ratio of
heats can only depend on those temperatures.
QC
QH
e  1
QC
QH
TC
 1
TH

TC
TH
15.11 Entropy
Any irreversible process increases
the entropy of the universe.
S universe  0
THE SECOND LAW OF THERMODYNAMICS STATED
IN TERMS OF ENTROPY
The total entropy of the universe does not change when a
reversible process occurs and increases when an irreversible
process occurs.
15.11 Entropy
Example 12 Energy Unavailable for Doing Work
Suppose that 1200 J of heat is used as input for an engine
under two different conditions (as shown on the right).
Determine the maximum amount of work that can be obtained
for each case.
ecarnot  1 
TC
TH
e
W
QH
15.11 Entropy
The maximum amount of work will be achieved when the
engine is a Carnot Engine, where
(a)
ecarnot  1 
TC
150 K
 1
 0.77
TH
650 K
W  ecarnot  QH  0.771200 J   920 J
(b)
ecarnot
TC
150 K
 1
 1
 0.57
TH
350 K
W  ecarnot  QH  0.571200 J   680 J
The irreversible process of heat through the copper
rod causes some energy to become unavailable.
15.12 The Third Law of Thermodynamics
THE THIRD LAW OF THERMODYNAMICS
It is not possible to lower the temperature of any system to absolute
zero in a finite number of steps.