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Phys2Ch4
 PROGRAM OF “PHYSICS”
Lecturer: Dr. DO Xuan Hoi
Room 413
E-mail : [email protected]
PHYSICS 2
(FLUID MECHANICS AND THERMAL PHYSICS)
02 credits (30 periods)
Chapter 1 Fluid Mechanics
Chapter 2 Heat, Temperature and the Zeroth
Law of Thermodynamics
Chapter 3 Heat, Work and the First Law of
Thermodynamics
Chapter 4 The Kinetic Theory of Gases
Chapter 5 Entropy and the Second Law of
Thermodynamics
References :
Halliday D., Resnick R. and Walker, J. (2005),
Fundamentals of Physics, Extended seventh edition.
John Willey and Sons, Inc.
Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley
Publishing Company
Hecht, E. (2000). Physics. Calculus, Second Edition.
Brooks/Cole.
Faughn/Serway (2006), Serway’s College Physics,
Brooks/Cole.
Roger Muncaster (1994), A-Level Physics, Stanley
Thornes.
http://ocw.mit.edu/OcwWeb/Physics/index.htm
http://www.opensourcephysics.org/index.html
http://hyperphysics.phyastr.gsu.edu/hbase/HFrame.html
http://www.practicalphysics.org/go/Default.ht
ml
http://www.msm.cam.ac.uk/
http://www.iop.org/index.html
.
.
.
Chapter 3
Heat, Work and the First Law of
Thermodynamics
1. Work and Heat in Thermodynamic
Processes
2. The First Law of Thermodynamics
3. Some Applications of The First Law of
Thermodynamics
4. Energy
transfer mechanisms
1. Work and Heat in
Thermodynamic Processes
 State of a system
 Description of the system in terms of state variables
►Pressure
►Volume
►Temperature
►Internal Energy
 A macroscopic state of an isolated system can be
specified only if the system is in internal thermal
equilibrium
 Work
►
►
►
Work is an important energy transfer
mechanism in thermodynamic systems
Heat is another energy transfer
mechanism
Example: gas cylinder with piston
 The gas is contained in a cylinder with a
moveable piston
 The gas occupies a volume V and exerts
pressure P on the walls of the cylinder
and on the piston
 Work done by a gas
Consider a gas contained in a cylinder
fitted with a movable piston.
► At equilibrium, the gas occupies a
volume V and exerts a uniform
pressure P on the cylinder’s walls and
on the piston. If the piston has a
cross-sectional area A, force exerted
by the gas on the piston is : F  PA
►
►
When the gas expands quasi-statically
(slowly enough to allow the system to remain essentially in
thermal equilibrium at all times)
As the piston moves up a distance dy, the work done by the
gas on the piston : dW  Fdy  PAdy  PdV
►
The work done by the gas on the piston :
dW  PdV
►
If the gas expands : dV > 0
 the work done by the gas : dW > 0
If the gas were compressed : dV < 0
 the work done by the gas (which can be
interpreted as work done on the gas) : dW < 0
When the volume remains constant
 No work is done on the gas
►
The work done by the gas on the piston :
dW  PdV
►
If the gas expands : dV > 0
 the work done by the gas : dW > 0
Suppose : P = const : isobaric process
(pronounced "eye-so-bear-ic")
►
P
P
i
f
P (Vf  Vi )
Vi
►
Vf
V
The work done by the gas equals the area under
the PV curve.
►
The work done by the gas on the piston :
dW  PdV
If the gas were compressed : dV < 0
 the work done by the gas (which can be interpreted
as work done on the gas) : dW < 0
W = P ΔV; W < 0
Work = Area under the curve
Work done on the gas
►
The work done by the gas on the piston :
dW  PdV
►
The total work done by the gas as its volume changes from
Vi to Vf is given by the integral:
W 
Vf
V PdV
i
►
The work done by a gas in the
expansion from an initial state to
a final state is the area under
the curve connecting the states
in a PV diagram
PV Diagrams
►
►
The curve on the diagram is called the path taken
between the initial and final states
The work done depends on the particular path
 Same initial and final states, but different amounts of work are
done
Question
Find work done by the gas in this cycle.
Work is equal to
the area :
P2
W   p2  p1 V2 V1 
P1
V1
V2
Work done by an ideal gas at constant temperature
The total work done by the gas as its volume changes from
Vf
V1 to Vf : W  PdV
V
i
Ideal gas : PV  nRT
Vf
nRT
W 
dV
V
Vi
Isothermal process: T  const
Vf
Vf
dV
W  nRT 
; W  nRT ln
Vi
Vi V
Also : PV
i i  PfVf
Vf
W  nRT ln
Vi
W  nRT ln
Pi
Pf
Vf  V i : W  0
When a system expands : work is positive.
When a system is compressed, its volume decreases and
it does negative work on its surroundings
Other Processes
► Isovolumetric
( or isochoric process - pronounced
"eye-so-kor-ic")
 Volume stays constant
 Vertical line on the PV diagram
► Isothermal
 Temperature stays the same
► Adiabatic
(pronounced "ay-dee-ah-bat-ic")
 No heat is exchanged with the surroundings
Q=0
Example:
Calculate work done by expanding gas of 1 mole if
initial pressure is 4000 Pa, initial volume is 0.2 m3, and
initial temperature is 96.2 K. Assume a process:
isobaric expansion to 0.3 m3, Tf = 144.3 K
Given:
Isobaric expansion:
n = 1 mole
3
3
Ti = 96.2 K W  P V  P Vf Vi   4000 Pa 0.3m  0.2m
Tf = 144.3 K
 400 J
3
Vi = 0.2 m
i
Vf = 0.3 m3
P = const
Find:
W=?


f
2. The First Law of Thermodynamics
 What is internal energy?
The internal energy of a system is the sum of the kinetic
energies of all of its constituent particles, plus the sum of
all the potential energies of interaction among these
particles.
During a change of state of the system, the internal energy
may change from an initial value U1 to a final value U2 .
The change in internal energy : U = U2 – U1.
► Consider
energy conservation in thermal
processes. Must include:
 Q
►Heat
►Positive
if energy is transferred to the system
 W
►Work
►Positive
if done by the system
 U
►Internal
energy
►Positive if the temperature increases
Add a quantity of heat Q to a system and the system
does no work during the process, the internal energy
increases by an amount equal to Q : U = Q.
► When a system does work W by expanding against
its surroundings and no heat is added during the
process, energy leaves the system and the internal energy
decreases.
►
►
When both heat transfer and work occur, the total change
in internal energy is
U  Q  W
(first law of thermodynamics)
►
This means that the change in internal energy of a system is
equal to the sum of the energy transferred across the system
boundary by heat and the energy transferred by work
EXAMPLE :
Notes About the First Law
► The
First Law is a general equation of
Conservation of Energy
► There is no practical, macroscopic, distinction
between the results of energy transfer by heat
and by work
► Q and W are related to the properties of state
for a system
Notes About the First Law
work W done by the system depends not
only on the initial and final states, but also on the
intermediate states - that is, on the path
► Like work, heat Q depends not only on the initial
and final states but also on the path
► While Q and W depend on the path, U = Q - W
is independent of path. The change in internal
energy of a system during any thermodynamic
process depends only on the initial and final
states, not on the path leading from one to the
other.
► The
Notes About the First Law
►
When a system undergoes an infinitesimal change in state
in which a small amount of energy dQ is transferred by
heat and a small amount of work dW is done, the internal
energy changes by a small amount :
dU  dQ  dW
(first law of thermodynamics for infinitesimal processes)
►
Because dQ and dW are inexact differentials
dU  Q  W
dU  Q  pdV
3. Some Applications
of The First Law of Thermodynamics
Kinds of Thermodynamic Processes :
 Adiabatic Process
An adiabatic process is defined as one with no heat
transfer into or out of a system : Q = O
(We can prevent heat flow either by surrounding the
system with thermally insulating material or by carrying
out the process so quickly that there is not enough time
for appreciable heat flow)
From the first law we find that for every adiabatic process :
U  U 2  U 1  W
From the first law we find that for every adiabatic process :
U  U 2  U 1  W
 When a system expands adiabatically :
W >0
(the system does work on its surroundings)
U < 0
(the internal energy decreases)
 When a system is compressed adiabatically :
W <0
(work is done on the system)
(work is done on the system by its surroundings)
U > 0
(the internal energy increases)
Kinds of Thermodynamic Processes :
 Isochoric Process
An isochoric process is a constant-volume process.
When the volume of a thermodynamic system is constant,
it does no work on its surroundings : W = 0
From the first law :
U  U 2  U 1  Q
In an isochoric process, all the energy added as heat
remains in the system as an increase in internal
energy.
Example : Heating a gas in a closed constant-volume
container
Kinds of Thermodynamic Processes :
 Isobaric Process
An isobaric process is a constant-pressure process.
In general, none of the three quantities U , Q , and W is
zero in an isobaric process
The work done by the gas is simply:
W  P (V2 V1 )
Kinds of Thermodynamic Processes :
 Cyclical Processes
A process that eventually returns a system to its initial
state is called a cyclic process. For such a process, the
final state is the same as the initial state
The total internal energy change must be zero :
U  U 2  U 1  0 ; U 2  U 1
From the first law : U  Q  W
W Q
If a net quantity of work W is done by the system during
this process, an equal amount of energy must have
flowed into the system as heat Q
Kinds of Thermodynamic Processes :
Cyclical Process in a PV Diagram
►
►
►
►
►
This is an ideal monatomic gas
confined in a cylinder by a
moveable piston
A to B : isovolumetric process
which increases the pressure
B to C : isothermal expansion
and lowers the pressure
C to A : isobaric compression
The gas returns to its original
state at point A
 Cyclical Process. Example :
The cyclic thermodynamic process of our body
(a thermodynamic system) every day
Example
Four different processes for a
constant amount of an ideal
gas, all starting at state a.
For the adiabatic process, Q
= 0; for the isochoric
process, W = 0; and for the
isothermal process,
U = O. The temperature
increases only during the
isobaric expansion.
A pV-diagram for four processes for a constant amount of an
ideal gas. The path followed in an adiabatic process (a to
1) is
called an adiabat. A vertical line (constant volume) is an
isochor, a horizontal line (constant pressure) is an isobar,
and a curve of constant temperature (shown as light blue
lines) is an isotherm.
The First Law and Human Metabolism
► The
First Law can be applied to living organisms
► The internal energy stored in humans goes into
other forms needed by the organs and into work
and heat
► The metabolic rate (ΔU / ΔT) is directly
proportional to the rate of oxygen consumption by
volume
 Basal metabolic rate (to maintain and run
organs, etc.) is about 80 W
Various Metabolic Rates
Fig. T12.1, p. 369
Slide 11
 Free expansions: These are adiabatic processes in
which no transfer of heat occurs between the system and
its environment and no work is done on or by the system.
Thus, Q = W = 0
The first law  U  U 2  U 1  0
PROBLEM 1
Suppose 1.00 g of water vaporizes isobarically
at atmospheric pressure (1.013  105 Pa). Its volume in the
liquid state is 1.00 cm3, and its volume in the vapor state is
1671 cm3. Find the work done in the expansion and the
change in internal energy of the system.
The heat of vaporization for water 2.26  106 J/kg
SOLUTION
PROBLEM 2
A 1.0-kg bar of copper is heated at atmospheric
pressure. If its temperature increases from 20°C to 50°C,
(a) what is the work done by the copper on the surrounding
atmosphere?
The density of copper is 8.92  103 kg/m3
SOLUTION
PROBLEM 2
A 1.0-kg bar of copper is heated at atmospheric
pressure. If its temperature increases from 20°C to 50°C,
(b) What quantity of energy is transferred to the copper by
heat?
The specific heat of copper is 387 J/kg0C
SOLUTION
PROBLEM 2
A 1.0-kg bar of copper is heated at atmospheric
pressure. If its temperature increases from 20°C to 50°C,
(c) What is the increase in internal energy of the copper?
SOLUTION
PROBLEM 3
A series of thermodynamic processes is shown in
the pV-diagram of Fig. 1. In process ab, 150 J of heat is added
to the system, and in process bd, 600 J of heat is added. Find
(a) the internal energy change in process ab
SOLUTION
PROBLEM 3
A series of thermodynamic processes is shown in
the pV-diagram of Fig. 1. In process ab, 150 J of heat is added
to the system, and in process bd, 600 J of heat is added. Find
(a) the internal energy change in process ab;
(b) the internal energy change in process abd (shown in light
blue)
SOLUTION
PROBLEM 3
A series of thermodynamic processes is shown in
the pV-diagram of Fig. 1. In process ab, 150 J of heat is added
to the system, and in process bd, 600 J of heat is added. Find
(a) the internal energy change in process ab;
(b) the internal energy change in process abd (shown in light
blue); and
(c) the total heat added in process acd (shown in dark blue).
SOLUTION
PROBLEM 4
A thermodynamic system is taken from state a to
state c in Fig. 1 along either path abc or path adc. Along path
abc, the work W done by the system is 450 J. Along path adc,
W is 120 J. The internal energies of each of the four states
shown in the figure are Ua = 150 J, Ub = 240 J, Uc = 680 J, and
Ud = 330 J. Calculate the heat flow Q for each of the four
processes ab, bc, ad, and de. In each process, does the system
absorb or liberate heat?
SOLUTION
PROBLEM 5
A gas in a cylinder is held at a constant pressure
of 2.30  105 Pa and is cooled and compressed from 1.70 m3 to
1.20 m3 . The internal energy of the gas decreases by
1.40  105 J.
(a) Find the work done by the gas.
(b) Find the absolute value of the heat flow into or out of the
gas, and state the direction of the heat flow.
SOLUTION
PROBLEM 6
A gas within a closed chamber undergoes the
cycle shown in the p-V diagram of Fig. 1.The horizontal scale
is set by VS = 4.0 m3. Calculate the net energy added to the
system as heat during one complete cycle.
SOLUTION
Cycle: UABCA = W – Q = 0; Q = W
W  W AB  WBC  WCA
VB
W AB   PdV
VA
V A  1 ; PA  10; VB  4 ; PB  30
W AB
WBC
4
10 
 20
   V   dV  60 J
3
3 
1
 PB (VB VC )  30(4  1)  90 J
WCA  0
20
10
P  V 
3
3
W  30 J ; Q  30 J
4. Energy transfer mechanisms
Methods of Heat Transfer
►
►
►
Need to know the rate at which energy is transferred
Need to know the mechanisms responsible for the transfer
Methods include
 Conduction
 Convection
 Radiation
4.1 Conduction
► The transfer can be viewed on an atomic scale
 It is an exchange of energy between microscopic
particles by collisions
 Less energetic particles gain energy during collisions
with more energetic particles
► Rate of conduction depends upon the
characteristics of the substance
Conduction example
►
►
►
►
The molecules vibrate about their
equilibrium positions
Particles near the flame vibrate
with larger amplitudes
These collide with adjacent molecules
and transfer some energy
Eventually, the energy travels entirely through
Conduction can occur
only if there is a
difference in temperature
between two parts of the
conducting medium
the rod
Law of thermal conduction
Consider a slab of material of thickness x and cross-sectional area
A. One face of the slab is at a temperature T1 , and the other
face is at a temperature T2.
► The slab allows energy to transfer from the
region of higher temperature to the region
of lower temperature
T 2  T1
Q
dT
P   kA
 kA
t
x
dx
Heat flow
►
►
temperature gradient
(the variation of temperature with position)
P is in Watts when Q is in Joules and t is in seconds
k is the thermal conductivity of the material
 Good conductors have high k values and good
insulators have low k values
Suppose that a long, uniform rod of length L is thermally insulated
so that energy cannot escape by heat from its surface except at the
ends.
One end is in thermal contact with an energy reservoir at
temperature T1 , and the other end is in thermal contact with a
reservoir at temperature T2 > T1
The rate of energy transfer by conduction
through the rod:
For a compound slab containing several materials of thicknesses
L1 , L2, . . . and thermal conductivities k1 , k2, . . .
(Home Insulation)
PROBLEM 7 Two slabs of thickness L1 and L2 and thermal
conductivities k1 and k2 are in thermal contact with each other, as
shown in figure. The temperatures of their outer surfaces are T1
and T2 , respectively, and T2 > T1 . Determine the temperature T
at the interface and the rate of energy transfer by conduction
through the slabs in the steady-state condition.
SOLUTION
The rate at which energy is transferred through slab 1:
The rate at which energy is transferred through slab 2:
When a steady state is reached, these two rates must be equal:
PROBLEM 8
A Styrofoam box used to keep drinks cold at a
picnic has total wall area (including the lid) of 0.80 m2 and
wall thickness 2.0 cm. It is filled with ice and water at 0°C.
What is the rate of heat flow into the box if the temperature
of the outside wall is 3O°C? How much ice melts in one day?
SOLUTION
The heat current (rate of heat flow):
The total heat flow Q in one day (86,400 s):
The heat of fusion of ice is 3.34 x 105 J/kg, so the quantity of ice
melted by this quantity of heat:
PROBLEM 9
A steel bar 10.0 cm long is welded end to end to
a copper bar 20.0 cm long. Both bars are insulated perfectly
on their sides. Each bar has a square cross section, 2.00 cm
on a side. The free end of the steel bar is maintained at
100°C and the free end of the copper bar is maintained at
0°C. Find the temperature at the junction of the two bars and
the total rate of heat flow.
SOLUTION
4.2 Convection
► Energy transferred by the movement of a substance
 When the movement results from differences in density,
it is called natural conduction
 When the movement is forced by a fan or a pump, it is
called forced convection
Convection example
►
►
►
►
Air directly above the flame is warmed and
expands
The density of the air decreases, and it rises
The mass of air warms the hand as it moves
by
Applications:
 Radiators
 Cooling automobile engines
4.3 Radiation
Radiation does not require physical contact
► All objects radiate energy continuously in the form of
electromagnetic waves due to thermal vibrations of the
molecules
► Rate of radiation is given by Stefan’s Law
►
Radiation example
►
►
The electromagnetic waves carry the
energy from the fire to the hands
No physical contact is necessary
Radiation equation
 The rate of energy transfer (the total energy radiated
from an object at temperature T per unit time) :
P   AeT 4
(Stefan-Boltzmann’s law)
 σ = 5.6696 x 10-8 W/m2 K4 (Stefan’s constant)
 A is the surface area of the object
e varies from 0 to 1
 e is a constant called the emissivity
 T is the temperature in Kelvins
►
The emitted energy increases markedly with
increasing temperature
PROBLEM 10 A cooler has a surface area of 0.5 m2 and an
average thickness of 2.0cm. How long will it take for 1.5 kg of ice to
melt in the cooler if the outside temperature is 30oC?
The thermal conductivity of the substance used to make the cooler
is 0.03 W/m-oC and the heat of fusion of ice is Lf = 3.34105J/kg
SOLUTION
dQ
dT
The rate of energy transfer is computed from: P 
 kA
dt
dx
For a surface of area A:
Q
30o C  0o C
Q
T
 22.5 W

 0.03  0.50 
 kA
t
x
t
0.02
Q L f m
Q  L f m  t 

22.5
22.5
3.34  105  1.5

 2.23  10 4 s
22.5
PROBLEM 11
A patient waiting to be seen by his physician is
asked to remove all his clothes in an examination room that is
at 16oC. Calculate the rate of his heat loss by radiation from
the patient, given that his skin temperature is 34oC and his
surface area is 1.2m2. Assume an emissivity of 0.80.
SOLUTION
From: P   eAT 4
, the rate of heat loss by radiation:
Pnet   eA(T 4  T04 )
Pnet  5.67 108  0.80 1.6  (307 4  2894 )
Pnet  140 W
PROBLEM 12 A black body is a body that completely absorbs all
the electromagnetic radiation falling on it. A small hole in the wall
of a cavity in an object behaves like a black body.
At what rate does radiation escape from a hole 10 cm2 in area in
the wall of a furnace whose interior is at temperature of 700oC?
SOLUTION
The absolute temperature of the interior:
T  700oC  273  973K
Since the hole acts as a black body,
its emissivity: e = 1
The hole area: 10-3m2
The hole radiates energy at the rate:
P  e AT 4 1 5.67108 103  9734  51W
PROBLEM 13 The Sun’s radius is given by Rs = 7108 m. The
average Sun-Earth distance is Rs = 1.51011 m. The power per unit
area from the Sun is measure at the Earth to be 1400W/m2. Assume
that the Sun is a perfect radiator: the emissivity is unit.
Estimate the surface temperature of the Sun.
SOLUTION
The emissivity is unit : P   AT
4
The conservation of energy:
 T 4 4 RS2  E (4 R 2 ) ;
 E ( R 2 ) 1 / 4
T 
2 
  RS 
T  5 800 K
Stefan-Boltzmann’s law for astrophysics
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