Download Q - Department of Applied Physics

Hong Kong Polytechnic University
Heat & Light (AP306)
Part I: Thermal Physics
Dr.Haitao Huang (黄海涛)
Department of Applied Physics, Hong Kong PolyU
Tel: 27665694; Office: CD613
Lecture Notes can be downloaded from http://ap.polyu.edu.hk/apahthua
Textbook:
Thermal Physics, 2nd Edition, C.B.P.Finn
Assessment Weighting:
Continuous Assessment:
Quiz (and midterm exam):
Final Exam:
65% (Minimum requirement: D)
Minimum requirement for overall grade: D
35%
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Some Concepts:
System; Surroundings;
Temperature
Wall
Piston
Gas system
Boundary / Wall
Surroundings
An equilibrium state is one in which all the bulk physical properties of the
system are uniform throughout the system and do not change with time.
State Variables / Thermodynamic Variables / Thermodynamic Coordinates
Stat Functions / State Properties
Two systems in thermal contact can finally reach thermal equilibrium.
Zeroth Law of Thermodynamics:
If each of two systems is in thermal equilibrium with a third, they are in thermal equilibrium
with one another.
A
A
B
B
C
A
C
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Temperature
Temperature:
A property of a system that determines whether or not that system is in thermal
equilibrium with other systems.
If two systems have the same temperature so that they are in thermal equilibrium,
this does not necessarily mean that they are in complete or thermodynamic
equilibrium. For this condition to hold, in addition to being in thermal equilibrium,
they would also have to be in mechanical equilibrium and chemical equilibrium.
Indicator Diagram:
P
T1
T2
T3
V
Equation of State:
All the bulk properties can be fixed by specifying two independent state variables.
f P,V , T   0
PV  nRT
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Temperature
Temperature Scale:
It is customary to choose a fixed point at which ice, water and water vapor coexist in
equilibrium; this is known as the triple point of water. The temperature at this point is
defined as 273.16K. The gas scale temperature is determined from,
Tgas  273.16
P
PTP
where Tgas is the temperature of an ideal gas with constant volume at pressure P.
PTP is the pressure of the constant volume gas at the triple point of water. The unit of
temperature on the ideal gas scale is K (Kelvin). The Celsius scale t is measured in
C and is related to the ideal gas scale T by,
 
t  C  T K   273.15
Homework:
exercises 1.1-1.3
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Reversible Process and Work
Process / End Points / Reversible Process:
P
P2, V2
Reversible processes are quasistatic processes where
no dissipative forces such as friction are present.
P1, V1
How to recognize reversible process?
Volume thermal expansivity

1  V 


V  T  P
Linear expansion coefficient

1  L 
 
L  T  F
Bulk modulus
and compressibility
Young’s modulus
V
  3
1
 P 
K  V 
 
 V T 
Y
L  F 


A  L T
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Reversible Process and Work
Example: Calculate the increase in tension F of a wire clamped between
two rigid supports, a distance L apart, when it is cooled from T1 to T2.
Known parameters: cross-section area, Young’s modulus and linear
expansion coefficient
 F 
 F 
 F 
dF  
dT

dL





 dT
 T  L
 L T
 T  L
F  F L, T 
 F 
F  F2  F1   
 dT
T1
 T  L
T2
 F   T   L 

   
  1
 T  L  L  F  F T
Y
L  F 


A  L T

1  L 
 
L  T  F
 F 
 F   L 

  
     AY
 T  L
 L T  T  F
T2  F 
T2
F   
dT


AY

dT   AY T2  T1 


T1
T
1
 T  L
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Reversible Process and Work
Example: Work done by an ideal gas in a cylinder with a
P
1
frictionless piston. (Work is path dependent.)
Suppose that the pressure is P and the balancing force is F at
the equilibrium state,
F  PA
3
2
V
Infinitesimal small work done by the gas
Total work:
Along 12:
V2
V2
V1
V1
dW  Fdx  PAdx  PdV
W   dW   PdV
W   PdV  nRT  dV V  nRT ln V2 V1 
Along 13  2:
V2
V2
V1
V1
W=P2(V2-V1)
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Reversible Process and Work
Sign convention for work:
When the surroundings do work on the system, that
work is positive; conversely, when the system does
work on the surroundings, that work is negative.
More examples on infinitesimal small work:
Movable bar
l
dx
Wire frame
The work done to stretch a wire at a tension F though a
infinitesimal distance dx is,
dW  Fdx
The work done to stretch a surface film (increase the
surface by dA) is,
dW  ldx  dA
In a reversible electrolytic cell, the work performed by
the external charging circuit is,
dW  dZ
For a uniformly magnetized material, the external work
required to increase the magnetic moment by dM in the
applied induction field B0 is,
dW  B0 dM
The work required to increase the overall dipole moment
of a uniformly polarized dielectric material by dP in an
electric field E is,
dW  EdP
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Reversible Process and Work
Example: Calculate the work done in changing the state of a compressible fluid from (P1,
T1) to (P2, T2) in a reversible process.
 P 
K  V 


V

T
V  V ( P, T )
V
 V 
 V 
dV  
 dP  
 dT   dP  VdT
K
 P T
 T  P
PV
dP  PVdT
K
dW   PdV 
2
W   dW  
1

P2
P1
1  V 


V  T  P
T2
PV
dP   PVdT
T1
K
In the case of an isothermal change of liquid: W  
P2
P1

PV
V
dP 
P22  P12
K
2K

Homework: exercises: 2.1-2.9
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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First Law of Thermodynamics
Joule’s Experiment
Determined the mechanical equivalent of heat, that is,
it required 4.2kJ of work to raise the temperature of 1kg
of water through one degree Kelvin.
First Law: If a thermally isolated system is brought from one equilibrium
state to another, the work necessary to achieve this change is independent
of the process used.
There must exist a state function whose difference between the two end points (1
and 2) is equal to the adiabatic work. Such a function is called the internal energy
U
Wadi  U 2  U1
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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First Law of Thermodynamics
If a system is not thermally isolated, the work done in taking the system between a
pair of equilibrium states depends on the path. The change of the internal energy
can be expressed as,
U  U 2  U1  W  Q
where Q is heat. This is the mathematical statement of the first law. This means
that the internal energy can be increased either by doing work on or by supplying
heat to the system. It is true for all processes whether reversible or irreversible.
In closed systems, heat is the non-mechanical exchange of energy between the
system and the surroundings because of their temperature difference.
The sign convention for Q is that it is positive when heat enters the system. For an
infinitesimal process, the first law takes the form,
dU=đW+đQ
In general, W and Q are path dependent. We call they are inexact differentials.
For a compressible fluid or gas where đW=-PdV for an infinitesimal reversible
process, the first law becomes,
đQ=dU+PdV
In differentiating between heat and work, it is very important to be clear as to what
constitutes the system.
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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First Law of Thermodynamics
Microscopic View of First Law:
Energy can be added to the system in the form of heat by increasing the
randomness of the motion of the constituent molecules. We can also increase the
energy of the system by performing work. In this way we displace the molecules in
an ordered way.
As a result of quantum mechanics, each of the N particles in the system can exist in
a series of discrete energy levels. If there is a population of ni particles in the ith
N
energy level i, the total internal energy of the system will be
U   ni  i
i 1
Now U can be changed in two ways: either the energies i can be changed, with the
populations remaining the same—this is work; or the populations ni can be changed
with the energies remaining the same—this is heat.
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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First Law of Thermodynamics
Heat Capacity C
the limiting ratio of the heat introduced reversibly into the system divided by the
temperature rise,
 Q  đQ
C  lim 

T 0 T

 dT
Specific Heat:
c
C 1 đQ

m m dT
Since there are a large number of possible reversible paths between the end points
with a temperature difference T, it follows that there are many possible heat
capacities.
đQ  U 
Heat Capacity at constant volume CV  V  

dT  T V
Heat Capacity at constant pressure CP 
dQP dU  PdV  H 



dT
dT

T

P
Enthalpy: H=U+PV
For ideal gas: U=U(T)
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Free Expansion:
First Law of Thermodynamics
partition
break
partition
gas
For an ideal gas, there will be no temperature change. Real gases decrease their
temperatures slightly when they undergo a free expansion.
 T 
Joule Coefficient:  J  

 V U
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First Law of Thermodynamics
Adiabatic Process:
P
dU
 U 
CV  


 T V dT
dU  CV dT
The first law becomes:
CP 
Adiabatic
đQ  CV dT  PdV
Isothermal
đQP  C  P V 


V
dT
V
 T  P
For ideal gas: CP  CV  nR
For a adiabatic process: 0  CV
After integration:
dT
dV
 nR
T
V
TV  1  const
van der Waals gas equation of state:
PV   const

n2a 
 P  2 V  nb   nRT
V 

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First Law of Thermodynamics
Throttling Process:
We do work on the left-hand side:
0
W    Pi dV  PiVi
Vi
The gas does work on the right-hand side:
Vf
W   Pf dV  Pf V f
0
Consider the gas on both sides as a whole
system
U f  U i  0  PiVi  Pf V f
U i  PiVi  U f  Pf V f
Hi  H f
This is a isenthalpic process.
Joule-Kelvin coefficient
 T 


P

H
 JK  
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First Law of Thermodynamics
Steady Flow Process:
We treat this unit mass as the system. Suppose the
shaft work done by the turbine is w, we can
summarize the following energy changes:
The internal energy changes by u2-u1.
The kinetic energy changes by ½(v22-v12).
The potential energy changes by g(z2-z1).
The net work done on the fluid is P1v1-P2v2-w.
The heat flow into the system is q.
KE  PE bulk  U  W  Q


1 2 v 22  v12  g z2  z1   u2  u1  P1v1  P2 v2  w  q


w  h1  h2  1 2 v12  v 22  g z1  z 2   q
This is the general equation for steady flow.
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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First Law of Thermodynamics
Steady Flow through a Nozzle:
for fast flow q=0
since w=0

w  h1  h2  1 2 v12  v 22

v12  v 22  2h2  h1 
Assuming an ideal gas and an adiabatic process

 T1 
P 
    1 
 T2 
 P2 
 1
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Second Law of Thermodynamics
Carnot Cycle:
hot body
Q1
E
W
Q2
cold body
Efficiency:
  W Q1
U  Q1  Q2  W  0
 1
W  Q1  Q2
Q2
Q1
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Second Law of Thermodynamics
The Kelvin-Planck statement:
It is impossible to construct a device that, operating in a cycle, will produce no effect
other than the extraction of heat from a single body at a uniform temperature and the
performance of an equivalent amount of work.
In a concise form: A process whose only effect is the complete conversion of
heat into work is impossible.
Key points:
There is no net change of the working system (substance).
The hot body is only a source of heat. It will not do work on the substance.
There is only a single hot body (or heat source).
The Clausius statement:
It is impossible to construct a device that, operating in a cycle, produces no effect
other than the transfer of heat from a colder to hotter body.
Q
reservo
ir at T
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Second Law of Thermodynamics
Schematic representation of the Kelvin-Planck (left) and Clausius (right) statements.
hot body
hot body
Q
Q
W=Q
R
Q
cold body
cold body
The statements are equivalent.
hot body
hot body
Q1
E
Q2
Q1+Q2
W=Q1
R
R
Q2
cold body
Q1
W=Q1-Q2
E
Q2
Q2
cold body
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Second Law of Thermodynamics
Carnot’s Theorem:
No engine operating between two reservoirs can be more efficient than a
Carnot engine operating between those same two reservoirs.
hot body
hot body
Q1
W
E
Q1
Q1
C
Q2
cold body
W
Q2=Q1-W
Q1
W
E
R
Q2=Q1-W
Q2=Q1-W
cold body
  C
All Carnot engine operating between the same two reservoirs have the same
efficiency.
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Second Law of Thermodynamics
Temperature Scale:
The efficiency of a Carnot engine operating between the two reservoirs is
dependent only on the temperatures of the reservoirs. This gives us a meaning of
defining a temperature scale, which is independent of any particular material. The
thermodynamic temperature T can be so defined that T1 and T2 for the two
reservoirs in a Carnot engine are related as,
C 
T1  T2
T
 1 2
T1
T1
 1
Q2
Q1
T1 Q1

T2 Q2
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Second Law of Thermodynamics
Temperature Scale:
For Carnot engine C12, T1 T2  Q1 Q2
T1
Q1
C12
For Carnot engine C23, T2 T3  Q2 Q3
T1 T3  Q1 Q3
Q2
Composite
engine C13
T2
Q2
C23
We use the symbol Tg for the gas scale temperature
and T for the thermodynamic temperature as just
defined. We will show that these two temperatures
are identical.
W12
W23
Q3
T3
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For isothermal bc:
Second Law of Thermodynamics
PV  nRT g1
đ Q  dU  PdV  PdV
since
we have
Vc
Vc
Vb
Vb
Q1   PdV  nRT g1 
Similarly,
dV
 nRT g1 ln Vc Vb 
V
Q2  nRT g 2 ln Vd Va 
Q1 T1 Tg1 ln Vc Vb 


Q2 T2 Tg 2 ln Vd Va 
For ab and cd adiabatics,
Tg1Vc 1  Tg 2Vd 1
Tg1Vb 1  Tg 2Va 1
Vc Vb  Vd Va
T1 Tg1

T2 Tg 2
Tg  const  T
Tg  T
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Efficiency:
C  1 
Second Law of Thermodynamics
Q2
T
1 2
Q1
T1
It is interesting to note that the efficiency would be 100% were we able to obtain a
lower temperature reservoir at absolute zero. This is forbidden by the third law.
Imagine now that the Carnot engine is run backwards to act as a refrigerator. The
coefficient of performance is defined as the heat extracted divided by the work
expended,
T1
Q
Q2
T2
CR  2 

W
Q1  Q2 T1  T2
Q1
One interesting application of the Carnot refrigerator is the
so-called heat pump. The efficiency of a Carnot heat pump
is,
Q
Q
T
1
HP
C 
1
W

1
Q1  Q2

1
T1  T2

C
W=Q1-Q2
Q2
T2
1  T2 T1
Since the efficiency of a Carnot heat pump rises as the
temperature difference between the reservoirs decreases,
heat pumps are best used in providing background heating.
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Otto-cycle:
ab:
bc:
cd:
da:
Second Law of Thermodynamics
P
c
TaV1 1  TbV2 1
Adiabatic
Q1  CV Tc  Tb 
b
Q1
Td V1 1  TcV2 1
d
a
Q2  CV Td  Ta 
 1
Td
V2
V1
Q2
V
T  Ta
Q2
1 d
Q1
Tc  Tb
 Ta V1 1  Tc  Tb V2 1
 V2 
  1   
 V1 
 1
1
1
rc 1
where rc is the compression ratio (=V1/V2). It is important to have as high a
compression ratio as possible in order to get a high efficiency.
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Entropy
đQ  0
T
Clausius inequality :
P
f
R2
For a reversible cycle, the equality sign holds.
R1
đ QR

R
T R

i
1
R1

i
f
đQR

T
R
2

R1 i
f
f
đ QR

T
R
2
đ QR
f T

i
f
đQR
0
T
i
V
i
đQR

T R
2

i
f
đQR
T
for reversible process R2:

f
R2 i
S  S f  S i 

R i
f
đ QR

T
R

2
i
f
đQR
T
đ QR
T
This state function is called entropy. For an infinitesimal reversible process,
đQR  TdS
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Entropy
Example:
Determine the entropy change of a beaker of water when it is heated at atmospheric
pressure between 20C and 100C.
Example:
Calculate the entropy change of an ideal gas undergoing a free expansion doubling
its volume.
Example:
A beaker of water is heated at atmospheric pressure from 20C and 100C by a
reservoir at 100C. Determine the total entropy change of water and reservoir.
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Entropy
Suppose now the path R1 is irreversible, according to the Clausius inequality,
f đQ
i đQ
đQ
 T  i T  R f T R  0

i
f
P
f
R2
đ Q  i đQR  f đQR  S  S
f
i

 T
T
R f T
R i
For an infinitesimal part of the process,
đQ  dS
T
R1
i
The equality sign holds if the process is reversible.
V
For an isolated system, dS  0
or for a finite process, S f  S i  S  0
Principle of increasing entropy:
The entropy of a thermally isolated system increases in any irreversible process and
is unaltered in a reversible process.
For a system thermally isolated from the surroundings: S approaches to a maximum;
For a system totally isolated from the surroundings: S approaches to a maximum
with U remains constant.
Heat & Light----by Dr.H.Huang, Department of Applied Physics
30
Hong Kong Polytechnic University
Entropy
The thermodynamic state of a system can be specified by any pair of independent
state functions. In particular, a state is equally well specified by the pair S and T as
by the pair P and V.
T
A Carnot cycle shown in a T-S diagram.
a
T1
For any reversible process
Q
R
Q1
adiabatic
 TdS
b
adiabatic
the net heat absorbed in a Carnot cycle is
given by the area shaded.
T2
d
Q2
c
The differential form of the first law is,
S
dU  đ Q  đ W
For an infinitesimal reversible process,
đ W   PdV
đ Q  TdS
dU  TdS  PdV
TdS  dU  PdV
This is the central equation of thermodynamics, or thermodynamic identity.
Example: Calculate the entropy of an ideal gas.
Heat & Light----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Thermodynamics Revision
Ideal Gas:
Equation of state: pV  nRT

Molar specific heat:


Heat capacity: Q  C T f  Ti  cm T f  Ti
Internal energy change:
c p  cV  R

U  CV T
Thermodynamics:
Work by the gas : đW=pdV
Entropy: dS  đQ
T
The first law: dU=đQ-đW
The second law: S  0
 U 
 S 
CV  
  T


T

T

V

V
 H 
 S 
CP  
  T

 T  P
 T  P
dU  TdS  PdV
Principle of increasing entropy
Efficiency of a thermal engine:  
W
Q
 1 2
Q1
Q1
Carnot cycle:   1 
Heat & Light----by Dr.H.Huang, Department of Applied Physics
T2
T1
32
Hong Kong Polytechnic University
Thermodynamics Revision
Special processes:
free expansion
U=0
throttling process
H=0
Basic concepts: state function; reversible process; irreversible process;
Thermal dynamic potentials (Legendre transformations):
H=U+PV;
F=U-TS;
G=H-TS
Maxwell Relations:
 T 
 P 






 V  S
 S V
 T 
 V 

 


P

S

S 
P
 P 
 S 

 


T

V

V 
T
 V 
 S 






 T  P
 P T
dU  TdS  PdV
dH  TdS  VdP
dF  PdV  SdT
dG  VdP  SdT
Cyclic Relation
Heat & Light----by Dr.H.Huang, Department of Applied Physics
33
Hong Kong Polytechnic University
Heat & Light----by Dr.H.Huang, Department of Applied Physics
34
Hong Kong Polytechnic University
Heat & Light----by Dr.H.Huang, Department of Applied Physics
35