Transcript bd Beach

Chapter 16 Thermodynamics
第十六章
熱學
What is heat?
Heat can flow from a hotter
body to a colder body.
In the eighteenth century, heat
was believed to be an invisible,
massless fluid called “caloric”.
The caloric theory assumes
that caloric is a conserved
quantity and that the particles
of caloric repel each other but
are attracted to the particles of
ordinary matter.
Heat flow
Heat is the energy transferred between a system and its environment
because of a temperature difference that exists between them.
Raising temperature
Heat capacity
In 1760 Joseph Black was the first person to realize that the rise in
temperature of a body could be used to determine the quantity of
heat absorbed by it.
Q
Heat capacity 
T
A commonly used (non-SI) unit of heat is the calorie, which is defined
as the quantity of heat required to raise the temperature of 1 g of
water from 14.5 °C to 15.5 °C.
1 calorie  4.186 J
A British unit for heat is Btu, that is the amount of heat required to
raise the temperature of 1 lb of water from 63 °F to 64 °F.
1 BTU  252.5 calorie  1057 J
Specific heat
The amount of heat required to raise a temperature change of ΔT of a
material is proportional to its mass.
Q  mcT
The proportional constant c is defined to be the specific heat of the
material.
1 Q
c
m T
The molar specific heat is defined as C = Mc, where M is the molar mass.
Q  nCT
where n is the number of moles in the material.
Specific heat of some materials
Specific heat of water
Example 1
During a bout with the flu an 80-kg man ran a fever of 39.0°C instead of
the normal body temperature of 37.0°C. Assuming that the human body is
mostly water, how much heat is required to raise his temperature by that
amount?
Q  mcT  80  4180  2 J  6.7 105 J  160 Cal
Determining specific heat
The method mixtures
Q1  Q2  0
m1c1T1  m2c2T2  0
T1  T f  T1
T2  T f  T2
Latent heat
Joseph Black also
recognize that adding
heat to a system does
not always result in a
change of its
temperature. This
happens when the
material in the system
undergoes a first-order
phase change.
The latent heat L is defined to be the heat needed to convert the material of
one unit mass from one phase to another at its phase transition temperature.
Q  mL is the heat needed for a sample of mass m.
Latent heats and phase transitions of
water
The surrounding air is at room temperature, but this ice–water (watervapor) mixture remains at 0°C (100°C) until all of the ice (water) has
melted (evaporated) and the phase change is complete.
Latent heats of some materials
Generation of heat by friction
In 1798 Benjamin Thompson noticed
that a large amount of heat was
generated during the process of
boring of cannons. It seemed to him
that heat could be continually
produced by mechanical work.
The mechanical equivalent of heat
James Joule first performed the mechanical work-heat
conversion experiment in 1845, and he repeated the
experiment several times. His final result after 40
years of work is 778 ft·lb of work is equivalent to 1 Btu.
The mechanical equivalent of heat
1Btu  778 ft  lb
 778  (12  2.54 /100)  (9.8 / 2.2)m  N
 1056 J
1Btu  252.5calorie
1calorie  4.183J
Modern day value:
1calorie  4.186J
Heat is energy transferred between two bodies as a consequence
of a temperature difference between them.
Signs for heat and work in
thermodynamics
Work done during volume change
dW  Fdx  pAdx  pdV
Work done during volume change
V2
W   pdV
V1
Example 2
An ideal gas undergoes an isothermal (constant-temperature)
expansion at temperature T, during which its volume changes from V1 to
V2. How much work does the gas do?
V2
W   pdV
V1
V2
W 
V1
pV  nRT
nRT
V2
dV  nRT n( )
V
V1
V2
p1
W  nRT n( )  nRT n( )
V1
p2
Path between thermodynamic states
Heat added in a thermodynamic process
Internal energy and the first law of
thermodynamics
First law of thermodynamics:
U  Q  W
Internal energy is a function of
thermodynamic states
U  Q  W
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 internal energy of a cup of coffee
depends on just its thermodynamic state—
how much water and ground coffee it
contains, and what its temperature is. It does
not depend on the history of how the coffee
was prepared—that is, the thermodynamic
path that led to its current state.
Cyclic processes and isolated systems
A cyclic process is a process
that eventually returns a
system to its initial state.
U  Q  W  0
Q  W
For an isolated system there
are no work done and no
heat exchange with its
surroundings.
Q  W  0 U  0
Infinitesimal change of states
U  Q  W
dW  pdV

dU  dQ  dW
dU  dQ  pdV
Example 3
You propose to eat a 900-calorie hot fudge sundae (with whipped
cream) and then run up several flights of stairs to work off the energy
you have taken in. How high do you have to climb? Assume that your
mass is 60.0 kg.
One food calorie equals 1000 cal, or 1 kcal.
1cal  4.190J
Q  W  mgh
Q
h
 6.41km
mg
Example 4
The figure below shows a pV-diagram for a cyclic process, one in which
the initial and final states are the same. It starts at point a and proceeds
counterclockwise in the pV-diagram to point b, then back to a, and the
total work is W = 500J. (a) Why is the work negative? (b) Find the
change in internal energy and the heat added during this process.
The work done equals the area under the
curve, with the area taken as positive for
increasing volume and negative for
decreasing volume. The positive work
done by the system from a to b is less
than the absolute value of the negative
work by the system from b to a.
U  0
Q  W  500 J
500 J of heat coming out of the system.
Example 5
A series of thermodynamic processes is shown in the pV-diagram of
the attached figure. 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).
U  Q  W
In process ab:
In process bd:
is valid for each process.
W  0  U ab  Qab
Wbd  pb (Vd  Vd )
 U bd  Qbd  Wbd
U ad  U ab  U bd
U bd  Qacd  U ad  Wac
Example 6
One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at
a constant pressure of 1 atm (1.013  105 Pa). The heat of vaporization at
this pressure is LV = 2.256  106 J/kg. Compute (a) the work done by the
water when it vaporizes and (b) its increase in internal energy.
W  p(V2  V1 )  1.013 105 1670 106 J  169 J
Q  mLV  2256 J
U  Q  W  2087 J
Types of thermodynamic processes
(a) adiabatic process: no heat transferred into or from the system.
Q0
U  W
(b) isochoric process: a constant-volume process.
W 0
 U  Q
(c) isobaric process: a constant pressure process.
W  p(V2  V1 ) U  Q  p(V2  V1 )
(d) Isothermal process: a constant temperature process.
T  0
(e) Adiabatic free expansion: a free expansion without exchanging
heat or doing work to the system’s surrounding.
Q W  0
while V increases.
What kind of process is it?
When the cork is popped on a bottle
of champagne,
When the partition is broken,
When the soup is heated in a
pot,
Ideal gases
The internal energy of an ideal gas depends only on its temperature,
not on its volume or pressure.
Molar specific heat at constant volume: CV
Since the work done by the gas is zero, we have
U  Q  nCV T
Molar specific heat at constant pressure: Cp
For n moles of a gas: Q  nC p T
Since the work done by the gas is pV, we have
U  Q  W  nC p T  pV
nCV T  nC p T  pV
nCV T  nC p T  nRT
C p  CV  R
Molar specific heats of some gases
The ratio of heat capacities, Cp/CV, is denoted by .

Cp
CV
Example 7
A typical dorm room or bedroom contains about 2500 moles of air.
Find the change in the internal energy of this much air when it is
cooled from 23.9C to 11.6C at a constant pressure of 1.00 atm. Treat
the air as an ideal gas with  = 1.400.
U  nCV T
C R

 V
CV
CV
Cp
R
 CV 
 1
Adiabatic processes for an ideal gas
In an adiabatic process:
dQ  0  dU   pdV
From PV = nRT, we have
pdV  Vdp  nRdT
CV ( pdV  Vdp)  RnCV dT   RpdV
(CV  R) pdV  CV Vdp  0
dV
CV  R


CV
CV
Cp

pV  const.
dp

0
V
p
or
TV  1  const.
nCV dT   pdV
Isothermal work and free expansion
Isothermal process: the system is kept in contact with a single heat
bath at temperature T when it expands or contracts.
Isothermal expansion
Free expansion
Isothermal work by ideal gas
dW  pdV
V2
W   pdV
V1
nRT
p
V
dV
W  nRT 
V1 V
V2
W  nRT n( )
V1
V2
Example 8
The compression ratio of a diesel engine is 15 to 1; this means that air
in the cylinders is compressed to 1/15 of its initial volume (Fig. 19.21).
If the initial pressure is 1.01  105 Pa and the initial temperature is 27C,
find the final pressure and the temperature after compression. Air is
mostly a mixture of diatomic oxygen and nitrogen; treat it as an ideal
gas with  = 1.40.
The piston compresses quickly, so
it is an adiabatic process.


p1V1  p2V2
 1
T1V1
 1
 T2V2
V1 
p2  p1 ( )
V2
V1  1
T2  T1 ( )
V2
Example 9
In Example 8, how much work does the gas do during the
compression if the initial volume of the cylinder is 1.00 L. Assume that
CV for air is 20.8 J/mole·K and  = 1.40.
V1 
p  p1 ( )
V
V2
W   pdV
V1
V2
W 
V1
V1 
 V2

p1 ( ) dV  p1V1  V dV
V1
V

W  p1V1
 1
2
V
 1
1
V
  1
p1V1  p2V2

 1
Speed of sound
v
dP
B  V
dV
B

Newton assumed that the compressions and rarefactions occur
isothermally:
dP
nRT
B  V
 V ( 2 )  P
dV
V
v
B
P



nM / V
RT
M
The correct process is adiabatic: B   P
v
 RT
M
The second law of thermodynamics tells us that heat
naturally flows from a hot body (such as a freshly cooked
ear of corn) to a cold one (such as a pat of butter). Is it ever
possible for heat to flow from a cold body to a hot one?
Irreversible processes
An egg is dropped onto a floor,
a pizza is baked,
a car is driven into a lamppost,
large waves erode a sandy beach,
—these one-way processes are irreversible, meaning that they cannot
be reversed by means of only small changes in their environment.
Reversible processes
Reversible processes are thus equilibrium processes, with the system
always in thermodynamic equilibrium.
The limit of a heat
engine
Industrial revolution of the West
Steam engine and Carnot’s analysis
Heat engines
Any device that transforms heat partly into work or mechanical energy is
called a heat engine.
Usually, a quantity of matter inside the engine undergoes inflow and
outflow of heat, expansion and compression, and sometimes change of
phase. We call this matter the working substance of the engine.
A cyclic process is a sequence of processes that eventually leaves the
working substance in the same state in which it started.
All heat engines absorb heat from a source at a relatively high
temperature, perform some mechanical work, and discard or reject
some heat at a lower temperature. A heat reservoir is a large
thermodynamic system that can exchange heat with the working
substance without changing the reservoir’s temperature.
Heat engines and thermal efficiency
The thermal efficiency  of a heat engine is defined as the work
output divided by the heat input:
W

QH
W  QH  QC  QH  QC
  1
QC
QH
Example 10
A gasoline engine in a large truck takes in 10,000 J of heat and delivers
2000 J of mechanical work per cycle. The heat is obtained by burning
gasoline with heat of combustion LC = 5.0  104 J/g. (a) What is the
thermal efficiency of this engine? (b) How much heat is discarded in
each cycle? (c) How much gasoline is burned in each cycle? (d) If the
engine goes through 25 cycles per second, what is its power output in
watts? In horsepower? (e) How much gasoline is burned per second?
Per hour?

W
2000 J

 0.2
QH 10000 J
QH
mgasoline 
 0.20 g
LC
QC  QH  W  8000 J
P  (25cycles / s)  (2000 J / cycle )  50kW
gas burned per second : (0.20 g / cycle )  (25cycle / s)  5.0 g / s
The gasoline engine (Otto Cycle)
The gasoline engine (Otto Cycle)
The six steps in the idealized Otto cycle:
1. Intake stroke (x to a): VxrV
2. Compression stroke (a to b): rVV
3. Ignition (b to c): V
4. Power stroke (c to d): V rV
x
5. Exhaust (d to a): rV
6. Exhaust stroke (a to x): rVVx
Efficiency of Otto cycle
QH  nCV (Tc  Tb )  0
QC  nCV (Ta  Td )  0
  1
QC
QH
Td  Ta
 1
Tc  Tb
For the two adiabatic processes ab and cd:
Ta (rV ) 1  TbV  1
  1
1
r
 1
Td (rV ) 1  TcV  1
for an ideal Otto cycle.
  0.56 for r  8 and   1.4, while   0.35 for real engines.
The Diesel engine
The four steps in the Diesel cycle:
1. Compression stroke (a to b): rVV
2. Fuel injection and ignition (b to c):
V Vc at constant Pb
3. Power stroke (c to d): VcrV
4. Cooling (d to a): rV
Typical theoretical efficiency of an
Diesel engine is 0.65 to 0.70.
Real efficiencies of heat engines
Gasoline engines: about 56%.
Diesel engines: about 65% to 70%.
Refrigerators
A refrigerator as a heat engine operating in reverse, i.e. it takes heat
from a cold place (the inside of the refrigerator) and gives it off to a
warmer place (usually the air in the room where the refrigerator is located).
The coefficient of performance K is
defined as
QC
K
W
QH  W  QC
QC
K
 QC  QH
Coefficient of performance (COP) of a
heat pump is defined as:
K
QH
W

QH
 QC  QH
Principle of the mechanical
refrigeration cycle.
An air conditioner
QC H
K

W
P
where H is the heat removal
rate and P is the power
needed, and K is typically 3.
EER is the coefficient of
performance with H given in
BTU/H and P in watt (W).
1W  3.413BTU / H
The second law of thermodynamics
It is impossible to build a heat engine that converts heat completely to
work—that is, an engine with 100% thermal efficiency.
The “engine” statement or the Kelvin-Planck statement of the second law:
It is impossible for any system to undergo a process in which it
absorbs heat from a reservoir at a single temperature and converts
the heat completely into mechanical work, with the system ending in
the same state in which it began.
The “refrigerator” statement or the Clausius statement of the second law:
It is impossible for any process to have as its sole result the transfer
of heat from a cooler to a hotter body.
The Kelvin-Planck statement of the
second law
It is impossible for a heat engine that
operates in a cycle to convert its
input heat completely into work.
The Clausius statement of the second law
It is impossible for a cyclic device to
transfer heat continuously from a
cold body to a hot body without the
input of work or other effect on the
environment.
Equivalence of the Kelvin-Planck and
Clausius statements
Equivalence of the Kelvin-Planck and
Clausius statements
Reversible and irreversible processes
For a process to be reversible three conditions must be satisfied:
(a) It must be quasistatic.
(b) There must be no friction.
(c) Any transfer of heat must occur at constant temperature,
or be associated with infinitesimal temperature difference.
Irreversible and reversible engines
The temperature of the firebox of a steam engine is much higher than the
temperature of water in the boiler, so heat flows irreversibly from firebox
to water. Carnot’s quest to understand the efficiency of steam engines led
him to the idea that an ideal engine with only reversible processes.
The Carnot Cycle
In 1824, Sadi Carnot devised a reversible cycle of operations, called Carnot
cycle.
Efficiency of the Carnot cycle
a to b:
c to d:
Vb
QH  nRTH n( )
Va
V
QC  nRTC n( c )
Vd
For an adiabatic process, we can show:
TV  1  const.
TH Vb 1  TCVc 1
Vb Vc

Va Vd
TH Va 1  TCVd 1
QC
QH
TC

TH
Carnot efficiency:  C  1 
TC
TH
Example 11
A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does
some work, and discards some heat to a reservoir at 350 K. How much
work does it do, how much heat is discarded, and what is the efficiency?
QC
QH
TC

TH
TC
QC 
QH  1400 J
TH
W  QH  QC  600 J
C  1
TC
 0.30
TH
Example 12
Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot
cycle with temperatures 227°C and 27°C. The initial pressure is 10.0  105
Pa, and during the isothermal expansion at the higher temperature the
volume doubles. (a) Find the pressure and volume at each of points a, b, c,
and d in the pV-diagram. (b) Find Q, W, and U for each step and for the
entire cycle. (c) Determine the efficiency directly from the results of part
(b), and compare it with the result from  C  1 TC / TH .
pV  nRT
Va  nRTa / pa  8.3110 4 m3
Vb  2Va  1.66 10 3 m3
pb  paVa / Vb  5.00 105 Pa
TV  1  const.
QH  nRTH n(Vb / Va )  0.200  8.314  500  n(2) J
Carnot refrigerator
K
QC
W
K Carnot

QC
QH  QC
TC

TH  TC
Example 13
If the cycle described in Example 3 is run backward as a refrigerator,
what is its coefficient of performance?
K Carnot
TC
300


 1.5
TH  TC 500  300
Carnot’s Theorem
All reversible engines operating between two given reservoirs have
the same efficiency.
No cyclical heat engine has a greater efficiency than a reversible
engine operating between the same two temperatures.
The Kelvin temperature scale
QC
QH
TC

TH

T
Q
QR
TR
If we choose the reference temperature to be 273.16 K at the triple
point of water, then it follows that the Kelvin and ideal-gas scales
are identical.
Entropy
The second law of thermodynamics, as we have stated it, is rather
different in form from many familiar physical laws. It is not an equation or
a quantitative relationship but rather a statement of impossibility.
However, the second law can be stated as a quantitative relationship
with the concept of entropy.
We have talked about several processes
that proceed naturally in the direction of
increasing disorder. Entropy provides a
quantitative measure of disorder.
Reversible cycles
Any reversible cycle can be approximated
by a series of Carnot cycles as shown in
the left figure.
For a Carnot cycle, we have:
QC
QH
TC

TH
or
QH QC

0
TH TC
Thus for a reversible cycle, we have:
dQ
Q
 T  T 0
Thermodynamic definition of entropy
b dQ
b dQ
dQR
R
R


 T a T a T  0
path I
path II
Thus we can define a state function S
as the potential energy from a
conservative force:
S  S f  Si  
i
or
dS 
dQR
T
f
dQR
T
Example 14
One kilogram of ice at 0C is melted and converted to water at 0C.
Compute its change in entropy, assuming that the melting is done
reversibly. The heat of fusion of water is Lf = 3.34  105 J/kg.
S  
i
f
dQR Q 3.34 105 J


 1.22 103 J / K
T
T
273K
Example 15
One kilogram of water at 0C is heated to 100C. Compute its change
in entropy.
S  
i
f
T f mcdT
Tf
dQR

 mcn( )
Ti
T
T
Ti
373
S  1.00kg  (4.19kJ / kg  K )  n(
)  1.31103 J / K
273
Reversible process for an ideal gas
The first law:
dQ  dU  dW
nRT
dQ  nCV dT  pdV  nCV dT 
dV
V
dQ
dT
dV
 nCV
 nR
T
T
V
f dQ
Tf
Vf
dT
dV
S  
  nCV
  nR
i
T
Vi
i
T
T
V
S  nCV n(
Tf
Ti
)  nRn(
Vf
Vi
)
Example 16
A gas expands adiabatically and reversibly. What is its change in
entropy?
In an adiabatic process, no heat enters or leaves the system.
Hence dQ = 0 and there is no change in entropy in this reversible
process.
Every reversible adiabatic process is a constant entropy process.
For this reason, reversible adiabatic processes are also called
isentropic processes.
The increase in disorder resulting from the gas occupying a greater
volume is exactly balanced by the decrease in disorder associated
with the lowered temperature and reduced molecular speeds.
Example 17
A thermally insulated box is divided by a partition into two compartments,
each having volume V. Initially, one compartment contains n moles of an
ideal gas at temperature T, and the other compartment is evacuated. We
then break the partition, and the gas expands to fill both compartments.
What is the entropy change in this free-expansion process?
Since W = 0, we have ΔU = 0.
For an ideal gas, it means that
Tf = Ti = T.
To figure out ΔS we need to find a
reversible process that links the
final state to the initial state. It is a
quasistatic isothermal expansion.
S g  nRn(
Vf
Vi
)
Note that there is no change
in its surrounding, so ΔSe = 0.
Su  S g  S e  0
Example 18
For the Carnot engine in Example 11, find the total entropy change
in the engine during one cycle.
QC  1400J
QH
S H 
 4J / K
TH
QC
S C 
 4 J / K
TC
STotal  S H  SC  0
Entropy in irreversible processes
In an idealized, reversible process involving only equilibrium states,
the total entropy change of the system and its surroundings is zero.
But all irreversible processes involve an increase in entropy.
Unlike energy, entropy is not a conserved quantity. The entropy of an
isolated system can change, it can never decrease.
Example 19
Suppose 1.00 kg of water at 100C is placed in thermal contact with 1.00
kg of water at 0C. What is the total change in entropy? Assume that the
specific heat of water is constant at 4190 J/kg·K over this temperature
range.
S H  
Tf
Ti
SC  
Tf
Ti
STotal
323mcdT
dQR
373

 mcn(
)
373
T
T
323
323 mcdT
dQR
323

 mcn(
)
273
T
T
273
323  323
 S H  SC  mcn(
)  102 J / K
273  373
Entropy and the second law
In a reversible process the entropy of an isolated system stays
constant; in an irreversible process the entropy increases.
S  0
Clausius statement leads to ΔS  0
and vice versa.
The availability of energy
One can convert work into heat completely, but cannot convert heat
into work completely. As matter of fact, convert some of the heat at
one temperature into work requires the existence of another
reservoir at a lower temperature.
We note that:
1. The degradation of energy is associated with the transition of the
system from an ordered to a disordered state.
2. In any natural (irreversible) process, some energy becomes
unavailable to do useful work.
Entropy and disorder
The second law requires that an isolated system tends to evolve
towards states of higher entropy.
A local decrease in entropy of a system can occur at the expense
of a greater increase in entropy in its environment.
The second law is often called “time’s arrow” because it tells us
that in (irreversible) natural processes an isolated system always
evolves toward states of higher entropy.
One prediction of the second law about the evolution of the
universe is that it will evolve toward a state of thermodynamic
equilibrium characterized by uniform density, temperature and
pressure. This gloomy state of maximum entropy is usually
referred to as the “heat death”.
Summary_1
Heat: a transfer of energy
Specific heat: Q  mcT  nC T
Latent heat: Q  mL
Work done in a quasistatic process: W 

Vf
Vi
PdV
First law of thermodynamics: U  Q  W
Internal energy of an ideal gas: U  nCv T
Relationship between molar specific heats: C p  Cv  R
Quasistatic adiabatic:
PV   const.
Summary_2
The thermal efficiency  of a heat engine is defined as the work
output divided by the heat input:
QL
W

 1
QH
QH
The second law of thermodynamics:
Kelvin-Plank: It is impossible for a heat engine that operates in
a cycle to convert its heat completely into work.
Clausius: It is impossible for a cyclic device to transfer heat
continuously from a cold body to a hot body without the input
of work or other effect on the environment.
In a reversible process the entropy of an isolated system stays
constant; in an irreversible process the entropy increases.
Summary_3
Entropy is a state function. The change in entropy S between a
initial state to a final state is defined by any reversible process that
links the two equilibrium states.
S  S f  Si  
i
f
dQR
T
Entropy is a measure of the disorder in a system. The second law
states that the natural (irreversible) processes tend to evolve to
states of greater disorder, or from states of low probability to states
of high probability.
Further reading & homework
• The Feynman lectures on Physics, Vol. 1,
Chapter 44, 45, 46
• Questions: 16.1, 16.5, 16.8, 16.10, 16.16
• Problems: 16.13, 16.17, 16.18, 16.20, 16.25,
16.30, 16.31, 16.35, 16.36