Transcript 幻灯片 1

Chapter 14 The Second Law of
Thermodynamics
Main Points of Chapter 14
• Engines and refrigerators
• Second law of thermodynamics
• Carnot cycle
• Other types of engines
• Entropy
14-1 Engines and Refrigerators
An engine is a device that cyclically transforms
thermal energy into mechanical energy.
1. An engine must work in
cycles if it is to be useful –
otherwise, when one cycle
does work the engine stops.
2. A cyclic engine must
include more than one
thermal reservoir.
In a cyclic process, it is impossible to avoid dumping
heat. The following situation is impossible:
Hot reservoir
Qh
Engine
W = Qh
Why is this impossible?
We shall see that, for a cyclic process, the Second
Law of Thermodynamics prevents perfect
heat-to-work conversion.
P
Qh
a
A
Ql
B
b
V
Every engine has an efficiency, which is a measure
of what fraction of the heat flow from the hotter
thermal reservoir is converted into work.
A Refrigerator is an engine run in reverse
P
Qh
a
A
Ql
B
b
V
Thermal energy will be taken from a cold
reservoir and released to a hotter reservoir
Work will be done on the system by surroundings.
P
Qh
Reservoir Th
a
A
Qabs
Reservoir Tc
The coefficient of performance
B
b
V
14-2 The Second Law of Thermodynamics
The First Law of Thermodynamics tells us that
energy is always conserved for a thermal system.
When ice cubes are put into hot tea, the ice cubes
melt and the tea cools.
But there are processes that do not occur
spontaneously even though energy would be
conserved, such as a lukewarm drink forming
ice cubes and becoming hot.
1. Spontaneous processes
A crate sliding over an ordinary surface
eventually stops.
When ice cubes are put into hot tea ,the ice cubes
melt and the tea cools.
If you puncture a helium-filled balloon in a closed
room, the helium gas spreads throughout the room.
These processes occur spontaneously.
These phenomena can be summarized as follows:
a. Not all the thermal energy in a thermal system
is available to do work.
b. Thermal systems spontaneously change only
in certain ways, and in particular, spontaneous
heat flow always goes from a body at higher
temperature to a body at lower temperature.
2. Statistical Interpretation of spontaneous processes
Spontaneous processes: compressed gas in a
bottle will expand when bottle is opened, but
won’t spontaneously go back in – why not?
Suppose a box contains four molecules of gas
There are five possible configurations.
There are different arrangements of the
molecules microstate for one configuration
All microstates are equally probable.
configuration
number of
microstates
L4，R0
1
L3，R1
4
L2，R2
6
L1，R3
L0，R4
4
Probability
1
L2,R2 is the most probable configuration
For large values of N there are
extremely large numbers
of microstates, but nearly all the
microstates belong to the
configuration in which the
molecules are divided equally
between the two halves of the box.
Free expansion of an ideal gas is a spontaneous process.
Its directionality is determined by what takes you to
the state with the highest probability.
It is a irreversible processes. The reverse process could
happen, but the probability is impossibly small
3 The Second Law of Thermodynamics
There are many ways of expressing the second law
of thermodynamics.
The Kelvin form. It is impossible to construct a cyclic
engine that converts thermal energy from a body into
equivalent amount of mechanical work without a
further change in its surroundings.
P
Thermal energy can be converted P1
into equivalent amount of
P2
mechanical work ,
but the engine is not a cyclic.
o V1
V2
V
If the Kelvin formulation were not true
Q
W
T air or ocean
It is the second law that limits the efficiencies of heat
engines to values less than 100%.
The Clausius form. It is impossible to construct a cyclic
engine whose only effect is to transfer thermal energy
from a colder body to a hotter body.
Refrigerator is a device that transfer thermal energy
from a colder body to a hotter body,
but work must be done on it.
Two formulations are equivalent
If the Kelvin formulation were not true
Hot reservoir
Q
E
W
W+ Ql
Ql
Ql
Ql
F
Cold reservoir
It shows that the Clausius formulation would be
not true
If the Clausius formulation were not true
Hot reservoir
Q Q
F
E
Q
W
W
Ql
Q-Ql
Cold reservoir
It shows that the Kelvin formulation would be
not true
14-3 The Carnot cycle
A minimal version of an engine has two reservoirs at
different temperatures Th and Tc, and follows a
reversible cycle known as the Carnot cycle.
The Carnot cycle has four steps:
1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression
How the cycle might be realized
Step 1:
Step 3:
Step 2:
p
A
Step 4:
Qh
1
4
B
2
C
D
3
Th
TC
QC
Va Vd
Vb Vc
V
p
A
Qh
1
4
B
2
C
D
3
Th
TC
QC
Va Vd
A Carnot cycle is more efficient if the two
temperatures are far apart
Vb Vc
V
ACT Three Carnot engines operate between
temperature limits of (a) 400 and 500 K, (b) 500 and
600 K, and (c) 400 and 600 K. Each engine extracts the
same amount of energy per cycle from the hightemperature reservoir. Rank the magnitudes of the
work done by the engines per cycle, greatest first.
(c), (a), (b),
The importance of Carnot engine
Two results:
1. All Carnot cycles that operate between the same
two temperatures have the same efficiency. The
efficiency of a Carnot cycle does not depend on
the use of an ideal gas.
2. The Carnot engine is the most efficient engine
possible that operates between any two given
temperatures.
To demonstrate the first result
We show if it were not true we would violate
the second law of dynamics
Suppose the efficiency of Carnot engine A is higher
than Carnot engine B. let B run in reverse.
Such arrangement
violates the Kelvin
form the second law
of thermodynamics
To demonstrate the second result
We show if an irreversible engine is more efficient
than a Carnot engine we would violate the second
law of dynamics. let Carnot engine run in reverse.
Th
QB
QA
Carnot
engine
W
irreversible
engine
The net effect is to have a
net heat flow from Th with
DW work done, which violates
the Kelvin statement.
Qc
Qc
Tc
The argument can not be reversed, because irreversible
engine is not reversible.
ACT The heat engine below is:
1) a reversible (Carnot) heat engine.
2) an irreversible heat engine.
3) a hoax.
4) none of the above.
 Example Water near the surface of a tropical ocean
has a temperature of 298.2 K (25.0 °C), while water
700 m beneath the surface has a temperature of 280.2 K
(7.0 °C). It has been proposed that the warm water be
used as the hot reservoir and the cool water as the cold
reservoir of a heat engine. Find the maximum possible
efficiency for such an engine.
Solution
The maximum possible efficiency is the efficiency
of a Carnot engine
Using TH = 298.2 K and TC = 280.2 K
 Example Prove that two reversible adiabatic paths
cannot intersect
P
Proof
dT=0
Assume that two
reversible adiabatic
paths intersect at point A
B
C
A
o
V
Draw an isotherm which intersect with adiabats
at B and C respectively.
This cycle would violate the second law of
thermodynamics. So two reversible adiabatic
paths cannot intersect.
ACT Is it possible to cool a house by leaving a
refrigerator door open? What would be the net
effect if you were to leave the door open?
Solution No
A refrigerator is essentially a heat engine operated in reverse.
It takes in a certain amount of work (W), in the form of
electrical energy from the power outlet, extracts a certain
amount of heat (Qc) from the low-temperature source (the
interior of the refrigerator), and released some heat (Qh) into
the high temperature environment (the house). According to
the first law of thermodynamics Qh = Qc + W. So, the net
effect is that the refrigerator consumes some electrical
energy and turns it into heat that is released into the house.
Keeping the refrigerator door open only makes the house
even warmer over all.
14-4 Other Types of Engines
Any closed curve in the P-V diagram represents a
reversible cycle
The Stirling Engine
1mol ideal gas
Otto cycle (Gasoline Engine) P
combustion
3
1mol ideal gas
adiabatic
exhaust
2
4
adiabatic
o
V2
1
V1
V
P
3
adiabatic
2
4
adiabatic
o
compression ratio
V2
1
V1
V
There are other possible cycles; here are a few:
 Example a reversible cycle
P A
Isotherms :AB,CD,EF
T1
B
C T
2
F
Adiabats: BC,DE,FA
Find
Solution
T3
o
G
D
E
V
To extrapolate Adiabat BC we have two Carnot cycles
Heat Pumps
Heat pumps and refrigerators are engines run in
reverse:
Refrigerator removes heat from cold reservoir, puts it
into surroundings, keeping food in reservoir cold.
Heat pump takes energy from cold reservoir and puts
it into a room or house, thereby warming it.
In either case, energy must be added!
Air conditioner
W
In summer
W
In winter
Heat pump
It pumps heat “uphill” from a lower temperature to a
higher temperature, just as a water pump forces water
uphill from a lower elevation to a higher elevation.
The coefficient of performance
It is easier to transfer thermal energy from the cold ground
to a warm house if the temperature difference is small
 Example There is a 70 W heat leak from a room at
temperature 22 °C into an ideal refrigerator. How
much electrical power is needed to keep the
refrigerator at -10 °C?
Solution
For the fridge, Qc must exactly compensate
the heat leak. So Qc = 70 J per second
For Carnot cycle
Power required = 8.5 W
 ACT An ideal or Carnot heat pump is used to
heat a house to a temperature of TH = 294 K (21 °C).
How much work must be done by the pump to deliver
QH = 3350 J of heat into the house when the outdoor
temperature TC is (a) 273 K (0 °C) and
(b) 252 K (-21 °C)?
Solution
For Carnot cycle
It is more difficult to transfer thermal energy from
the cold ground to a warm house if the temperature
difference is larger.
 Example Work from a hot brick
Heat a brick to 400 K. Connect it to a Carnot Engine.
How much work can we extract if the cold reservoir is
300 K? The brick has a constant heat capacity of C = 1
J/K.
Solution Did you use the relation: W = Qh(1 -Tc/Th) ?
If so, you missed that the brick was cooling
during the process.
Suppose the temperature of the brick was T
dW = dQh (1 -Tc/T) =-CdT (1 -Tc/T)
14-5 Entropy and the Second Law
1. Entropy as a Thermodynamic Variable
In a Carnot cycle
To generalize to any reversible cycle
Clausius’ equation
The integral
depends only on the initial
rev
and final states, and not on the path
There must be a function S that depends only on
the state of the gas and not on how it got that way;
this is called entropy.
Definition of Entropy:
rev
 Example Calculate the entropy change of 1 mol of
ideal gas that undergoes an isothermal transformation
from an initial state to a final state
Solution
rev
 Act Calculate the entropy change of 1 mol of ideal
gas that undergoes an reversible adiabatic transformation
 Example a reversible cycle
P A
Isotherms :AB,CD,EF
T1
B
C T
2
F
Adiabats: BC,DE,FA
Find
Solution
Clausius’ equation
T3
o
G
D
E
V
2. Entropy of an Ideal Gas
To calculate the entropy change of 1mol ideal gas
from initial state (pi.Vi,Ti) to final state (pf,Vf,Tf)
We design a reversible process that takes the
system from the initial state to the final state.
isochore
isotherm
(pf,Vf,Tf)
(pi,Vi,Ti)
(p,Vf,Ti)
The general form of the entropy for ideal gas
P
T
S
V
ACT Point i in Figure below represents the initial state
of an ideal gas at temperature T. Taking algebraic signs
into account, rank the entropy changes that the gas
undergoes as it moves, successively and reversibly, from
point i to points a, b, c, and d, greatest first.
b, a, c, d
T-S diagram
For a Carnot cycle
T
TH
Q=W
TC
S1
S2
S
3. How Entropy Changes for Irreversible or
Spontaneous Processes
To find the entropy change for an irreversible process
occurring in an isolated system, replace that process with
any reversible process that connects the same initial
and final states. Calculate the entropy change for this
reversible process.
 Example Free expansion for an ideal gas
One mole of nitrogen gas is confined to the left side of
the container . You open the stopcock and the volume of
the gas doubles. What is the entropy change of the gas
for this irreversible process? Treat the gas as ideal.
Solution
We know that the temperature of the gas does not
change in the free expansion.
So we use the reversible process of isothermal expansion
to connect the same initial and final states.
Free expansion is neither adiabatic nor isothermal
expansion, even though DT and Q are zero.
Free expansion is an irreversible process --- the gas
molecules have virtually no chance of returning to the
original state.
 Example Figure below shows two identical copper
blocks of mass m = 1.5 kg: block L at temperature TiL =
60°C and block R at temperature TiR = 20°C. The
blocks are in a thermally insulated box and are
separated by an insulating shutter. When we lift the
shutter, the blocks eventually come to the equilibrium
temperature Tf = 40°C .What is the net entropy
change of the two-block system during this irreversible
process? The specific heat of copper is 386 J/kg·K.
Solution
To calculate the entropy change, we must find a
reversible process that takes the system from the initial
state to the final state.
For such a reversible process we need a thermal
reservoir whose temperature can be changed slowly (say,
by turning a knob). We then take the blocks through the
following two steps.
ACT A quantity of 0.20 mol of argon gas
contained in a volume of 5.0L mixed with 0.50 mol
of neon gas contained in a volume of 12.5L, making
a total volume of 17.5L.Both gases are at the same
temperature and can be regarded as ideal. Each
volume is thermally isolated. What is the change in
entropy ?
Solution
It is a irreversible process
Entropy of mixing
4. The Entropy of an Isolated System Never Decreases
If a process occurs in an isolated system, the
entropy of the system increases for irreversible
processes and remains constant for reversible
processes. It never decreases.
The second law of thermodynamics
in terms of entropy
The greater-than sign applies to irreversible processes
The equals sign to reversible processes
A spontaneous process in an isolated system
increases the system’s entropy
For an isolated system the state of maximum
entropy is the state of stable equilibrium
To demonstrate that the entropy of an isolated
system never decrease
Irreversible processes (e.g. processes with friction )
From the second result of Carnot cycle:
irreversible elements reduce the efficiency
To generalize to any irreversible cycle
Clausius’ inequality
Consider an irreversible cycle 1a2b1
P
1
a
b
2
V
irrev
For isolated system
irreversible processes
reversible processes
5. Engines and Entropy
The principle that entropy never decreases is a
way of formulating the second law of
thermodynamics that emphasizes the
directionality of thermal processes
We show how the entropy formulation of the
second law of thermodynamics applies to engines
and leads to the Kelvin formulation.
Can we construct an engine that in a single cycle
extracts thermal energy from the reservoir and
Totally converts into work?
Suppose we could do so.
The change of the entropy of an isolated system
that consists of the engine and the reservoir is
This process violates the entropy formulation of the
second law of thermodynamics.
So we should introduce the second reservoir (TC)
The change of the entropy of the isolated system
that consists of the engine and two reservoirs is
Here QC must be positive, heat flow into the reservoir
The entropy formulation of the second law of
thermodynamics leads to the two results of Carnot cycle
14-6 The Meaning of Entropy
If an irreversible process occurs in a isolated system,
the entropy of the system increases.
Example 1: Free expansion of an ideal gas
The initial state is more order than the final state
because the individual molecules are more localized
The initial state has lower probability than the
final state
Example 2: the melting of a block of ice
a block of ice with each of its H2O molecules
fixed rigidly in place in a highly structured
and ordered arrangement.
the puddle of water into which the ice melts is
disordered and unorganized, for the molecules
in a liquid are free to move from place to place
Example 3
With the aid of 100 pounds of
dynamite, demolition experts
caused this hotel-casino in Las
Vegas to go from an ordered
state (lower entropy) to a
disordered state (higher
entropy).
Entropy is a measure of the disorder of a system.
For an isolated system, the highest entropy is one
with the greatest number of microstates and thus
the highest probability.
In1877, Austrian physicist Lugwig Boltzmann
derived the relation between the entropy S and
the number of microstates W of
that configuration.
This famous formula is
engraved on Boltzmann’s
tombstone.
Order, Entropy, Probability, and the Arrow of Time
Entropy can be interpreted in terms of order
and disorder.
Many familiar processes increase entropy –
shuffling cards, breaking eggs, and so on.
We never see these processes spontaneously
happening in reverse – a movie played backwards
looks silly. This directionality is referred to as the
arrow of time.
You can judge the time sequence by the evolution of
an isolated system of large number of particles.
Spontaneous processes proceed in one direction,
in which entropy increases.
 Example When one mole of an ideal gas doubles its
volume in a free expansion, the entropy increase
from the initial state i to the final state f is
Derive this result with statistical mechanics.
Solution
*The degradation of energy
What does it mean to conserve energy if energy is
always conserved ?
Some forms of energy are more useful for doing
work than others.
If an irreversible process occurs, energy is
conserved, but some of the energy becomes
unavailable to do work and is “wasted” .
The hot-to-cold heat flow is irreversible
The lost of heat flow is the lost of
the ability to do work
Energy is always conserved. But
the ability to do work
reduces.
The change of the entropy
Energy Unavailable for Doing Work
The irreversible process of heat flow through the copper
rod causes energy to become unavailable for doing work
in the amount of Wunavailable = 240 J.
In an irreversible process, energy becomes
unvailable to do work
T0 is the Kelvin temperature of the coldest heat reservoir.
Since irreversible processes cause the entropy of
the universe to increase, they cause energy to be
degraded, in the sense that part of the energy
becomes unavailable for the performance of work.
Act A crate sliding over an ordinary surface
eventually stops. Calculate the energy unavailable
for doing work for this irreversible process.
Solution
Suppose the total kinetic energy K is converted
into heat and is transferred to a heat reservoir (T).
A Carnot engine operates between the hot reservoir
(T) and a cold reservoir(T0). It absorbs heat K from
the hot reservoir.
The work done by the engine
the energy unavailable for doing work
Summary of Chapter 14
• Cyclic engines do work in repeated cycles by
extracting heat flow from hotter systems. Efficiency:
• Carnot cycle is the most efficient between two
reservoirs:
Summary of Chapter 14, cont.
• Entropy is a measure of disorder in a system; only
changes in entropy have physical significance.
•For infinitesimal reversible transformations:
• For an ideal gas ( 1 mol ):
Summary of Chapter 14, cont.
• For a reversible cycle:
• For an irreversible process:
irrev