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Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Chapter 20: (REVIEW) Heat Engines
The concept of a heat engine is crucial to modern civilization. Heat engines,
beginning with James Watt’s invention of the steam engine, are the basis of
the mechanical age where machines do work. The vast majority of machines
which produce useful work do so ultimately by burning fuel and converting the
chemical potential energy stored in the fuel into heat and then converting that
heat into work.
Thermodynamic Limits to Heat Engines
The First Law of Thermodynamics states the equivalence of all forms of
energy, and the equivalence of heat and work as different forms of energy which
must be conserved. In principle, according to the First Law of Thermodynamics,
it should be possible to convert all the heat energy into useful work. In reality,
this is found to be impossible. There are fundamental limits to how much work
can be extracted from a given amount of heat in any real device, and the rest of
the energy must unfortunately be wasted. This is the Second Law of Thermodynamics.
Ultimate Efficiency of Heat Engines
All real heat engines perform work by extracting heat Qh from a hot source at
temperature Th , converting part of that heat into work W , and then discarding
an amount of heat Qc into a cold sink at temperature Tc . According to the
First Law W = Qh − Qc , and according to the Second Law, the efficiency
² ≡ W/Qh is limited according to the range of absolute temperatures in use:
First Law of Thermodynamics: ² ≡
W
Qh − Q c
=
Qh
Qh
TH − T c
Th
The Second Law of thermodynamics states that all heat engines must expel a
minimum amount of heat Qc into a cold sink at temperature Tc . Since Tc > 0,
typically by a significant amount, then all real engines are (much) less than 100%
efficient. A good example of a real engine in practical use is that of a car engine,
where the “hot source” is the burning gasoline, and the “cold sink” is just the
exhaust into the atmosphere.
Second Law of Thermodynamics: ² ≤
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Chapter 20: The Second Law of Thermodynamics
Things that you will never see happen
If you want to make yourself some tea, the first thing you have to do is boil some
water. So you put some water in tea kettle on the burner of a stove. Lastly,
you turn on the burner of the stove. You do this because of the Second Law of
Thermodynamics. If there were no Second Law, and one had only the First Law
(conservation of energy), there would be nothing to prevent the burner of the
stove from getting colder and the kettle getting hotter. According to the First
Law, then, heat could spontaneously be transferred from cold material to hot
material, without violating conservation of energy. In reality, we instinctively
know that this is not true. Hot substances will lose their heat to cold substances,
never the reverse.
Principles of a Heat Engine
The concept of the Second Law of Thermodynamics can be made more
practical by considering the general form of the heat engine, which is any device
which converts heat into work. Heat energy Qh is extracted from a hot source
at Th , part of that energy is converted into work W by the heat engine, and the
remainder of the heat Qc = Qh −W is expelled into the cold sink at Tc . The heat
engine operates in some continuous cycle, always returning to its initial state.
The first Law of Thermodynamics states that the work done is the difference
in the extracted and the expelled heats: W = Qh − Qc . The Second Law of
Thermodynamics states that there must always be a certain amount of expelled
(wasted) heat Qc in proportion to the temperature difference T h −Tc . The Second
Law sets an unavoidable limit on the efficiency of any heat engine operating
between any two temperatures Th and Tc . All real engines have actual efficiencies
less than this limit.
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Realistic Model of a Gasoline Engine
The Otto Cycle
The Otto Cycle is an idealized model of how a real gasoline engine works. As
with any engine model, we show the cycle in a pV pressure-Volume diagram
(figure 20.6, page 679). The Otto Cycle has four lines in such a pV diagram.
1) A gas-air mixture in a piston is compressed adiabatically from point a with
(pa , Va ) to point b with (pb , Vb ). The compression occurs so rapidly that
effectively no heat enters or leaves the piston. The compression ratio is
r = Va /Vb which is typically 8 in an ordinary gasoline engine.
2) At point b the gas-air mixture is ignited by a spark plug. The ignition of
the gas-air mixture occurs extremely rapidly, liberating a great deal of heat
energy, raising the temperature of the mixture. The gas-air mixture goes to
a new point c in the pV diagram, but the volume remains constant during
this very short amount of time. So at point c we have (pc , Vc ) with pc > pb
and Vc = Vb .
3) The hot gas-air mixture expands adiabatically to a new point d, where the
volume is the original piston volume and the pressure and temperature are
lowered. The point d has (pd , Vd ) with Vd = Va . This is the power stroke to
the drive shaft of the engine.
4) The last stroke is the exhaust stroke, where the spent gas-fuel mixture
is expelled from the piston’s volume. In this stroke the volume remains
constant while the pressure and temperature drop to the initial values at
point a in the pV diagram.
The efficiency of the Otto cycle is examined in detail on page 670. The final
result is the following simple formula
1
Efficiency of an Otto cycle gas engine: ² = 1 − γ−1 ≈ 0.56 = 56%
r
where γ = 1.40 for the gas-air mixture, and r is the compression ratio of the
piston. You can see that higher values of r will lead to better efficiencies. Unfortunately, higher values of r also lead to more air pollution and need more
expensive gasoline grades. So ordinary modern day cars, as distinct from race
cars, have a relatively low ratio value r ≈ 8. Nonetheless, you can see that a
real gasoline engine with a 20% efficieny is much lower than the ultimate limit
for a gasoline engine. The most efficient gas engines are at about 35%.
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Realistic Model of a Diesel Engine
The Diesel Cycle
The diesel cycle (Fig. 20.7, page 680) is quite similar to the gasoline engine
cycle, with one important difference. At the end of the adiabatic compression
stroke of a diesel engine, there is no ignition by a spark plug. There is also no
fuel in the piston during the compression stroke. Instead, after the compression
stroke is introduced, and the temperature has become very high (see page 4 of
these notes, or page 697 in the text), then diesel fuel is carefully injected into
the piston over a very brief period of time. The pressure stays approximately
constant while the fuel is ignited by the high temperature of the air. After the
fuel is ignited, there is a power stroke as in the gasoline engine.
One can work out (but the textbook does not), that the efficiency of a diesel
engine higher than that of a gasoline engine. With a compression ratio r =
15 − 20, the theoretical efficiency of a diesel engine is between 65% and 75%,
compared to the theoretical efficiency of 56% of a gasoline engine. Just as with
the gasoline engines, a real diesel engine will have a significantly lower efficiency,
but one can always hope to get better efficiencies with improved engine design
and materials.
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Chapter 20: The Second Law and Refrigerators
We all know what a refrigerator (or an air conditioner) is, but we may not know
how it works. Fundamentally, a refrigerator is a heat engine running in reverse.
An amount of heat Qc is extracted from a cold source Tc (the inside of the
refrigerator), then an amount of work W is done (by the compressor), and finally
an amount of heat Qh = Qc + W is expelled into the hot sink (the surrounding
kitchen). Just as it is impossible for a kettle to start boiling by simply sitting on
a cold burner, it is also impossible to make a operate a refrigerator unless work
W is done. (Is it a good idea on a hot day to leave the refrigerator door open in
the hope of cooling down the kitchen ? Why do window air conditioner units,
or central air conditioner units, always have their fans blowing outside ?)
REVIEW: The Efficiency of a Heat Engine
The efficiency of a heat engine is given by the amount of work W generated
divided by the amount of heat extracted Qh .
Example Find the efficiency of a heat engine that produces 2000 J of energy in
the combustion phase, and loses 1500 J in the exhaust and in friction.
The work done by this engine is W = Qh − Qc = 2000 − 1500 = 500 J. So the
efficiency is
W
500
= 25%
=
²≡
Qh
2000
The efficiency ² can also be written in terms of the ratio Qc /Qh :
²≡
Qh − Q c
Qc
W
≡
≡1−
Qh
Qh
Qh
The Carnot Cycle
To get perfect efficiency (² = 1) would require that Qc = 0 meaning no heat is
expelled into the cold sink. But according to the Second Law of Thermodynamics
this is impossible if the cold sink has any temperature Tc > 0. This was the
discovery of the French physicist Carnot in the early nineteenth century. Carnot
devised a hypothetical machine which we now call the Carnot Engine which has
the highest possible theoretical efficiency.
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Chapter 20: The Carnot Cycle
Historical Background
During the early 1800s, just after the invention of the steam engine by James
Watt in England, there was an enormous interest in Europe in trying to improve
the efficiencies of engines. This interest led to the development of the branch of
physics called Thermodynamics, and one of the early pioneers in that field was
a young Frenchman called Sadi Carnot.
Analysis of The Efficiency of a Carnot Engine
Carnot devised the most efficient engine possible, which we now call the
Carnot Cycle. This engine, a hypothetical gas expanding and contracting, is
operated between two temperature reservoirs Th and Tc , and goes through 4
steps A → B → C → D → A in completing its cycle. The first step AB is
an isothermal expansion from volume VA to volume VB . In this step heat Qh is
extracted and work WAB is done on the gas.
Qh = WAB = nRTh ln (
VB
)
VA
In the second stage, there is an adiabatic expansion from volume VB to VC . In
the third stage, there is an isothermal contraction from VC to VD , which expels
heat Qc into the low temperature reservoir Tc . In this stage, work is done on the
gas W < 0
VC
Qc = −WCD = nRTc ln ( )
VD
The ratio Qc /Qh determines the efficiency of the Carnot cycle
Qc
Tc ln (VC /VD )
Tc
=
=⇒
Qh
Th ln (VB /VA )
Th
Ã
!
(adiabatic expansion law)
So the efficiency of the Carnot Cycle is given by
²Carnot = 1 − Qc /Qh = 1 − Tc /Th = (Th − Tc )/Th
The above equation is fundamentally important, as it sets an unbreakable efficiency limit to any real heat engine.
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Entropy and the Second Law of Thermodynamics
Qualitative Description of Entropy
The thermodynamics physical quantity called Entropy will be the last physical
quantity which we introduce into this semester’s introductory physics course.
Entropy is closely associated with the Second Law of Thermodynamics. In fact,
one qualitative statement of the Second Law of Thermodynamics is that in any
closed system, such as a cyclic heat engine, then the entropy can never decrease.
Only changes in entropy, like changes in potential energy or changes in velocity
over a period of time, are physically significant.
On one qualitative level, entropy is a measure of the disorder in a system. So
the Second Law of Thermodynamics is a statement that in a closed system, disorder can never decrease. For real systems this disorder parameter will typically
increase. So there is the explanation of why your room will get messy over time,
after originally been cleaned up and put in good order.
Quantitative Calculation of Thermodynamics Entropy (pages 691–94)
The amount of entropy in a system is symbolized by the letter S. We can
calculate the changes in S by first giving an example of an isothermal expansion
of a system when an infinitesimal amount of heat dQ is added to the system.
We know that adding such an amount of heat isothermally will cause the system
to expand and do an infinitesimal amount of work dW . By the first law
dV
dQ
nRT
dV =⇒
=
dQ = dW = pdV =
V
V
nRT
We can infer that the gas molecules have become more disordered, as they occupy
a greater volume and are more random in their positions at any given time. We
quantify this increase disorder by dS according to the following calculation
dQ
20.17
Increase in entropy for a reversible process: dS =
T
For an isothermal expansion when a total amount of heat Q is added to the gas,
it is easy to calculate the change in entropy
Z S
Z dQ
Q
2
∆S for isothermal process:
dS = S2 − S1 =
=
20.18
S1
T
T
Because T is constant during an isothermal process, then we can integrate dQ
to be Q. You can see in this derivation that entropy is in units of Joules/K, or
calories/K.
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Lecture 25: The Second Law of Thermodynamics, Preview of Final
Examples of Entropy Change
Melting 1 kg of ice
Initially entropy can seem like a mysterious, hard to understand concept, especially when non-gas systems are included. However, some simple examples
of entropy calculations should make the concept less intimidating. We can
easily calculate how much entropy is added to one kilogram of ice if the ice
melts at a constant temperature of 0o C. The latent heat of fusion for ice is
Lf = 3.34 × 105 J/kg. Since the melting occurs at a constant temperature,
we can integrate for the entropy change, just as we did for an isothermal gas
expansion
Entropy change for melting 1 kg ice:
Z S
2
S1
dS = S2 − S1 =
Z
dQ Q
=
T
T
3.34 × 105 J/kg ∗ 1 kg
= 1.22 J/K
Q = mLf =⇒ S2 − S1 =
273 K
It is important to realize that the T in the entropy change formula is in the
Kelvin scale.
Raising 1 kg of water from 0o C to 100o C
Next we consider the 1 kg of water to be raised from 0o C to 100o C, in
what is called a reversible manner. We will discuss the concept of a reversible
process next. For now we have for mass m = 1 kg and water’s specific
heat c = 4910 J/kg-K, then the heat change will be
dQ = mc dT
In turn the entropy change from T = T1 = 0o C to T = T2 = 100o C will be
Entropy change for 1 kg water:
Z S
2
S1
dS = S2 − S1 =
Z T
2
T1
dQ Z 373 mc dT
=
273
T
T
373
S2 − S1 = 1.0 ∗ 4190 ∗ ln
= 1.31 × 103 J/K
273
You should realize that in this calculation the temperature T was not constant.
However, as long as you know the relation between the heat differential dQ and
the absolute temperature T , then you can complete the integral to get an actual
numerical value for the entropy change.
Ã
!
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Reversible Changes and Entropy Definition
Reversible and Irreversible Changes
If a system proceeds from one thermodynamics state, say (p1 , V1 , T1 ) to a second thermodynamic state (p2 , V2 , T2 ), we can characterize that change is either
reversible or irreversible. You can think of a reversible change is one that happens gradually enough that the system is in a well-defined thermodynamic state
(P, V, T ) at all times during the change. At any time during the change, the process might be reversed and the system could retrace it steps back to (p 1 , V1 , T1 ).
In contrast, an irreversible change occurs suddenly or violently enough, that the
system is not in a well-defined thermodynamic state except at the beginning and
at the end. It will not be possible for the system to retrace its path in the same
exact reversed steps.
Examples for a gas doubling its volume
A good comparison of reversible and irreversible changes is to consider a gas in
a thermodynamic state (p1 , V1 , T1 ). We want to double the volume of the gas.
One way to do this is through an adiabatic expansion governed by the adiabatic
ideal gas law
p1 V1γ = p2 V2γ
If the gas volume is kept insulated from any heat exchange with the outside
world, and the gas gradually expands to double its initial volume by pushing on
a piston cap, then we have a good example of a reversible expansion. At any
time, we could reverse the process by pushing down slightly on the piston cap.
It should be clear that work is done by the gas in this reversible expansion.
A second way of doing the expansion, also adiabatically, is simply to pull the
piston cap out very rapidly such that the gas undergoes a free expansion with
no work done. One could also imagine punching a hole in the piston cap, with a
second piston cap positioned at double the gas volume and a vacuum between the
two caps. The gas expands in a chaotic manner with no equilibrium temperature
and pressure until the gas finally settles down, so to speak, in its now doubled
volume. It should be clear to you that in this kind of irreversible expansion, no
work was done by the expanding gas. The change is still adiabatic because the
gas volume was insulated at all times during the expansion.
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Entropy Changes For the Expansion of a Gas
Entropy changes for a gas reversibly doubling its volume
The entropy change equation
Z 2 dQ
∆S =
1 T
is true only for reversible processes. In the reversible, adiabatic expansion described above, can you calculate the entropy change? This is a trivial calculation
because an adiabatic, reversible change involves no heat transfer dQ at any time.
Hence, no matter what the temperature during the expansion, we have dQ = 0
at all times. Hence for an adiabatic reversible change, the entropy change is
exactly zero. Processes which have no entropy changed are called isentropic.
Entropy changes for a gas irreversibly doubling its volume
For n moles of a gas irreversibly doubling its volume adiabatically, we also have
dQ = 0 at all times. However, we cannot use the formula
Z 2 dQ
Entropy change for a reversible process: ∆S =
1 T
because this formula works only for reversible processes. However, we can figure
out the change in entropy ∆S because the change in entropy only depends
upon the initial and final states. We know in this case that the final volume is
double the initial volume, and that the final temperature is the same is the initial
temperature. That is, we know that the free expansion process is isothermal from
the first law of thermodynamics relating the heat added to the change in internal
energy plus the work done Q = ∆U + W Since Q = 0 because of adiabaticity
and W = 0 since this is a free expansion, then ∆U = 0. Moreover, since U the
internal energy depends only on temperature T , a ∆U = 0 process is always
isothermal. We know the initial and final thermodynamic states completely:
Initial: p1 , V1 , T1
;
Final: p2 , V2 = 2V1 , T2 = T1
Clearly, from the ideal gas law, p2 = p1 /2. When the volume doubles isothermally, the pressure must be halved. We can figure out the entropy change by
imagining an isothermal reversible expansion process between these two states
Ã
!
Z 2 dW
Z 2V p dV
Z 2V dV
Z 2 dQ
2V1
1
1
∆S =
=
=
= nR
= nR ln
= nR ln(2)
V1
V1
1 T
1 T
T
V
V1
The free expansion of a gas to twice its initial volume results in entropy change
∆S = nR ln(2). Does ∆S > 0 have a simple physics interpretation?
Lecture 25: The Second Law of Thermodynamics, Preview of Final
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Physics Meaning of Entropy Changes
Review of results for doubling the volume of n moles of an ideal gas
Process
Initial p, V, T
Final p, V, T
∆S
γ
γ−1
Reversible adiabatic expansion
p 1 , V 1 , T1
(0.5) p1 , 2V1 , (0.5) T1
0
p1
Irreversible free expansion
p 1 , V 1 , T1
nR ln(2)
2 , 2V1 , T1
The above table shows the initial and final thermodynamics states for 1) doubling
the volume of a gas reversibly, or 2) irreversibly. The reversible expansion is done
adiabatically, and we can uses the adiabatic ideal gas laws to compute the final
pressure and temperature in terms of the initial pressure and pressure, as written.
The γ parameter is the ratio of Cp /CV which is 1.67 or 1.40 for monatomic or
diatomic gases, respectively.
In the reversible, adiabatic expansion, there is no entropy change since dQ = 0 at
all times during the expansion. So the reversible adiabatic change has ∆S = 0.
In the irreversible free expansion, which is also adiabatic, we cannot use the
R
dQ/T formula. However, we can compute the entropy change by looking a
reversible expansion between the same initial and final thermodynamic states as
shown in the table. We can see that for the irreversible free expansion volume
doubling, there is a positive entropy change ∆S = nR ln(2)
Qualitative interpretation of a non-zero entropy change
In these two expansions, the reversible expansion produced useful mechanical
work. The amount of work, from the formula shown in the previous lecture for
a reversible adiabatic expansion is
W =
1
1
p1 V 1
(p1 V1 − p2 V2 ) =
(p1 V1 − (0.5)γ p1 2V1 ) =
(1 − 2(0.5)γ )
γ−1
γ−1
γ−1
For the irreversible free expansion no work was done. Even worse, the potential
which existed for doing the work in the reversible expansion is lost forever. An
increase in entropy for a closed thermodynamics system means that the potential
for doing work by that system has forever been lost1 .
Finally, you can easily prove to yourself that the entropy change for one cycle of
a Carnot engine is zero. That is the secret of its maximum efficiency.
1
You should prove to yourself that the factor in the equation for W above, (1 − 2(0.5) γ ), is a positive number.