Transcript Engines and the Second Law of Thermodynamics

Lecture PowerPoints
Chapter 20
Physics for Scientists and
Engineers, with Modern
Physics, 4th edition
Giancoli
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Copyright © 2009 Pearson Education, Inc.
Chapter 20
Second Law of
Thermodynamics
Copyright © 2009 Pearson Education, Inc.
Units of Chapter 20
• The Second Law of Thermodynamics—
Introduction
• Heat Engines
• Reversible and Irreversible Processes; the
Carnot Engine
• Refrigerators, Air Conditioners, and Heat
Pumps
• Entropy
• Entropy and the Second Law of
Thermodynamics
Copyright © 2009 Pearson Education, Inc.
Units of Chapter 20
• Order to Disorder
• Unavailability of Energy; Heat Death
• Statistical Interpretation of Entropy and the
Second Law
• Thermodynamic Temperature; Third Law of
Thermodynamics
• Thermal Pollution, Global Warming, and
Energy Resources
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20-1 The Second Law of
Thermodynamics—Introduction
The first law of thermodynamics tells us that
energy is conserved. However, the absence of
the process illustrated above indicates that
conservation of energy is not the whole story. If it
were, movies run backwards would look
perfectly normal to us!
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20-1 The Second Law of
Thermodynamics—Introduction
The second law of thermodynamics is a
statement about which processes occur
and which do not. There are many ways
to state the second law; here is one:
Heat can flow spontaneously from a hot
object to a cold object; it will not flow
spontaneously from a cold object to a hot
object.
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20-2 Heat Engines
It is easy to produce thermal energy using
work, but how does one produce work using
thermal energy?
This is a heat engine;
mechanical energy can
be obtained from
thermal energy only
when heat can flow from
a higher temperature to
a lower temperature.
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20-2 Heat Engines
We will discuss only engines that run in a
repeating cycle; the change in internal energy
over a cycle is zero, as the system returns to
its initial state.
The high-temperature reservoir transfers an
amount of heat QH to the engine, where part of
it is transformed into work W and the rest, QL,
is exhausted to the lower temperature
reservoir. Note that all three of these quantities
are positive.
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20-2 Heat Engines
A steam engine is one type of heat engine.
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20-2 Heat Engines
The internal combustion engine is a type of heat
engine as well.
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20-2 Heat Engines
Why does a heat engine need a temperature
difference?
Otherwise the work done on the system in
one part of the cycle would be equal to the
work done by the system in another part,
and the net work would be zero.
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20-2 Heat Engines
The efficiency of the heat engine is the ratio of
the work done to the heat input:
Using conservation of energy to eliminate W,
we find:
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20-2 Heat Engines
Example 20-1: Car efficiency.
An automobile engine has an efficiency of
20% and produces an average of 23,000 J of
mechanical work per second during
operation.
(a) How much heat input is required, and
(b) How much heat is discharged as waste
heat from this engine, per second?
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20-2 Heat Engines
No heat engine can have an efficiency of
100%. This is another way of writing the
second law of thermodynamics:
No device is possible whose sole effect is to
transform a given amount of heat completely
into work.
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20-3 Reversible and Irreversible
Processes; the Carnot Engine
The Carnot engine was created to examine
the efficiency of a heat engine. It is idealized,
as it has no friction. Each leg of its cycle is
reversible.
The Carnot cycle consists of:
• Isothermal expansion
• Adiabatic expansion
• Isothermal compression
• Adiabatic compression
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20-3 Reversible and Irreversible
Processes; the Carnot Engine
For an ideal reversible engine, the efficiency can
be written in terms of the temperature:
From this we see that 100% efficiency can be
achieved only if the cold reservoir is at absolute
zero, which is impossible.
Real engines have some frictional losses; the
best achieve 60–80% of the Carnot value of
efficiency.
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20-3 Reversible and Irreversible
Processes; the Carnot Engine
Example 20-2: A phony claim?
An engine manufacturer makes the following
claims: An engine’s heat input per second is
9.0 kJ at 435 K. The heat output per second is
4.0 kJ at 285 K. Do you believe these claims?
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20-3 Reversible and Irreversible
Processes; the Carnot Engine
Automobiles run on the Otto cycle, shown
here, which is two adiabatic paths alternating
with two constant-volume paths. The gas
enters the engine at point a and is ignited at
point b. Curve cd is the power stroke, and da is
the exhaust.
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20-3 Reversible and Irreversible
Processes; the Carnot Engine
Example 20-3: The Otto cycle.
(a) Show that for an ideal gas as working
substance, the efficiency of an Otto cycle
engine is
e = 1 – (Va/Vb)1-γ
where γ is the ratio of specific heats (γ = CP/CV)
and Va/Vb is the compression ratio.
(b) Calculate the efficiency for a compression
ratio Va/Vb = 8.0 assuming a diatomic gas like
O2 and N2.
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20-4 Refrigerators, Air Conditioners,
and Heat Pumps
These appliances are essentially heat engines
operating in reverse.
By doing work, heat is
extracted from the cold
reservoir and exhausted
to the hot reservoir.
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20-4 Refrigerators, Air Conditioners,
and Heat Pumps
This figure shows more details of a typical
refrigerator.
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20-4 Refrigerators, Air Conditioners,
and Heat Pumps
Refrigerator performance is measured by the
coefficient of performance (COP):
Substituting:
For an ideal refrigerator,
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20-4 Refrigerators, Air Conditioners,
and Heat Pumps
Example 20-4: Making ice.
A freezer has a COP of 3.8 and uses
200 W of power. How long would it
take this otherwise empty freezer to
freeze an ice-cube tray that contains
600 g of water at 0°C?
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20-4 Refrigerators, Air Conditioners,
and Heat Pumps
A heat pump can heat a house in the winter:
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20-4 Refrigerators, Air Conditioners,
and Heat Pumps
A heat pump is evaluated on a
different type of “Coefficient of
Performance” COP:
COP (heat pump) = QH/W
vs.
COP (refrigerator) = Qc/W
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Example 20-5: Heat pump.
A heat pump has a coefficient of performance
of 3.0* and is rated to do work at 1500 W.
(a)How much heat can it add to a room per
second?
(b)
COP = 3.0 = Qh/W =>
Qh = 3.0 x W = 4500 W = 4500 Joule/Sec
so…
4500 J per second, or ~4.3 BTU
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Example 20-5: Heat pump.
A heat pump has a coefficient of performance
of 3.0* and is rated to do work at 1500 W.
(b) If the heat pump were turned around to act
as an air conditioner in the summer, what
would you expect its coefficient of
performance to be, assuming all else stays
the same?
Qc = Qh – W = 4500 J – 1500 J = 3000 J
And COP (refrigerator) = Qc/W = 3000/1500 = 2
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20-5 Entropy
You can’t measure the “heat” of something!
You CAN measure how much heat must be
added to change something’s temperature.
Consider isothermal process (T constant):
How much heat
must you add as
the gas expands to
keep it at T?
P
V
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20-5 Entropy
How much heat must you add as the gas
expands to keep it at T?
DU = Qin – Wout = 0 (isothermal)
Q in = W out = nRT*ln(Vf/Vi)
P
V
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Define Entropy Change
DS = Q/T
For this reversible
isothermal process,
DS = nR*ln(Vf/Vi)
20-5 Entropy
Entropy = “S”
A *state* variable for a gas (P, V, T)
A measure of “disorder” of a system
An indicator of the likelihood (probability)
of reaction directions (towards higher S)
Measured only when it *changes*
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20-5 Entropy
Definition of the change in entropy S when
an amount of heat Q is added:
if the process is reversible and the
temperature is constant. Otherwise,
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20-5 Entropy
IF the temperature is NOT constant:
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20-5 Entropy
Consider Carnot cycles:
TH/TL = QH/QL
So…
QH/TH = QL/TL
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20-5 Entropy
Since for any Carnot cycle QH/TH + QL/TL = 0,
we see that the integral of dQ/T around a
closed path is zero. This means that entropy
is a state variable—the change in its value
depends only on the initial and final states.
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20-5 Entropy
Any reversible cycle can be written as a
succession of Carnot cycles; therefore, what
is true for a Carnot cycle is true of all
reversible cycles.
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20-5 Entropy
Since for any Carnot cycle QH/TH + QL/TL = 0,
if we approximate any reversible cycle as an
infinite sum of Carnot cycles, we see that the
integral of dQ/T around a closed path is zero.
This means that entropy is a state variable—
the change in its value depends only on the
initial and final states.
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20-6 Entropy and the Second Law of
Thermodynamics
Example 20-6: Entropy change
when mixing water.
A sample of 50.0 kg of water at
20.00°C is mixed with 50.0 kg of
water at 24.00°C. Estimate the
change in entropy.
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20-6 Entropy and the Second Law of
Thermodynamics
The total entropy always increases when
heat flows from a warmer object to a colder
one in an isolated two-body system. The
heat transferred is the same, and the cooler
object is at a lower average temperature
than the warmer one, so the entropy gained
by the cooler one is always more than the
entropy lost by the warmer one.
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20-6 Entropy and the Second Law of
Thermodynamics
Example 20-7: Entropy changes in a free
expansion.
Consider the adiabatic free expansion of n
moles of an ideal gas from volume V1 to
volume V2, where V2 > V1. Calculate the
change in entropy
(a) of the gas and
(b) of the surrounding environment.
(c) Evaluate ΔS for 1.00 mole, with V2 = 2.00 V1.
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20-6 Entropy and the Second Law of
Thermodynamics
Example 20-8: Heat transfer.
A red-hot 2.00-kg piece of iron at
temperature T1 = 880 K is thrown into a
huge lake whose temperature is T2 = 280 K.
Assume the lake is so large that its
temperature rise is insignificant. Determine
the change in entropy
(a) of the iron and
(b) of the surrounding environment (the
lake).
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20-6 Entropy and the Second Law of
Thermodynamics
The fact that after every interaction the
entropy of the system plus the environment
increases is another way of putting the
second law of thermodynamics:
The entropy of an isolated system never
decreases. It either stays constant
(reversible processes) or increases
(irreversible processes).
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20-7 Order to Disorder
Entropy is a measure of the disorder of a
system. This gives us yet another statement of
the second law:
Natural processes tend to move toward a state of
greater disorder.
Example: If you put milk and sugar in your
coffee and stir it, you wind up with coffee that
is uniformly milky and sweet. No amount of
stirring will get the milk and sugar to come
back out of solution.
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20-7 Order to Disorder
Another example: When a tornado hits a
building, there is major damage. You never see
a tornado approach a pile of rubble and leave a
building behind when it passes.
Thermal equilibrium is a similar process—the
uniform final state has more disorder than the
separate temperatures in the initial state.
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20-8 Unavailability of Energy; Heat
Death
Another consequence of the second law:
In any natural process, some energy becomes
unavailable to do useful work.
If we look at the universe as a whole, it seems
inevitable that, as more and more energy is
converted to unavailable forms, the ability to do
work anywhere will gradually vanish. This is
called the heat death of the universe.
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Entropy vs. Enthalpy
Enthalpy = Internal Energy PLUS Energy in the
surrounding system:
The enthalpy of a system is defined as:
H = U + pV
And the change in enthalpy =
DH = DU + D (pV)
For constant pressure processes:
DH = DU + pDV = DU +Work = Q!
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Entropy vs. Enthalpy
Enthalpy is sometimes described as the heat content
of a system under a given pressure, Such a
visualization assumes no energy exchange with the
environment other than heat or expansion work.
Given such restrictions, it can be shown that:
The enthalpy is the total amount of energy which the
system can emit through heat,
Adding or removing energy through heat is the only
way to change the enthalpy, and
The amount of change in enthalpy is equal to the
amount of energy added through heat.
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Entropy vs. Enthalpy
ENTHALPHY: Is the energy content of a
process (chemical, thermodynamic,
mechanical, etc) that can be recovered. It is
also described as useful energy.
ENTROPY: Is the energy content of a process
(chemical, thermodynamic, mechanical, etc)
that CANNOT be recovered. It is also
described as chaos.
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20-9 Statistical Interpretation of
Entropy and the Second Law
Microstate: a particular configuration of atoms
Macrostate: a particular set of macroscopic
variables
This example uses coin tosses:
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20-9 Statistical Interpretation of
Entropy and the Second Law
Similarly, the most probable distribution
of velocities in a gas is Maxwellian:
The most probable state
is the one with the
greatest disorder, or the
greatest entropy. With k
being Boltzmann’s
constant and W the
number of microstates,
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20-9 Statistical Interpretation of
Entropy and the Second Law
Example 20-9: Free expansion—statistical
determination of entropy.
Determine the change in entropy for the
adiabatic free expansion of one mole of a
gas as its volume doubles. Assume W,
the number of microstates for each
macrostate, is the number of possible
positions.
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20-9 Statistical Interpretation of
Entropy and the Second Law
In this form, the second law of
thermodynamics does not forbid
processes in which the total entropy
decreases; it just makes them
exceedingly unlikely.
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20-10 Thermodynamic Temperature;
Third Law of Thermodynamics
Since the ratio of heats exchanged between
the hot and cold reservoirs in a Carnot
engine is equal to the ratio of temperatures,
we can define a temperature scale using the
triple point of water:
T = (273.16K)(Q/Qtp).
Here, Q and Qtp are the heats exchanged by
a Carnot engine with reservoirs at
temperatures T and Ttp.
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20-10 Thermodynamic Temperature;
Third Law of Thermodynamics
Also, since the maximum efficiency of a
heat engine is
there is no way to achieve a temperature of
absolute zero. This is the third law of
thermodynamics:
It is not possible to reach absolute zero in
any finite number of processes.
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20-11 Thermal Pollution, Global
Warming, and Energy Resources
Over 90% of the energy used in the U.S. is
generated using heat engines to drive
turbines and generators—even nuclear
power plants use the energy generated from
fission heat water for a steam engine. The
thermal output QL of all these heat engines
contributes to warming of the atmosphere
and water. This is an inevitable
consequence of the second law of
thermodynamics.
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Summary of Chapter 20
• Heat engine changes heat into useful work
• Efficiency: work/heat input
• Maximum efficiency: 1 – TL/TH
• Refrigerators and air conditioners are heat
engines, reversed; COP = heat removed/work
• Heat pump: COP = heat delivered/work
• Second law of thermodynamics: Natural
processes always tend to increase entropy
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Summary of Chapter 20
• Entropy change in reversible process:
• Change in entropy gives direction to “arrow
of time”
• As time goes on, energy becomes degraded.
• Heat engines cause thermal pollution.
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