First Law of Thermodynamics - Derry Area School District

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Transcript First Law of Thermodynamics - Derry Area School District

AP Physics
Chapter 12
Thermodynamics
Chapter 12: Thermodynamics
12.1 Thermodynamic Systems, States, and
Processes
12.2 The First Law of Thermodynamics
12.3 The Second Law of Thermodynamics and
Entropy
12.4 Heat Engines and Thermal Pumps
12.5 The Carnot Cycle and Ideal Heat Engines
Learning Objectives
A. Ideal Gases
1.
Students will know how to apply the ideal gas law and thermodynamic
principles, so they can:
a. Relate the pressure and volume of a gas during isothermal
expansion or compression.
b. Relate the pressure and temperature of a gas during constantvolume heating or cooling, or the volume and temperature during
constant-pressure heating or cooling.
c. Calculate the work performed on or by a gas during an expansion or
compression at constant pressure.
d. Understand the process of adiabatic expansion or compression of a
gas.
e. Identify or sketch on a PV diagram the curves that represent each of
the above processes.
Learning Objectives
B. Laws of Thermodynamics
1.
Students will know how to apply the first law of thermodynamics, so they
can:
a. Relate heat absorbed by a gas, the work performed by the gas, and
the internal energy change of the gas for any of the processes above.
b. Relate work performed by a gas in a cyclic process to the area
enclosed by a curve on a PV diagram.
2. Students will understand the second law of thermodynamics, the concept
of entropy, and heat engines and the Carnot cycle, so they can:
a. Determine when entropy will increase, decrease, or remain the same
during a particular situation.
b. Compute the maximum possible efficiency of a heat engine operating
between two given temperatures.
c. Compute the actual efficiency of a heat engine.
d. Relate the heats exchanged at each thermal reservoir in a Carnot
cycle to the temperatures of the reservoirs.
Homework for Chapter 12
• Read Chapter 12
• HW 12.A: p. 412: 5, 6, 12-18, 22.
• HW 12.B: p. 414-415: 29, 30, 47-49, 50-52, 54, 56, 57.
• HW 12.C: Handout and p.416: 72-74; 76.
12.1: Thermodynamic Systems,
States, and Processes
Warmup: Are You a Conservationist?
Physics Warmup #94
Conservation laws are the cornerstones of physics. When we say a quantity is
conserved, we mean that the sum total of that quantity in a system never changes.
Quantities that are known to be conserved include momentum, energy, angular
momentum, and electric charge. Each has its own conservation law.
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Solve this anagram to identify one of the conservation laws in physics.
an uncommon, softer motive
Answer: conservation of momentum
12.1 Thermodynamic Systems, States, and Processes
thermodynamics
deals with the transfer or the actions
(dynamics) of heat.
- the study grew out of efforts to develop a engine that could
convert heat to mechanical work (such as the steam engine.)
system
- a definate quantity of matter enclosed by boundaries
or surfaces.
ex: a real boundary, such a the gas contained in a pistoncylinder of an engine.
ex: an imaginary boundary, such as a cubic meter of air in
this room.
thermally isolated system
out of the system.
- no heat transfer into or
heat reservoir
- a system with unlimited heat capacity so that
the addition or removal of heat causes no change in its temperature.
12.1 Thermodynamic Systems, States, and Processes
equations of state
- describe the conditions of
thermodynamic systems.
- ex: ideal gas law: pV = NkBT is an equation of state.
- state variables are p, V, T for a given mass
process
variables of a system.
- a change in the state, or the thermodynamic
reversible process
- one with a known path through
equilibrium states.
- usually slow; path can be retraced.
irreversible process
- one for which the intermediate steps
are nonequilibrium states.
- usually fast
- Irreversible does not mean that the system can’t be taken back
to the initial state; it only means the process path can’t be retraced
because of nonequilibrium conditions.
Reversible vs. Irreversible Processes
• If a process goes quickly from state 1 to state 2, the process is irreversible since
we do not know the “path”.
• However, if the gas is taken through many closely spaced states (as in going
from state 3 to state 4), the process is reversible, at least in principle.
• Reversible means “ exactly retraceable”.
12.2: The First Law of
Thermodynamics
Warmup: The Thrill of Conservation
Physics Warmup #96
Roller Coasters are an excellent example of the conservation of energy. Work is
done to raise the cars to the top of the first hill and then gravity transforms the
energy back and forth between potential and kinetic for the rest of the ride.
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Cedar Point Amusement Park in Sandusky, OH is the self-proclaimed roller
coaster capital of the world. One of its many coasters is the Magnum XL-200. It
stands out above the horizon with a incredible drop of 59.3 meters on the first hill!
Disregarding any loss of energy to friction, calculate the speed of the cars at the
bottom of that first hill.
Answer: 34.1 m/s
12.2 The First Law of Thermodynamics
Disclaimer:
The conventions we are using for WORK
are NOT the same as our textbook uses.
They are the conventions aligned with the
AP exam.
12.2 The First Law of Thermodynamics
The First Law of Thermodynamics is a statement of energy
conservation for thermodynamic systems.
∆U = Q + W
On Gold Sheet
Q : heat
∆U : change in internal energy
W: work
Sign Convention (Very Important!)
The system is the gas, fluid, etc. you are analyzing.
+Q means heat added to the system
-Q means heat removed from the system
+W means work done on the system (compression)
-W means work done by the system (expansion)
12.2 The First Law of Thermodynamics
Example: A gas is kept in a cylinder that can be compressed by pushing down
on a piston. You add 2500 J of heat to the system, and then you push the piston
1.0 m down with a constant force of 1800 N. What is the change in the gas’s
internal energy?
Solution:
Q = + 2,500 J
Work is done ON the gas to compress it, so W is positive.
Since W = Force x distance
= 1800 N x 1.0 m
W = + 1800 J
∆U=Q+W
= 2,500 J + 1800 J
∆ U = 4,300 J
12.2 The First Law of Thermodynamics
Four Reversible Processes You Need to Know:
1) isobaric
– constant pressure
• At contant pressure, the work done by/on the gas is given by
W = -pV = -p (V2 – V1)
•
•
On Gold Sheet
The work done is positive when V is negative, during compression.
The work done is negative when V is positive, during expansion.
In an isobaric process,
the temperature of the
gas increases as the
volume increases, but
the pressure doesn’t
change.
Since V/T is constant, as
volume increases, so
must temperature.
Heat is added to the
gas. The gas expands.
Work is negative.
W = - p∆V
= - p (V2 – V1)
12.2 The First Law of Thermodynamics
Four Reversible Processes You Need to Know:
2) isochoric
– constant volume (aka isometric, isovolumetric)
• V = 0, hence W = 0.
• Hence Q = U (all the supplied heat goes into increasing the internal
energy of the system.)
In an isometric process,
the volume remains
constant.
Since P/T must remain
constant, as pressure
increases, so must
temperature.
The change in
volume is zero, so
the work is zero.
Q=∆U
All the heat added
to the gas goes
towards increasing
the internal energy
of the gas.
12.2 The First Law of Thermodynamics
Four Reversible Processes You Need to Know:
3) isothermal
– constant temperature
• U = 0, hence Q = -W.
4) adiabatic
– thermally insulated; no heat is exchanged
• Q = 0, hence U = W.
In an isothermal
process, the
temperature stays
constant, so
∆U = 0 and
Q = -W
Since PV is
constant (Gas
Laws), as one
increases the other
must decrease.
Heat is added to
the piston. The
volume of the gas
increase.
Work was done by
the gas, so Work is
negative.
In an adiabatic
process, no heat is
added or removed
from the system.
Q=0
and
∆U = W
The change in
internal energy of
the gas is due to
the work done on
the gas.
Here, work is done
by the gas during
expansion. Work is
negative. Therefore,
the internal energy
and temperature
decreases.
W = - pV = - area
Work done by a gas =
area under the curve on
a PV diagram.
-
+
12.2 The First Law of Thermodynamics
Example 12.1: A system consists of 3.0 kg of water at 80°C. 25 J of work is done
on the system by stirring with a paddle wheel, while 63 J of heat is removed.
What is the change in internal energy of the system?
12.2 The First Law of Thermodynamics
Example 12.2: During an adiabatic process, an ideal gas gained 5.0 J of
internal energy.
a) What is the heat involved in the process?
b) What is the work done in the process?
12.2 The First Law of Thermodynamics
Example 12.3: In an isometric process, the internal energy of a system
decreases by 50 J.
a) What is the work done?
b) What is the heat exchange?
12.2 The First Law of Thermodynamics
Example 12.3.1) 0.25 mole of gas is kept at 1150 K. The gas undergoes
adiabatic expansion, reaching a final temperature of 400 K. How much work
was done on or by the gas?
P
a
1150 K isotherm
b
400 K isotherm
V
12.2 The First Law of Thermodynamics: Check for Understanding
1. There may be an exchange of heat with the surroundings for
a) a thermally isolated system
b) a completely isolated system
c) a heat reservoir
d) none of these
Answer: c
12.2 The First Law of Thermodynamics: Check for Understanding
2. On a PV diagram, a reversible process is one
a) whose path is known
b) whose path is unknown
c) for which the intermediate steps are nonequilibrium states
d) none of these
Answer: a
12.2 The First Law of Thermodynamics: Check for Understanding
3. There is no exchange of heat in an
a) isothermal process
b) adiabatic process
c) isobaric process
d) isometric process
Answer: b
12.2 The First Law of Thermodynamics: Check for Understanding
4. Positive work is work done
a) on the system
b) by the system
c) for the system
d) of the system
Answer: a (compression)
12.2 The First Law of Thermodynamics: Check for Understanding
5. If heat is added to a system of ideal gas during an isothermal
expansion process,
a) work is done by the system
b) the internal energy decreases
c) the effect is the same as for an isometric process
d) none of these
Answer: a
Homework for Chapter 12.1-2
• HW 12.A: p. 412: 5, 6, 12-18, 22.
12.3: The Second Law of
Thermodynamics and Entropy
Warmup: Perpetual Motion I
Physics Warmup #98
Perpetual motion machines are machines that, once working, would keep on going
by themselves. There are actually two types of these theoretical perpetual motion
machines. A perpetual motion machine of the first kind is one in which the machine
produces more energy than it uses while running.
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Use the laws of physics to explain why a perpetual motion machine of the first kind
is theoretically impossible to build.
Answer: The law of conservation of energy and the first law of thermodynamics
do not allow for the creation of energy.
12.3 The Second Law of Thermodynamics and Entropy
• The second law of thermodynamics specifies the direction in which a process
can naturally or spontaneously take place.
•This law can be stated in the following three different forms:
1. Clausius statement: (No perfect refrigerator)
Heat, by itself, cannot flow from a body at lower temperature to a body at
higher temperature.
2. Kelvin-Planck Statement: (No perfect heat engine)
No heat engine working in a cyclic process will absorb heat from a single
reservoir and convert it completely to work.
3. Entropy:
In any thermodynamic process that proceeds from one equilibrium state to
another, the entropy of a (system + environment) either remains
unchanged or increases.
12.3 The Second Law of Thermodynamics and Entropy
entropy
- a measure of the disorder of a thermodynamic system
and is related to heat and temperature.
• Entropy is a thermodynamic quantity. It is a state variable i.e. its value
depends on the state and not on the process by which the system is brought
to that state.
• Other state variables for a thermodynamic system include temperature,
pressure, volume, and internal energy. The amount of heat Q and the work
done are not state variables. Their values depend on the process connecting
two states.
• Entropy is related to randomness, chaos, or disorder of a thermodynamic
system. An increase in the randomness, chaos, or disorder of a
thermodynamic system implies an increase in the entropy of the system.
• All isolated systems tend toward a state of greater disorder.
Example: A glass falls off a shelf and breaks. To put that glass back together,
and decrease the glass’s entropy, someone has to do the work.
Understanding Entropy: Macrostates and Microstates
• Take out 5 identical coins. There are 6 possible outcomes of heads and tails
when the coins are thrown, as long as you don’t care which particular coin is a
head or tail. These outcomes are called macrostates, and in this case they are
5H
4H1T
3H2T
2H3T
1H4T
5T
• If you look at the 5H macrostate, you can see there’s only 1 way this macrostate
can be achieved: All the coins must be heads.
• If you look at the 4H1T macrostate, there are 5 ways to achieve it depending on
which of the 5 coins is a head. You can say this macrostate has 5 microstates,
and it has a higher entropy then the 5H macrostate.
• The 3H2T and 2H3T macrostates each have 10 ways that they can be achieved;
these macrostates each have 10 microstates, so they have a higher entropy than
the 5H state or the 4H1T state.
• When you throw the coins, each of the total of 32 = 25 microstates is equally
likely, and the macrostate with the greatest number of microstates, the
macrostate with the highest entropy, will be the most probable.
Understanding Entropy: Macrostates and Microstates
• The last sentence of the previous slide is really the essence of the second law.
When you look at large systems like the ideal gas with 1023 particles, the most
likely macrostate – described by p, V, and T and obeying the ideal gas law – has
so many microstates associated with it that it’s the only one you have any chance
of observing.
• When you allow two systems at different temperatures to exchange energy with
each other, the final macrostate of the system as a whole will be the one with the
most number of microstates, maximizing entropy.
• Now you can say why heat spontaneously flows from hot to cold: It’s
overwhelmingly more likely than the alternative. In terms of entropy, the second
law is:
A closed system, or a system that won’t transfer energy to or from its
surroundings, will always develop to maximize the entropy of the system.
The entropy will never decrease.
12.3 The Second Law of Thermodynamics and Entropy
• The entropy of an isolated system increases for every natural process. However,
if a system is not isolated it may undergo a decrease in entropy.
example for an isolated system: freezing ice. Water at room temperature will not
spontaneously make ice. To freeze ice, we need to remove heat. This creates a
decrease in entropy, or more order.
example for a non-isolated system: Put an ice cube tray full of water in the freezer.
The water will freeze as the ice cubes are made, and entropy is decreased.
However, there will be a larger increase in entropy somewhere else in the
environment (such as heat expelled by the refrigerator.)
Question: As a computer performs its calculations, it is organizing information.
Therefore, the computer’s entropy is decreasing. Is this a violation of the second
law of thermodynamics?
Answer: No. Order is increased by doing work on the closed system (the
computer). In doing this, however, the computer releases heat, increasing the
entropy of the rest of the world.
12.3 The Second Law of Thermodynamics and Entropy
• For a system, the total change in entropy is the addition of the changes in entropy
of the objects in the system.
• For any thermodynamic process,  entropy (system + environment)  0
• For irreversible processes,  entropy (system + environment) > 0
• For reversible processes,  entropy (system + environment) = 0
• For a reversible cycle,  entropy (cycle) = 0
• For free expansion,  entropy (system) > 0
• Adding heat (Q) to a system at a constant temperature increases its entropy.
12.3 Summary
• Entropy is one of the arrows of time. The entropy of the universe is always
increasing. This may lead to what is called the “heat death” of the universe.
• Energy of the universe is constant; entropy is not!
• Energy obeys the conservation law; entropy does not!
First Law of Thermodynamics: YOU CAN’T WIN.
- Energy is conserved. You can’t get energy out of nothing.
Second Law of Thermodynamics: YOU CAN’T WIN; YOU CAN’T BREAK
EVEN.
- All the available energy cannot be converted into work.
Third Law of Thermodynamics:
- It is not possible to lower the temperature of any system to
absolute zero. Doing so would violate the second law.
12.3 The Second Law of Thermodynamics and Entropy: Check for
Understanding
1. The second law of thermodynamics
a) describes the state of a system
b) applies only when the first law is satisfied
c) precludes perpetual motion machines
d) does not apply to an isolated system.
Answer: c
12.3 The Second Law of Thermodynamics and Entropy
2. In any natural process, the change in entropy of the universe is
a) negative
b) positive
c) zero
d) the same as the change in thermal energy
Answer: b
12.3 The Second Law of Thermodynamics and Entropy
3. Do the entropies of the systems in the following processes increase or
decrease?
a) ice melts
increases
b) water vapor condenses decreases, since heat is removed
c) water is heated on the stove increases, since heat is added
d) food is cooled in a refrigerator decreases, since heat is removed
12.4: Heat Engines and Thermal
Pumps
Warmup: Perpetual Motion II
Physics Warmup #99
Although the law of conservation of energy and the first law of thermodynamics do
not allow for the possibility of a perpetual motion machine of the first kind, they do
not forbid the possibility of a perpetual motion machine of the second kind. This
machine would simply create an amount of energy equal to that required to run it,
thus opening the possibility of its running itself forever, once started.
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Although the laws of physics referred to above do not prohibit such a machine,
another law of physics does (the second law of thermodynamics). Use what you
know about friction to explain why a perpetual motion machine of the second kind
could not work.
Answer: Friction between the parts would convert some of the energy to heat,
which could not be totally retrieved.
Perpetual motion machines don’t exist, but perpetual motion does.
Example: planets orbit the sun.
12.4 Heat Engines and Thermal Pumps
Review of the Second Law of Thermodynamics
• Clausius Statement: Heat does not flow spontaneously from a colder
body to a warmer body. This does not mean heat can not flow from a
colder body to a warmer body, it just can not take place spontaneously.
• Kevin-Planck Statement: In a thermal cycle, heat energy cannot be
completely transformed to mechanical work. This does not mean energy
is not conserved – it just means that 100% of heat energy cannot be
converted to mechanical work in a cycle.
• It is impossible to construct an operational perpetual motion machine.
• The total entropy of the universe increases in every natural process.
12.4 Heat Engines and Thermal Pumps
heat engine
- any cyclic device
that converts heat to
work.
It absorbs heat (Qhot
or Qin) from a high
temperature reservoir
, does net work
(Wnet), and exhausts
heat (Qcold or Qout) to
a low-temperature
reservoir.
Note the width of the
arrow for Qin is equal
to the combined
widths of Qout and
Wnet, reflecting the
conservation of
energy.
12.4 Heat Engines and Thermal Pumps
This cyclic process
consists of 2 isobars and 2
isomets. The net work
output per cycle is the area
of the rectangle formed by
the process paths.
For a general heat engine
cycle, the net work per
cycle is still represented by
the area inside the cycle’s
separate paths.
After one cycle, every
thermodynamic variable
returns to its original value,
including thermal energy U,
so ∆U = 0.
12.4 Heat Engines and Thermal Pumps
Example: The Otto Cycle (p. 404 in text)
- the theoretical process cycle developed by Geman engineer
Nickolas Otto (1832-1891). He used his design to build one of the first
successful gasoline engines.
12.4 Heat Engines and Thermal Pumps
thermal efficiency (εth)
ratio of its work output and its heat input.
εth = Wnet
Qin
Since Wnet
engine is:
= Qhot – Qcold,
- of a heat engine is defined as the
On Gold Sheet
the thermal efficiency per cycle of a heat
εth = Wnet = Qhot – Qcold = 1 - Qcold
Qhot
Qhot
Qhot
“Efficiency is what you get out for what you put in.”
12.4 Heat Engines and Thermal Pumps
Example 12.6: A heat engine has an efficiency of 25% and extracts 120 J of heat
from a hot reservoir per cycle.
a) How much net work does it perform in each cycle?
b) How much heat does it exhaust in each cycle?
12.4 Heat Engines and Thermal Pumps
thermal pump
- a device that transfers heat energy from a low
temperature reservoir to a high temperature reservoir.
To do this, work must
be done, according to
the second law of
thermodynamics, since
it will not happen on its
own.
Notice the width of the
arrows, and how they
represent the
conservation of
energy.
12.4 Heat Engines and Thermal Pumps
An air conditioner is an
example of a thermal
pump.
With work input, it
transfers heat (Qcold)
from a low–temperature
reservoir (inside the
house) to a hightemperature reservoir
(outside).
12.4 Heat Engines and Thermal Pumps
Another example of a thermal pump is a
refrigerator. (p.406 in text(
12.4-5 Heat Engines and Thermal Pumps
• In a refrigerator, heat (Qcold) is carried away from the interior by the
refrigerant (used to be Freon, but now we use other substances). The
refrigerant has a relatively low boiling point. The heat causes the refrigerant
to boil and is carried away as latent heat of vaporization.
• The vapor is drawn into the compressor. The compression increases the
temperature of the gas, which is discharged from the compressor as
superheated vapor.
• This vapor condenses in the cooler condenser unit, where circulating air
carries away the latent heat of fusion and the heat of compression and
discharges it to the surroundings (Qhot).
heat pump
- a common commercial device used to cool homes in
the summer and heat them in the winter.
12.4 Heat Engines and Thermal Pumps
Example 12.7: A small container of gas undergoes a thermodynamic cycle. The
gas begins at room temperature. First, the gas expands isobarically until its
volume has doubled. Second, the gas expands adiabatically. Third, the gas is
cooled isobarically; finally, the gas is compressed adiabatically until it returns to
its original state.
a) The initial state of the gas is indicated on the PV diagram below. Sketch this
process on the graph.
P
initial state
1
2
4
3
V
b) Is the temperature of the gas greater right before of right after the adiabatic
expansion? Justify your answer.
The temperature is greater right before the expansion. By definition, in an
adiabatic process, no heat is added or removed. But because the gas expanded,
work was done by the gas, meaning the W term in the first law of thermodynamics
is negative. So, by U = Q + W, Q = 0 and W is negative; so U is negative as
well. Internal energy is directly related to temperature. Therefore, because internal
energy decreases, so does temperature.
c) Is heat added to or removed from the gas in one cycle?
In a full cycle, the gas begins and ends at the same state. So the total change in
internal energy is zero. Now consider the total work done on or by the gas. When
the gas expands in the first and second process, there’s more area under the
graph than when the gas compresses in the second and third process. So, the
gas does more work expanding than compressing; the net work is thus done by
the gas, and is negative. To get no change in internal energy, the Q term in the
first law of thermodynamics must be positive; heat must be added to the gas.
d) Does this gas act as a refrigerator or a heat engine?
This is a heat engine. Heat is added to the gas, and net work is done by the gas;
that’s what a heat engine does. (In a refrigerator, net work is done on the gas, and
heat is removed from the gas.)
12.4-5 Heat Engines and Thermal Pumps: Summary
• For a cyclical process, at the end of one cycle the system is back to the same
state and hence temperature. Therefore, for one cycle
U = 0 and Q = -W
• If W is negative and Q is positive, work is being done by the system and heat is added to
the system. The cycle represents a heat engine (clockwise cycle).
• If W is positive and Q is negative, work is being done on the system and heat is removed
from the system. The cycle represents a heat pump/refrigerator (counterclockwise cycle).
• A heat engine absorbs QH amount of heat form a hot reservoir at temperature TH,
converts an amount W into work and rejects the rest QC to the cold reservoir at
temperature TC.
Hence, QH = W + QC
(absolute values of quantities)
• The efficiency (e) for a heat engine is given by
e = W or
QH
e = QH – Q C
QH
or
e = 1 - QC
QH
12.4 Heat Engines and Thermal Pumps Check for Understanding
1. For a cyclic heat engine,
a) εth > 1
b) Qhot = Wout
c) ∆U = Wout
d) Qhot > Qcold
Answer: d
12.4 Heat Engines and Thermal Pumps Check for Understanding
2. A thermal pump
a) is rated by thermal efficiency
b) requires work input
c) is not consistent with the 2nd law of thermodynamics
d) violates the 1st law of thermodynamics
Answer: b
12.4 Heat Engines and Thermal Pumps Check for Understanding
3. Is leaving a refrigerator door open a practical way to air condition a room?
Explain.
Answer: No. As a matter of fact, the room will be heated. The heat expelled
to the room by the refrigerator is more than the heat removed by the
refrigerator from the room.
Homework for Chapter 12.3 & 12.4
• HW 12.B: p. 414-415: 29, 30, 47-49, 50-52, 54, 56, 57.
12.5: The Carnot Cycle and Ideal
Heat Engines
Warmup: A Clean Start
Physics Warmup #179
In 2003, a company in Germany created an interactive washing machine. It
understands spoken-word commands, such as, “Prewash, then hot wash at 95
degrees, then spin at 1,400 revolutions.” It can also give verbal instructions on
how to use the machine! Of course, this all comes with a price.
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The inventors of the interactive washing machine have been quoted as saying that
it is needed because electronic appliances have become too complicated for
consumers to use. However, some critics have suggested that this may be a
misplaced attempt to aid the technology challenged consumer. Choose a different
appliance in which a similar voice-recognition system would provide a greater
benefit to its user and explain how it would help.
12.5 The Carnot Cycle and Ideal Heat Engines
Fact: Due to the second law of thermodynamics, any cyclic heat engine must
exhaust some heat energy. Therefore, no real engine can be 100% efficient.
Question: What is the maximum possible efficiency of a heat engine?
Sadi Carnot (1796-1832), a French engineer, studied this very question.
Carnot cycle
the ideal for a heat engine,
consists of two isothermal and
two adiabatic processes.
Heat is absorbed during the
isothermal expansion and
exhausted during the isothermal
compression.
12.5 The Carnot Cycle and Ideal Heat Engines
Carnot efficiency (εc)
- gives the upper limit of efficiency
(unattainable in the real world).
εc = 1 – Tcold =
Thot
Thot – Tcold
Thot
On Gold Sheet
where Thot and Tcold are the Kelvin temperatures of the low-temperature and
high-temperature reservoirs, respectively.
•The greater the difference in temperatures of the heat reservoirs, the greater
the Carnot efficiency.
example: If Thot is twice Tcold, then the Carnot efficiency will be:
εc = 1 – Tcold =
1 – 0.50 = 0.50 = 50%
Thot
however, if Thot is four times Tcold, then the Carnot efficiency will be:
εc = 1 – Tcold =
Thot
1 – 0.25 = 0.75 = 75%
12.5 The Carnot Cycle and Ideal Heat Engines
relative efficiency
- the ratio of the thermal efficiency to the Carnot
efficiency.
εrel = εth
εC
example: A gasoline engine has a thermal efficiency of 20% and a Carnot
efficiency of 30%. What is its relative efficiency?
εrel = εth = 0.20 = 67%
εC 0.30
12.5 The Carnot Cycle and Ideal Heat Engines
Example 12.8: What is the Carnot efficiency of a gasoline engine which
operates between 450 K and room temperature, 300K?
12.5 The Carnot Cycle and Ideal Heat Engines
the third law of thermodynamics
never been observed experimentally.
- absolute zero has
• If it were, this would in effect violate the second law. Reducing the entropy of
a system requires increasing the entropy of its surroundings.
• In cryogenic experiments, scientist have come close to absolute zero – to
about 2.0 x 10-8 K) – but have never reached it.
12.5 The Carnot Cycle and Ideal Heat Engines: Summary
• A Carnot engine is an ideal engine which has maximum possible efficiency
for given temperature of the cold and hot reservoirs. For the Carnot engine
QC = TC
QH
TH
where the temperatures are measured in kelvin.
• The efficiency of the Carnot engine can be written as
e = 1 – TC
TH
12.5 The Carnot Cycle and Ideal Heat Engines: Check for
Understanding
1. The Carnot cycle consists of
a) two isobaric and two isothermal processes
b) two isometric and two adiabatic processes
c) two adiabatic and two isothermal processes
d) four arbitrary processes that return the system to its initial state.
Answer: c
12.5 The Carnot Cycle and Ideal Heat Engines: Check for
Understanding
2. Which of the following temperature reservoir relationships would have the
highest efficiency for a Carnot engine?
a) Tcold = 0.15 Thot
b) Tcold = 0.25 Thot
c) Tcold = 0.50 Thot
d) Tcold = 0.90 Thot
Answer: a
12.5 The Carnot Cycle and Ideal Heat Engines: Check for
Understanding
3. Diesel engines are more efficient than gasoline engines. Which type of
engine runs hotter? Why?
Answer: Diesel engines run hotter as diesel fuel burns at a higher
temperature. According to Carnot efficiency, the higher the hot reservoir
temperature, the higher the efficiency.
Homework for Chapter 12.5
• HW 12.C: Handout and p.416: 72-74; 76.
Formulas for Chapter 12
U = Q + W
First Law of Thermodynamics
+ Q is heat added to the system
-Q is heat removed from the system
+ W is work done on the system
- W is work done by the system
W = - p V
Calculates the work done by an expanding gas at constant
pressure. It is negative.
=W
c
QH
Defines and computes the thermal efficiency of a heat
engine.
= TH – TC
TH
Defines and calculates the Carnot efficiency of an ideal
heat engine.