Ch11/12 . Thermodynamics
1. Thermodynamics is the branch of physics that is built upon
the fundamental laws that heat and work obey.
In thermodynamics the collection of objects on which
attention is being focused is called the system, while everything
else in the environment is called the surroundings. The system
and its surroundings are separated by walls of some kind.
Walls that permit heat to flow through them, such as those of
the engine block, are called diathermal walls. Perfectly
insulating walls that do not permit heat to flow between the
system and its surroundings are known as adiabatic walls.
1
2. THERMODYMANICS
Thermodynamics is the
study of the motion of heat
energy as it is transferred
from the system to the
surrounding or from the
surrounding to the system.
The transfer of heat could be due to
a physical change or a chemical
change.
There are three laws of chemical
thermodynamics.
CHEMICAL
THERMODYMANICS
3. The first law of thermodynamics:
Energy and matter can be neither created nor destroyed; only
transformed from one form to another. The energy and matter
of the universe is constant.
4. The second law of thermodynamics:
In any spontaneous process there is always an increase in the
entropy of the universe. The entropy is increasing.
5. The third law of thermodynamics:
The entropy of a perfect crystal at 0 K is zero.
There is no molecular motion at absolute 0 K.
6. HEAT
The energy that flows into or out of a
system because of a difference in
temperature between the thermodynamic
system and its surrounding.
Symbolized by "q".
A. When heat is evolved by a system,
energy is lost and "q” is negative (-).
B. When heat is absorbed by the system,
the energy is added and "q" is positive (+).
HEAT FLOW
Heat can flow in one of two directions:
a. Exothermic
To give off heat; energy is lost from the
system: (-q)
b. Endothermic
To absorb heat; energy is added to the
system: (+q)
If the heat transfer involves a chemical reaction then q is called:
HEAT OF REACTION
The heat energy (DH; enthalpy) required to
return a system to the given temperature
at the completion of the reaction.
q = DH
at constant pressure
The heat of reaction can be specific to a reaction like:
HEAT OF COMBUSTION
The quantity of heat energy given off when
a specified amount of substance burns in
oxygen.
UNITS: kJ/mol (kilojoules per mole) or kcal/mol (kilocalories per mole)
HEAT CAPACITY &
SPECIFIC HEAT
7. HEAT CAPACITY: The quantity of heat
needed to raise the temperature of a
substance one degree Celsius (or one
Kelvin).
q = Cp DT
8. SPECIFIC HEAT: The quantity of heat
required to raise the temperature of one
gram of a substance by one degree
Celsius (or one Kelvin).
q = s x m x DT
s= c- specific heat
Both Cp & s are chemical specific constants found in the textbook or CRC
Handbook.
UNITS for HEAT ENERGY
9. Heat energy is usually measured in
either Joules, given by the unit (J), and
kilojoules (kJ) or in calories, written
shorthand as (cal), and kilocalories
(kcal).
10. 1 cal = 4.184 J
NOTE: This conversion correlates to the specific heat of water
which is 1 cal/g oC or 4.184 J/g oC.
SPECIFIC HEAT
• 11. Determine the energy (in kJ) required
to raise the temperature of 100.0 g of
water from 20.0 oC to 85.0 oC?
m = 100.0 g
q = m x s x DT
DT = Tf -Ti = 85.0 - 20.0 oC = 65.0 oC
s (H2O) = 4.184 J/ g - oC
q = (100.0 g) x (4.184 J/g-oC) x (65.0oC)
q = 27196 J (1 kJ / 1000J) = 27.2 kJ
• 12. Determine the specific heat of an
unknown metal that required 2.56 kcal of
heat to raise the temperature of 150.00 g
from 15.0 oC to 200.0 oC?
S = 0.0923 cal /g -oC
13. LAW OF CONSERVATION OF
ENERGY
• The law of conservation of energy (the
first law of thermodynamics), when
related to heat transfer between two
objects, can be stated as:
The heat lost by the hot object = the heat
gained by the cold object
-qhot = qcold
-mh x sh x DTh = mc x sc x DTc
where DT = Tfinal - Tinitial
LAW OF CONSERVATION OF ENERGY
• 14. Assuming no heat is lost, what mass of
cold water at 0.00oC is needed to cool 100.0
g of water at 97.6oC to 12.0 oC?
-mh x sh x DTh = mc x sc x DTc
- (100.0g) (1 cal/goC) (12.0-97.6oC) = m (1 cal/goC) (12.0 - 0.0 oC)
8560 cal = m (12.0 cal/g)
m = 8560 cal / (12.0 cal/g)
m = 713 g
• Calculate the specific heat of an unknown metal if a 92.00 g
piece at 100.0oC is dropped into 175.0 mL of water at 17.8 oC.
The final temperature of the mixture was 39.4oC.
s (metal) = 0.678 cal/g oC
CW 1: Heat PROBLEMS
1. Iron metal has a specific heat of 0.449 J/goC. How much heat is
transferred to a 5.00 g piece of iron, initially at 20.0 oC, when it is
placed in a beaker of boiling water at 1 atm?
180. J
2. How many calories of energy are given off to lower the temperature of
100.0 g of iron from 150.0 oC to 35.0 oC?
3
1.23 x 10 cal
3. If 3.47 kJ were absorbed by 75.0 g H2O at 20.0 oC, what would be the
final temperature of the water?
31.1 oC
4. A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g
of lead shots (at 100 oC) was added to the calorimeter and the final
temperature of the mixture was 34.4 oC. What is the specific heat of
lead?
o
1.28 J/g C
5. A 17.9 g sample of unknown metal was heated to 48.31 oC. It was then
added to 28.05 g of water in an insulted cup. The water temperature
rose from 21.04 oC to 23.98oC. What is the specific heat of the metal
in J/goC?
0.792 J/goC
HW HEAT PROBLEM
• _____1. A 250.0 g metal bar requires 5.866 kJ to change its
temperature from 22.0oC to 100.0oC. What is the specific heat
of the metal in J/goC?
• _____2. How many joules are required to lower the temperature
of 100.0 g of iron from 75.0 oC to 25.0 oC?
• _____ 3. If 40.0 kJ were absorbed by 500.0 g H2O at 10.0 oC,
what would be the final temperature of the water?
• _____ 4. A 250 g of water at 376.3 oC is mixed with 350.0 mL of
water at 5.0 oC. Calculate the final temperature of the mixture.
• _____5. A 400 g piece of gold at 500.0oC is dropped into 15.0 L
of water at 22.0oC. The specific heat of gold is 0.131 J/goC or
0.0312 cal/goC. Calculate the final temperature of the mixture
assuming no heat is lost to the surroundings.
The study of heat and its
transformation into mechanical
energy is called
thermodynamics. The word
thermodynamics stems from
Greek words meaning
“movement of heat.” The
foundation of thermodynamics is
the conservation of energy and
the fact that heat flows from hot
to cold. It provides the basic
theory of heat engines.
Absolute Zero
As the thermal motion of atoms in a substance
approaches zero, the kinetic energy of the atoms
approaches zero, and the temperature of the
substance approaches a lower limit.
Absolute Zero
15. Absolute zero is the temperature at which no more energy
can be extracted from a substance.
At absolute zero, no further lowering of its temperature is
possible.
This temperature is 273 degrees below zero on the Celsius scale.
Absolute Zero
The absolute temperatures of various
objects and phenomena.
Absolute Zero
think!
A sample of hydrogen gas has a temperature of 0°C. If the
gas is heated until its molecules have doubled their average
kinetic energy (the gas has twice the absolute temperature),
what will be its temperature in degrees Celsius?
24.1 Absolute Zero
think!
A sample of hydrogen gas has a temperature of 0°C. If the
gas is heated until its molecules have doubled their average
kinetic energy (the gas has twice the absolute temperature),
what will be its temperature in degrees Celsius?
Answer:
At 0°C the gas has an absolute temperature of 273 K. Twice
as much average kinetic energy means it has twice the
absolute temperature. This would be 546 K, or 273°C.
24.1 Absolute Zero
What happens to a substance’s temperature as the
motion of its atoms approaches zero?
The Zeroth Law of Thermodynamics
21
Thermal equilibrium: Two systems are said to be in thermal
equilibrium if there is no net flow of heat between them
when they are brought into thermal contact.
Temperature is the indicator of thermal equilibrium in the
sense that there is no net flow of heat between two systems in
thermal contact that have the same temperature.
THE ZEROTH LAW OF THERMODYNAMICS
Two systems individually in thermal equilibrium with a third
system* are in thermal equilibrium with each other.
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16. First Law of Thermodynamics
The first law of thermodynamics states that
whenever heat is added to a system, it transforms to
an equal amount of some other form of energy.
First Law of Thermodynamics
HISTORY :
In the eighteenth century, heat was thought to be an invisible fluid called
caloric, which flowed like water from hot objects to cold objects.
In the 1840s, James Joule demonstrated that the flow of heat was nothing
more than the flow of energy itself.
The caloric theory of heat was gradually abandoned.
First Law of Thermodynamics
As the weights fall, they give up potential energy and warm the water
accordingly. This was first demonstrated by James Joule, for whom the unit
of energy is named.
First Law of Thermodynamics
Today, we view heat as a form
of energy.
Energy can neither be created
nor destroyed.
16. The first law of thermodynamics is
the law of conservation of energy applied
to thermal systems.
First Law of Thermodynamics
If we add heat energy to a system, the added energy does one or both of two
things:
• increases the internal energy of the system if it remains in the system
• does external work if it leaves the system
16. So, the first law of thermodynamics states:
Heat added = increase in internal energy + external work done
by the system
Q = ΔU + W
ΔU = Q-W
First Law of Thermodynamics
Work
Adding heat is not the only way to increase the internal energy of a
system.
If we set the “heat added” part of the first law to zero, changes in internal
energy are equal to the work done on or by the system.
First Law of Thermodynamics
think!
17. a. If 10 J of energy is added to a system that does
no external work, by how much will the internal energy
of that system be raised?
24.2 First Law of Thermodynamics
think!
If 10 J of energy is added to a system that does no
external work, by how much will the internal energy of
that system be raised?
Answer:
ΔU = Q- W = 10J -0 = 10J
The First Law of Thermodynamics
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THE FIRST LAW OF THERMODYNAMICS
The internal energy of a system changes from an initial
value Ui to a final value of Uf due to heat Q and work.
Q is positive when the system gains heat and negative
when it loses heat.
W is positive when work is done by the system and
negative when work is done on the system.
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CW Internal Energy
Positive and Negative Work
33
17. B The figure illustrates a system and its surroundings. In ist
part , the system gains 1500 J of heat from its surroundings, and
2200 J of work is done by the system on the surroundings.
17. C In the second part , the system also gains 1500 J of heat,
but 2200 J of work is done on the system by the surroundings. In
each case, determine the change in the internal energy of the
system.
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18. An Ideal Gas
The temperature of three moles of a monatomic ideal gas is
reduced from Ti = 540 K to Tf = 350 K by two different
methods. In the first method 5500 J of heat flows into the gas,
while in the second, 1500 J of heat flows into it. In each case
find (a) the change in the internal energy and (b) the work
done by the gas.
(a)
(b)
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st
1
CW 2: on
Law of
Thermodynamics
• #1 if 30 Joules is added to the system that
10 Joules of external work is done how
much internal energy of that system be
raised?
• #2 If 30 joules is added to the system ,
how much will the internal system be
raised if no external work is done?
Check Your Understanding 1
A gas is enclosed within a chamber that is fitted with a
frictionless piston. The piston is then pushed in, thereby
compressing the gas. Which statement below regarding this
process is consistent with the first law of thermodynamics?
a. The internal energy of the gas will increase.
b. The internal energy of the gas will decrease.
c. The internal energy of the gas will not change.
d. The internal energy of the gas may increase, decrease, or
remain the same, depending on the amount of heat that the gas
gains or loses.
37
Check Your Understanding 1
A gas is enclosed within a chamber that is fitted with a
frictionless piston. The piston is then pushed in, thereby
compressing the gas. Which statement below regarding this
process is consistent with the first law of thermodynamics?
a. The internal energy of the gas will increase.
b. The internal energy of the gas will decrease.
c. The internal energy of the gas will not change.
d. The internal energy of the gas may increase, decrease, or
remain the same, depending on the amount of heat that the gas
gains or loses.
(d)
38
First Law of Thermodynamics
What does the first law of thermodynamics state?
19. Thermal Processes
quasi-static means that it occurs slowly enough that a
uniform pressure and temperature exist throughout all
regions of the system at all times.
a. An isobaric process is one that occurs at constant pressure.
40
The substance in
the chamber is
expanding
isobarically
because the
pressure is held
constant by the
external
atmosphere and
the weight of the
piston and the
block.
41
20.
Isobaric Expansion of Water
One gram of water is placed in the cylinder in above figure, and
the pressure is maintained at 2.0 × 105 Pa. The temperature of the
water is raised by 31 C°. In one case, the water is in the liquid
phase and expands by the small amount of 1.0 × 10–8 m3. In
another case, the water is in the gas phase and expands by the
much greater amount of 7.1 × 10–5 m3. For the water in each case,
find (a) the work done and (b) the change in the internal energy.
Cliquid water = 4186 J/(kg·C°)
Cgas water = 2020 J/(kg·C°).
42
(a)
(b)
43
a. For an isobaric process, a pressure-versus-volume plot is a
horizontal straight line, and the work done [W = P(V f – V i)]
is the colored rectangular area under the graph.
44
b. The substance in the
chamber is being heated
isochorically because the
rigid chamber keeps the
volume constant.
c. The pressure-volume plot
for an isochoric process is a
vertical straight line. The
area under the graph is
zero, indicating that no
work is done.
c. isochoric process, one that occurs at constant
volume.
45
d.. isothermal process, one that takes place at constant
temperature. (when the system is an ideal gas.)
e. There is adiabatic process, one that occurs without the transfer
of heat . Since there is no heat transfer, Q equals zero, and the
first law indicates that D U = Q – W = –W. Thus, when work is
done by a system adiabatically, W is positive and the internal
energy of the system decreases by exactly the amount of the
work done. When work is done on a system adiabatically, W is
negative and the internal energy increases correspondingly.
46
21. Adiabatic Processes
When work is done on a gas by adiabatically
compressing it, the gas gains internal energy and
becomes warmer.
Adiabatic Processes
When a gas is compressed or expanded so that no heat enters
or leaves a system, the process is said to be adiabatic.
Adiabatic Processes
Do work on a pump by
pressing down on the
piston and the air is
warmed.
Adiabatic Processes
Blow warm air onto your hand from your wide-open mouth. Now reduce
the opening between your lips so the air expands as you blow. Adiabatic
expansion—the air is cooled.
Adiabatic Processes
A common example of a near-adiabatic process is the compression and
expansion of gases in the cylinders of an automobile engine.
Compression and expansion occur in only a few hundredths of a
second, too fast for heat energy to leave the combustion chamber.
For very high compressions, like those in a diesel engine, the
temperatures are high enough to ignite a fuel mixture without a spark
plug.
Diesel engines have no spark plugs.
CW 3 : Explain the How the
Gasoline engine works.
• P 358 Fig 24.4
• Flow Chart And
Report – 50 points
• Use markers and
white paper
• Discuss where
Adiabatic process
takes place in the
engine’s system ?
• Discuss where the
1st Law is Applied
in the engine’s
system ?
Adiabatic Processes
One cycle of a four-stroke internal combustion engine.
24.3 Adiabatic Processes
One cycle of a four-stroke internal combustion engine.
24.3 Adiabatic Processes
One cycle of a four-stroke internal combustion engine.
24.3 Adiabatic Processes
One cycle of a four-stroke internal combustion engine.
24.3 Adiabatic Processes
One cycle of a four-stroke internal combustion engine.
Adiabatic Processes
When a gas adiabatically expands, it does work on its surroundings
and gives up internal energy, and thus becomes cooler.
Adiabatic Processes
Heat and Temperature
Air temperature may be changed by adding or subtracting heat,
by changing the pressure of the air, or by both.
Heat may be added by solar radiation, by long-wave Earth radiation,
by condensation, or by contact with the warm ground.
Heat may be subtracted by radiation to space, by evaporation of rain
falling through dry air, or by contact with cold surfaces.
Adiabatic Processes
For many atmospheric processes, the amount of heat added or
subtracted is small enough that the process is nearly adiabatic.
In this case, an increase in pressure will cause an increase in
temperature, and vice versa.
We then have the adiabatic form of the first law:
Change in air temperature ~ pressure change
Adiabatic Processes
Pressure and Temperature
Adiabatic processes in the atmosphere occur in large masses of air
that have dimensions on the order of kilometers.
We’ll call these large masses of air blobs.
As a blob of air flows up the side of a mountain, its pressure lessens,
allowing it to expand and cool.
The reduced pressure results in reduced temperature.
Explain : Pressure
1.High Altitude
2. Low Altitude
24.3 Adiabatic Processes
19. The temperature of a blob of dry air
that expands adiabatically changes by
about 10°C for each kilometer of
elevation.
24.3 Adiabatic Processes
An example of this adiabatic warming is the chinook—a warm, dry wind
that blows from the Rocky Mountains across the Great Plains.
Cold air moving down the slopes of the mountains is compressed into a
smaller volume and is appreciably warmed.
Communities in the paths of chinooks experience relatively warm weather
in midwinter.
24.3 Adiabatic Processes
A thunderhead is the result of the
rapid adiabatic cooling of a rising
mass of moist air. Its energy
comes from condensation and
freezing of water vapor.
24.3 Adiabatic Processes
What is the effect of adiabatic compression on a
gas?
22. Second Law of Thermodynamics
The second law of thermodynamics states
that heat will never of itself flow from a
cold object to a hot object.
Second and Third Laws of Thermodynamics
Examples :
a.If a hot brick is next to a cold brick, heat flows from the hot brick to the
cold brick until both bricks arrive at thermal equilibrium.
b.In winter, heat flows from inside a warm heated home to the cold air
outside.
c.In summer, heat flows from the hot air outside into the home’s cooler
interior.
(Heat can be made to flow the other way, but only by imposing external
effort—as occurs with heat pumps.)
23. Absolute ZERO Possible – 3rd Law
Absolute ZERO Possible?There is also
a third law of thermodynamics: no
system can reach absolute zero.
As investigators attempt to reach this
lowest temperature, it becomes more
difficult to get closer to it.
Physicists have been able to record
temperatures that are less than a millionth
of 1 kelvin—but never as low as 0 K.
Second and Third Laws of Thermodynamics
What does the second law of thermodynamics state
about heat flow?
CW4: HEAT Engine
and Boiler 50pts
• Draw the Boiler on
the right -10pts
• Draw the Generic
Heat Engine on the
left of the paper.10pts
• Connect the Heat
engine Part to the
Boiler Part
Discuss in what part does
adiabatic process takes
place? Explain why. 10pts
Discuss where the 1st,2nd,3rd
Laws of Thermodynamics
apply in these engines.
Explain why ? 10pts
24.5 Heat Engines and the Second Law
According to the second law of
thermodynamics, no heat engine can convert all
heat input to mechanical energy output.
24.5 Heat Engines and the Second Law
Heat Engine Mechanics
24. A heat engine is any device that changes internal energy into
mechanical work.
The basic idea behind a heat engine is that mechanical work can
be obtained as heat flows from high temperature to low
temperature.
Some of the heat can be transformed into work in a heat engine.
24.5 Heat Engines and the Second Law
24 . Every heat engine will
• increase its internal energy by absorbing heat from a reservoir of
higher temperature,
• convert some of this energy into mechanical work, and
• expel the remaining energy as heat to some lower-temperature
reservoir.
24.5 Heat Engines and the Second Law
24.. When heat energy flows in any
heat engine from a high-temperature
place to a low-temperature place, part
of this energy is transformed into
work output.
24.5 Heat Engines and the Second Law
Heat Engine Physics
25. A steam turbine engine demonstrates the role of temperature difference
between heat reservoir and sink.
24.5 Heat Engines and the Second Law
25. The second law states that when work is done by a heat engine running
between two temperatures, Thot and Tcold, only some of the input heat at Thot
can be converted to work.
The rest is expelled as heat at Tcold.
24.5 Heat Engines and the Second Law
There is always heat exhaust, which may be desirable or
undesirable.
Hot steam expelled in a laundry on a cold winter day may be quite
desirable.
. The same steam on a hot summer day is something else. When
expelled heat is undesirable, we call it thermal pollution.
24.5 Heat Engines and the Second Law
26. Heat Engine Efficiency
French engineer Sadi Carnot carefully analyzed the heat engine and
made a fundamental discovery:
The upper fraction of heat that can be converted to useful work,
even under ideal conditions, depends on the temperature difference
between the hot reservoir and the cold sink.
26 . Heat Engines and the Second Law
The Carnot efficiency, or ideal efficiency, of a heat engine is the
ideal maximum percentage of input energy that the engine can
convert to work.
Thot is the temperature of the hot reservoir.
Tcold is the temperature of the cold.
26.. Heat Engines and the Second Law
Ideal efficiency depends only on the temperature difference
between input and exhaust.
When temperature ratios are involved, the absolute temperature
scale must be used, so Thot and Tcold are expressed in kelvins.
The higher the steam temperature driving a motor or
turbogenerator, the higher the efficiency of power production.
24.5 Heat Engines and the Second Law
think!
27. What is the ideal efficiency of an engine if both its hot
reservoir and exhaust are the same temperature—say, 400 K?
The equation for ideal efficiency is as follows:
24.5 Heat Engines and the Second Law
think!
What is the ideal efficiency of an engine if both its hot reservoir
and exhaust are the same temperature—say, 400 K? The
equation for ideal efficiency is as follows:
Answer:
Zero efficiency; (400 - 400)/400 = 0. This means no work
output is possible for any heat engine unless a temperature
difference exists between the reservoir and the sink.
24.5 Heat Engines and the Second Law
28. For example, when the heat reservoir in a steam turbine is 400
K (127°C) and the sink is 300 K (27°C), calculate the ideal
efficiency .
24.5 Heat Engines and the Second Law
28. For example, when the heat reservoir in a steam turbine is 400
K (127°C) and the sink is 300 K (27°C), the ideal efficiency is
Under ideal conditions, 25% of the internal energy of the steam can
become work, while the remaining 75% is expelled as waste.
Increasing operating temperature to 600 K yields an efficiency of
(600 — 300)/600 = 1/2, twice the efficiency at 400 K.
CW5 : Carnot Ideal Efficiency
1.Calculate the ideal
efficiency of a heat
engine that takes in
energy at 400K and
expels heat to a
reservoir at 300K . If the
operating temperature
is increased from 400K
to 600K, what will be its
ideal efficiency?
• 2. Calculate an
ideal efficiency of a
steam turbine that
has a hot reservoir
of 127 C high
pressure steam
and a sink of 27C .
24.5 Heat Engines and the Second Law
23. For example, when the heat reservoir in a steam turbine is 400
K (127°C) and the sink is 300 K (27°C), the ideal efficiency is
Under ideal conditions, 25% of the internal energy of the steam can
become work, while the remaining 75% is expelled as waste.
Increasing operating temperature to 600 K yields an efficiency of
(600 — 300)/600 = 1/2, twice the efficiency at 400 K.
24.5 Heat Engines and the Second Law
By condensing the steam, the pressure on the back sides is greatly reduced.
With confined steam, temperature and pressure go hand in hand—increase
temperature and you increase pressure.
The pressure difference is directly related to the temperature difference
between the heat source and the exhaust.
24.5 Heat Engines and the Second Law
Carnot’s equation states the upper
limit of efficiency for all heat engines.
The higher the operating temperature
(compared with exhaust temperature)
of any heat engine, the higher the
efficiency.
Only some of the heat input can be
converted to work—even without
considering friction.
24.5 Heat Engines and the Second Law
think!
24. What is the ideal efficiency of an engine if both its hot
reservoir and exhaust are the same temperature—say, 400 K?
The equation for ideal efficiency is as follows:
24.5 Heat Engines and the Second Law
How does the second law of thermodynamics apply to
heat engines?
Read 24.6 -24.7
Order to Disorder
Very Important
Facts that affects
The environment
and people’s lives
Vocabulary :
New words and
definition
24.6 Order Tends to Disorder
Natural systems tend to proceed toward a
state of greater disorder.
24.6 Order Tends to Disorder
The first law of thermodynamics states that energy can be neither created
nor destroyed.
DISORDER
The second law adds that whenever energy transforms, some of it
degenerates into waste heat, unavailable to do work.
DISORDER
Another way to say this is that organized, usable energy degenerates into
disorganized, nonusable energy.
It is then unavailable for doing the same work again.
24.6 Order Tends to Disorder
EXAMPLE of DISORDER or WASTE of ENERGY
Push a heavy crate across a rough floor and all your work will go into
heating the floor and crate. Work against friction turns into disorganized
energy.
24.6 Order Tends to Disorder
EXAMPLE of DISORDER – WASTE ENERGY
Organized energy in the form of electricity that goes into electric lights in
homes and office buildings degenerates to heat energy.
The electrical energy in the lamps, even the part that briefly exists in the
form of light, turns into heat energy.
This energy is degenerated and has no further use.
24.6 Order Tends to Disorder
EXAMPLE of PUTTING ORDER TO
DISORDER
The Transamerica® Pyramid and some
other buildings are heated by electric
lighting, which is why the lights are on
most of the time.
24.6 Order Tends to Disorder
We see that the quality of energy is lowered with each
transformation.
Organized energy tends to disorganized forms.
24.6 Order Tends to Disorder
Imagine that in a corner of a room sits a
closed jar filled with argon gas atoms.
When the lid is removed, the argon atoms
move in haphazard directions, eventually
mixing with the air molecules in the room.
The system moves from a more ordered
state (argon atoms concentrated in the jar)
to a more disordered state (argon atoms
spread evenly throughout the room).
24.6 Order Tends to Disorder
What happens to the orderly state of any
natural system?
24.7 Entropy
According to the second law of thermodynamics, in
the long run, the entropy of a system always
increases for natural processes.
24.7 Entropy
25. Entropy is the measure of the amount of disorder in a
system.
Disorder increases; entropy increases.
24.7 Entropy
When a physical system can distribute its energy freely, entropy increases
and energy of the system available for work decreases.
24.7 Entropy
This run-down house demonstrates entropy. Without continual
maintenance, the house will eventually fall apart.
24.7 Entropy
For the system “life forms plus their
waste products” there is still a net
increase in entropy.
Energy must be transformed into the
living system to support life. When it is
not, the organism soon dies and tends
toward disorder.
24.7 Entropy
Even the most improbable states may occur, and entropy spontaneously
decrease:
• haphazard motions of air molecules could momentarily become
harmonious in a corner of the room
• a barrel of pennies dumped on the floor could
show all heads
• a breeze might come into a messy room and
make it organized
The odds of these things occurring are infinitesimally small.
24.7 Entropy
The motto of this contractor—“Increasing entropy is our
business”—is appropriate because by knocking down the
building, the contractor increases the disorder of the structure.
24.7 Entropy
The laws of thermodynamics are sometimes put this way:
• You can’t win (because you can’t get any more energy out of a
system than you put in).
• You can’t break even (because you can’t even get as much
energy out as you put in).
• You can’t get out of the game (entropy in the universe is always
increasing).
24.7 Entropy
What always happens to the entropy of
systems?
Experiment :
Heat
I. Purpose :
•To determine the total amount of Heat
involved in the process of Mixture using Q=
mcΔT
•Q= mhV or Q= mhf using
II. Materials :
graduated cylinder
10 ml of water
ice cube
plates
beaker
thermometer
Triple beam balance
III. Procedures
IV. Data and Results
V=1
0
mL
Mass,
g
M1
Water
Mass,
g
M2
Ice
cube
Ti – initial
temperature.
water
Tf – final
temperature,
Mixture when
The ice melted
Q= m1cΔT
water
Q= m2hf
ice
Q = m2cΔT
Melted ice
Total Q
V. Calculations
Using equations
Q= mcΔT
Q= mhf
Analysis and Conclusions
• 1. Explain and Discuss the whole
experiment .
• 2. Explain how you calculate the total heat
of the system .
The area under a
pressure-volume graph
is the work for any kind
of process.
The colored area gives
the work done by the
gas for the process
from X to Y.
116
Example 4. Work and the Area Under
a Pressure-Volume Graph
Determine the work for the process in
which the pressure, volume, and
temperature of a gas are changed
along the straight line from X to Y in
the figure.
= +180 J
117
Check Your Understanding 2
The drawing shows a
pressure-versusvolume plot for a
three-step process: A to
B, B to C, and C to A.
For each step, the
work can be positive,
negative, or zero.
Which answer below
correctly describes the
work for the three
steps?
118
A
B
B
C
C
A
a.
Positive
Negative
Negative
b.
Positive
Positive
Negative
c.
Negative
Negative
Positive
d.
Positive
Negative
Zero
e.
Negative
Positive
Zero
(b)
119
Thermal Processes Using an Ideal Gas
120
ISOTHERMAL EXPANSION OR COMPRESSION
P = nRT/V
W = P DV = P(Vf – Vi)
121
Example 5.
Isothermal Expansion of an Ideal Gas
Two moles of the monatomic gas argon expand isothermally at
298 K, from an initial volume of Vi = 0.025 m3 to a final volume of
Vf = 0.050 m3. Assuming that argon is an ideal gas, find (a) the
work done by the gas, (b) the change in the internal energy of the
gas, and (c) the heat supplied to the. gas.
(a)
(b)
(c)
122
ADIABATIC EXPANSION OR COMPRESSION
123
[Ti = PiVi/(nR)]
[Tf = PfVf/(nR)].
124
Type of Thermal
Process
Work Done
Isobaric (constant
pressure)
W = P(Vf – Vi)
Isochoric (constant
volume)
W = 0 J
Isothermal
(constant
temperature)
Adiabatic (no heat
flow)
First Law of
Thermodynamics
(DU = Q – W)
(for an ideal gas)
(for a monatomic ideal gas)
125
Specific Heat Capacities
where the capital letter C refers to the molar specific heat
capacity in units of J/(mol·K).
126
127
128
The Second Law of Thermodynamics
THE SECOND LAW OF THERMODYNAMICS:
THE HEAT FLOW STATEMENT
Heat flows spontaneously from a substance at a
higher temperature to a substance at a lower
temperature and does not flow spontaneously in the
reverse direction.
129
130
Heat Engines
A heat engine is any device that uses heat to perform work. It
has three essential features:
1.Heat is supplied to the engine at a relatively high input
temperature from a place called the hot reservoir.
2.Part of the input heat is used to perform work by the working
substance of the engine, which is the material within the engine
that actually does the work (e.g., the gasoline-air mixture in an
automobile engine).
3.The remainder of the input heat is rejected to a place called
the cold reservoir, which has a temperature lower than the input
temperature.
131
These three symbols refer
to magnitudes only,
without reference to
algebraic signs. Therefore,
when these symbols appear
in an equation, they do not
have negative values
assigned to them.
132
Efficiencies are often quoted as percentages obtained by
multiplying the ratio W/QH by a factor of 100.
133
Example 6. An Automobile Engine
An automobile engine has an efficiency of 22.0% and produces
2510 J of work. How much heat is rejected by the engine?
134
Carnot's Principle and the Carnot
Engine
A reversible process is one in which both the system and its
environment can be returned to exactly the states they were in
before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE
STATEMENT OF THE SECOND LAW OF
THERMODYNAMICS
No irreversible engine operating between two reservoirs at
constant temperatures can have a greater efficiency than a
reversible engine operating between the same temperatures.
Furthermore, all reversible engines operating between the
same temperatures have the same efficiency.
135
A Carnot engine is a
reversible engine in which
all input heat QH
originates from a hot
reservoir at a single
temperature TH, and all
rejected heat QC goes into
a cold reservoir at a single
temperature TC. The work
done by the engine is W.
136
where the temperatures TC and TH must be expressed in
kelvins .
137
Example 7.
A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a temperature
of 298.2 K (25.0 °C), whereas 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.
TH = 298.2 K and TC = 280.2 K
138
Conceptual Example 8. Limits on the
Efficiency of a Heat Engine
Consider a hypothetical engine that receives 1000 J of heat as
input from a hot reservoir and delivers 1000 J of work, rejecting
no heat to a cold reservoir whose temperature is above 0 K. Decide
whether this engine violates the first or the second law of
thermodynamics, or both.
It is the second law, not the first law, that limits the
efficiencies of heat engines to values less than 100%.
139
Refrigerators, Air Conditioners, and
Heat Pumps
In the refrigeration process, work W is used to remove heat QC
from the cold reservoir and deposit heat QH into the hot reservoir.
140
In a
refrigerator,
the interior
of the unit is
the cold
reservoir,
while the
warmer
exterior is
the hot
reservoir.
141
A window air
conditioner
removes heat from
a room, which is
the cold reservoir,
and deposits heat
outdoors, which is
the hot reservoir.
142
Conceptual Example 9.
You Can’t Beat the Second Law of
Thermodynamics
Is it possible to cool your kitchen by leaving the refrigerator
door open or cool your bedroom by putting a window air
conditioner on the floor by the bed?
Rather than cooling the kitchen, the open
refrigerator warms it up. The air conditioner
actually warms the bedroom.
143
144
In a heat pump the
cold reservoir is the
wintry outdoors, and
the hot reservoir is the
inside of the house.
145
This conventional
electric heating
system is delivering
1000 J of heat to the
living room.
QH = W + QC and QC/QH = TC/TH
146
Example 10. A Heat Pump
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)?
(a)
(b)
147
148
Check Your Understanding 3
Each drawing represents a hypothetical heat engine or a
hypothetical heat pump and shows the corresponding heats and
work. Only one is allowed in nature. Which is it?
(c)
149
Entropy
In general, irreversible processes cause us to lose some, but not
necessarily all, of the ability to perform work. This partial loss
can be expressed in terms of a concept called entropy.
Reversible processes do not alter the total entropy of the
universe.
150
Although the
relation D S =
(Q/T)R applies
to reversible
processes, it
can be used as
part of an
indirect
procedure to
find the
entropy change
for an
irreversible
process.
151
Example 11. The Entropy of the
Universe Increases
1200 J of heat flow
spontaneously through a
copper rod from a hot
reservoir at 650 K to a
cold reservoir at 350 K.
Determine the amount by
which this irreversible
process changes the
entropy of the universe,
assuming that no other
changes occur.
152
Any irreversible process increases the entropy of the universe.
153
THE SECOND LAW OF THERMODYNAMICS STATED
IN TERMS OF ENTROPY
The total entropy of the universe does not change when a
reversible process occurs ( DSuniverse = 0 J/K) and does
increases when an irreversible process occurs ( D Suniverse >
0 J/K).
154
Example 12.
Energy Unavailable for Doing Work
155
Suppose that 1200 J of heat is used as input for an engine under two
different conditions. In Figure part a the heat is supplied by a hot
reservoir whose temperature is 650 K. In part b of the drawing, the heat
flows irreversibly through a copper rod into a second reservoir whose
temperature is 350 K and then enters the engine. In either case, a 150-K
reservoir is used as the cold reservoir. For each case, determine the
maximum amount of work that can be obtained from the 1200 J of heat.
156
where T0 is the Kelvin temperature of the coldest heat
reservoir.
157
A block of ice is an example of an ordered system
relative to a puddle of water.
158
Example 13. Order to Disorder
Find the change in entropy that results when a 2.3-kg block of
ice melts slowly (reversibly) at 273 K (0 °C).
159
Check Your Understanding 4
Two equal amounts of water are mixed together in an insulated
container. The initial temperatures of the water are different,
but the mixture reaches a uniform temperature. Do the energy
and the entropy of the water increase, decrease, or remain
constant as a result of the mixing process?
(d)
Energy of the
Water
Entropy of the
Water
a.
Increases
Increases
b.
Decreases
Decreases
c.
Remains constant
Decreases
d.
Remains constant
Increases
160
The Third Law of Thermodynamics
THE THIRD LAW OF THERMODYNAMICS
It is not possible to lower the temperature of any system to
absolute zero (T = 0 K) in a finite number of steps.
161
Concepts & Calculations
Example 14. The Sublimation of Zinc
The sublimation of zinc (mass per mole = 0.0654 kg/mol) occurs
at a temperature of 6.00 × 102 K, and the latent heat of
sublimation is 1.99 × 106 J/kg. The pressure remains constant
during the sublimation. Assume that the zinc vapor can be
treated as a monatomic ideal gas and that the volume of solid
zinc is negligible compared to the corresponding vapor. What is
the change in the internal energy of the zinc when 1.50 kg of
zinc sublimates?
Q = DU + W,
Q = mLs,
W = nRT
162
163
Concepts & Calculations Example 15.
The Work–Energy Theorem
Each of two Carnot engines
uses the same cold reservoir
at a temperature of 275 K
for its exhaust heat. Each
engine receives 1450 J of
input heat.
164
The work from either of these engines is used to drive a pulley
arrangement that uses a rope to accelerate a 125-kg crate from
rest along a horizontal frictionless surface. With engine 1 the
crate attains a speed of 2.00 m/s, while with engine 2 it attains a
speed of 3.00 m/s. Find the temperature of the hot reservoir for
each engine.
165
166
Conceptual Question 14
REASONING AND SOLUTION The efficiency of a Carnot engine is
given by Equation 15.15 efficiency  1  ( TC / TH ). Three reversible
engines A, B, and C, use the same cold reservoir for their exhaust
heats. They use different hot reservoirs with the following
temperatures: (A) 1000 K; (B) 1100 K; and (C) 900 K. We can
rank these engines in order of increasing efficiency according to the
following considerations. The ratio TC / THis inversely proportional
to the value of TH. The ratio TC / TH will be smallest for engine B;
therefore, the quantity 1 – TC / TH will be largest for engine B. Thus,
engine B has the largest efficiency. Similarly, the ratio TC / THwill be
largest for engine C; therefore, the quantity 1 – TC / TH will be
smallest for engine C. Thus, engine C has the smallest efficiency.
Hence, the engines are, in order of increasing efficiency: engine C,
engine A, and engine B.
167
Conceptual Question 15
REASONING AND SOLUTION The efficiency of a Carnot engine
is given by Equation 15.15: efficiency  1  ( TC / TH )
a. Lowering the Kelvin temperature of the cold reservoir by a
factor of four makes the ratio TC / THone-fourth as great.
b. Raising the Kelvin temperature of the hot reservoir by a factor of
four makes the ratio TC / THone-fourth as great.
c. Cutting the Kelvin temperature of the cold reservoir in half and
doubling the Kelvin temperature of the hot reservoir makes the
ratio TC / TH one-fourth as great.
Therefore, all three possible improvements have the same effect on
the efficiency of a Carnot engine.
168
Problem 51
REASONING AND SOLUTION The efficiency of the engine
is e = 1 - (TC/TH) so
(i) Increase TH by 40 K; e = 1 - [(350 K)/(690 K)] = 0.493
(ii) Decrease TC by 40 K; e = 1 - [(310 K)/(650 K)] = 0.523
The greatest improvement is made by lowing the temperature
of the cold reservoir.
169
Problem 63
REASONING AND SOLUTION Let CP denote the coefficient
of performance. By definition (Equation 15.16), CP = QC/W,
so that
7.6 104 J
W

 3.8  104 J
CP
2.0
QC
Thus, the amount of heat that is pumped out of the back of
the air conditioner is
QH  W  QC  3.8 104 J + 7.6 104 J = 1.14 105 J
170
The temperature rise in the room can be found as follows:
QH = CV n DT
QH
QH
1.14 105 J
DT 

5
 1.4 K
5R n
CV n
[8.31 J/(mol  K)](3800 mol)
2
2
 
171
Problem 64
REASONING AND SOLUTION The amount of heat removed
from the ice QC is
QC = mLf = (2.0 kg)(33.5 * 104 J/kg) = 6.7 * 105 J
The amount of heat leaving the refrigerator QH is therefore
QH = QC(TH/TC) = (6.7 * 105 J)(300 K)/(258 K) = 7.8 * 105 J
The amount of work done by the refrigerator is therefore,
W = QH - QC = 1.1 * 105 J
At $0.10 per kWh (or $0.10 per 3.6 * 106 J), the cost is
$0.1
5
3
(
1
.
1

10
J
)

$
3
.
0

10
 0.3 cents
6
3.6 10 J
172