Physics 121,Lecture 25,Pg 1
Physics 121,Lecture 25
Today’s Agenda
? Announcements
?No more homeworks !
?About midterm 2,
?Final,Monday Dec,12 @ 3:30PM Room P-38.
? Today’s topics
?Heat and energy
?Laws of thermodynamics
?Work in thermodynamics
?First Law of thermodynamics and applications
?Heat engines and Second Law of thermodynamics
?Reversible/irreversible processes and Entropy
Physics 121,Lecture 25,Pg 2
Chap,14,Energy in Thermal Processes
? Solids,liquids or gases have internal energy
?Kinetic energy from random motion of molecules
? translation,rotation,vibration
?At equilibrium,it is related to temperature
? Heat,transfer of energy from one object to another as a
result of their different temperatures
? Thermal contact,energy can flow between objects
T1 T2
U1 U2
>
Physics 121,Lecture 25,Pg 3
Heat
? Heat,Q = C ? T
? Q = amount of heat that must be supplied to raise
the temperature by an amount ? T,
? [Q] = Joules or calories.
? energy to raise 1 g of water from 14.5 to 15.5 oC
? James Prescott Joule found mechanical equivalent
of heat.
? C, Heat capacity
1 Cal = 4.186 J
1 kcal = 1 Cal = 4186 J
? Q = c m ? T
? c,specific heat (heat capacity per units of mass)
? amount of heat to raise T of 1 kg by 1oC
? [c] = J/(kg oC)
? Sign convention,+Q, heat gained- Q, heat lost
Physics 121,Lecture 25,Pg 4
Specific Heat, examples
? You have equal masses of aluminum and copper at the
same initial temperature,You add 1000 J of heat to each of
them,Which one ends up at the higher final temperature?
a) aluminum
b) copper
c) the same
Substance c in J/(kg-C)
aluminum 900
copper 387
iron 452
lead 128
human body 3500
water 4186
ice 2000
Physics 121,Lecture 25,Pg 5
Latent Heat
? L = Q / m
? Heat per unit mass
[L] = J/kg
? Q = ? m L
+ if heat needed (boiling)
- if heat given (freezing)
? Lf, Latent heat of fusion
solid ? liquid
? Lv, Latent heat of vaporization
liquid ? gas
? Latent heat,amount of energy needed to add or to remove
from a substance to change the state of that substance.
?Phase change,T remains constant but internal energy changes
?heat does not result in change in T (latent =,hidden”)
? e.g., solid ? liquid or liquid ? gas
heat goes to breaking chemical bonds
Lf (J/kg) Lv (J/kg)
water 33.5 x 104 22.6 x 105
Physics 121,Lecture 25,Pg 6
Latent Heats of Fusion and Vaporization
Energy added (J)
T (oC)
120
100
80
60
40
20
0
-20
-40 Water
Water
+
Ice
Water + Steam Steam
62.7 396 815 3080
Physics 121,Lecture 25,Pg 7
Energy transfer mechanisms
? Thermal conduction (or conduction):
?Energy transferred by direct contact.
?E.g.,energy enters the water through
the bottom of the pan by thermal
conduction,
?Important,home insulation,etc.
? Rate of energy transfer
?through a slab of area A and
thickness ?x,with opposite faces at
different temperatures,Tc and Th
?k, thermal conductivity
?x
Th TcA
Energy
flow
? =Q/?t = k A (Th - Tc ) / ?x
Physics 121,Lecture 25,Pg 8
Thermal Conductivities
Aluminum 238 Air 0.0234 Asbestos 0.25
Copper 397 Helium 0.138 Concrete 1.3
Gold 314 Hydrogen 0.172 Glass 0.84
Iron 79.5 Nitrogen 0.0234 Ice 1.6
Lead 34.7 Oxygen 0.0238 Water 0.60
Silver 427 Rubber 0.2 Wood 0.10
J/s m 0C J/s m 0C J/s m 0C
Physics 121,Lecture 25,Pg 9
Energy transfer mechanisms
? Convection:
?Energy is transferred by flow of substance
?E.g., heating a room (air convection)
?E.g., warming of North Altantic by warm waters
from the equatorial regions
?Natural convection,from differences in density
?Forced convection,from pump of fan
? Radiation:
?Energy is transferred by photons
?E.g., infrared lamps
?Stephan’s law
?s = 5.7?10-8 W/m2 K4,T is in Kelvin,and A is the surface area
?e is a constant called the emissivity
? = sAe T4, Power
Physics 121,Lecture 25,Pg 10
Resisting Energy Transfer
? The Thermos bottle,also called a
Dewar flask is designed to minimize
energy transfer by conduction,
convection,and radiation,The
standard flask is a double-walled
Pyrex glass with silvered walls and
the space between the walls is
evacuated.
Vacuum
Silvered
surfaces
Hot or
cold
liquid
Physics 121,Lecture 25,Pg 11
Chap,15,the Laws of Thermodynamics
0) If two objects are in thermal equilibrium with a third,they are in
equilibrium with each other.
1) There is a quantity known as internal energy that in an isolated
system always remains the same.
2) There is a quantity known as entropy that in a closed system
always remains the same (reversible) or increases (irreversible).
Physics 121,Lecture 25,Pg 12
Zeroth Law of Thermodynamics
? Thermal equilibrium,
when objects in
thermal contact cease
heat transfer
?same temperature
T1 T2
U1 U2
=
If objects A and B are separately in thermal equilibrium
with a third object C,then objects A and B are in
thermal equilibrium with each other.
A
C B
Physics 121,Lecture 25,Pg 13
1st Law,Work & Heat
? Two types of variables
?State variables,describe the system
(e.g,T,P,V,U).
?Transfer variables,describe the process
(e.g,Q,W).
=0 unless a process occurs
? change in state variables.
? Work done on gas
?W = F d cos? = -F ?y
= - PA ?y = - P ?V
?valid only for isobaric processes
(P constant)
?If not,use average force or calculus,
W = area under PV curve
PV diagram
Physics 121,Lecture 25,Pg 14
1st Law,Work & Heat
? Work:
?Depends on the path taken in the PV-diagram
?Same for Q (heat)
Physics 121,Lecture 25,Pg 15
b) = |W1|a) > |W1| c) < |W1|
i
f
p
V
2
1
? Consider the two paths,1 and 2,
connecting points i and f on the pV
diagram.
?The magnitude of the work,|W2|,
done by the system in going from i
to f along path 2 is
Lecture 25,ACT 1
Work
Physics 121,Lecture 25,Pg 16
First Law of Thermodynamics
? Isolated system
?No interaction with surroundings
?Q = W = 0 ??U = 0.
?Uf = Ui, internal energy remains constant.
? First Law of Thermodynamics
?U = Q + W
variation of internal energy
heat flow,in” (+) or,out” (-)
work done,on” the system
?Independent of path in PV-diagram
?Depends only on state of the system (P,V,T,…)
?Energy conservation statement ?only U changes
Physics 121,Lecture 25,Pg 17
Other Applications
? Cyclic process:
?Process that starts and ends at the state
?Must have ?U = 0 ? Q = -W,
? Adiabatic process:
?No energy transferred through heat ? Q = 0.
?So,?U = W,
?Important for
?expansion of gas in combustion engines
?Liquifaction of gases in cooling systems,etc.
? Isobaric process,(P is constant)
?Work is simply
?
W ? ? P d V ? ? P (V 2V
1
V 2? ? V
1 )
Physics 121,Lecture 25,Pg 18
Other Applications (continued)
? Isovolumetric process:
?Constant volume ? W =0.
?So ?U = Q ? all heat is transferred into internal energy
?e.g,heating a,can” (no work done).
? Isothermal process:
?T is constant
?Using PV=nRT,we find P= nRT/V.
?Work becomes,
?
W ? ? P d V ? ?V
1
V 2? n R T dV
V ?V 1
V 2? n R T ln V 1
V 2
?
? ?
?
? ?
?PV is constant.
?PV-diagram,isotherm.
Physics 121,Lecture 25,Pg 19
p
V
? Identify the nature of paths A,B,C,and D
?Isobaric,isothermal,isovolumetric,and adiabatic
Lecture 25,ACT 2
Processes
A
B
C
D
T1
T2
T3T4
Physics 121,Lecture 25,Pg 20
Heat Engines
? We now try to do more than just raise the temperature of an
object by adding heat,We want to add heat to get some work
done!
? Heat engines:
?Purpose,Convert heat into work using a cyclic process
?Example,Cycle a piston of gas between hot and cold
reservoirs* (Stirling cycle)
1) hold volume fixed,raise temperature by adding heat
2) hold temperature fixed,do work by expansion
3) hold volume fixed,lower temperature by draining heat
4) hold temperature fixed,compress back to original V
Physics 121,Lecture 25,Pg 21
Heat Engines
? Example,the Stirling cycle
Gas
T=TH
Gas
T=TH
Gas
T=TC
Gas
T=TC
V
P
TC
TH
Va Vb
1 2
3
4
We can represent this
cycle on a P-V diagram:
1 1 2
34
*reservoir,large body whose
temperature does not change
when it absorbs or gives up heat
Physics 121,Lecture 25,Pg 22
? Identify whether
?Heat is ADDED or REMOVED from the
gas
?Work is done BY or ON the gas for each
step of the Stirling cycle:
V
P
TC
TH
Va Vb
1 2
3
4
ADDED
REMOVED
BY
ON
1
HEAT
WORK
step
ADDED
REMOVED
BY
ON
2
ADDED
REMOVED
BY
ON
3
ADDED
REMOVED
BY
ON
4
QU0W ???? QW0U ???? QU0W ???? QW0U ???
Heat Engines
Physics 121,Lecture 25,Pg 23
Heat Engines and the 2nd Law of
Thermodynamics
? A schematic representation of a
heat engine,The engine receives
energy Qh from the hot reservoir,
expels energy Qc to the cold
reservoir,and does work W.
? If working substance is a gas
Hot reservoir
Cold reservoir
Engine
Qh
Qc
Weng
V
P
Area = Weng
Physics 121,Lecture 25,Pg 24
Heat Engines and the 2nd Law of
Thermodynamics
? A heat engine goes through a
cycle
?1st Law gives
?U = Q + W =0
? So Qnet=|Qh| - |Qc| = -W = Weng
Hot reservoir
Cold reservoir
Engine
Qh
Qc
Weng
Physics 121,Lecture 25,Pg 25
Efficiency of a Heat Engine
? How can we define a,figure of merit” for a heat engine?
? Define the efficiency e as:
It is impossible to construct a heat engine that,operating
in a cycle,produces no other effect than the absorption
of energy from a reservoir and the performance of an
equal amount of work
?
e ? W e n gQ
h
? Q h ? Q cQ
h
? 1 ? Q cQ
h
Physics 121,Lecture 25,Pg 26
Heat Engines and the Second law of
Thermodynamics
Reservoir
Engine
Qh
Weng
It is impossible to
construct a heat engine
that,operating in a cycle,
produces no other effect
than the absorption of
energy from a reservoir
and the performance of
an equal amount of work
Physics 121,Lecture 25,Pg 27
? Consider two heat engines:
?Engine I,
?Requires Qin = 100 J of heat added to system to get
W=10 J of work
?Engine II:
?To get W=10 J of work,Qout = 100 J of heat is exhausted
to the environment
? Compare eI,the efficiency of engine I,to eII,the efficiency of
engine II.
A) eI < eII B) eI > eII C) Not enough data to determine
Lecture 25,ACT 3
Efficiency
Physics 121,Lecture 25,Pg 28
Reversible/irreversible processes
? Reversible process:
?Every state along some path is an equilibrium state
?The system can be returned to its initial conditions along the
same path
? Irreversible process;
?Process which is not reversible !
? Most real physical processes are irreversible
?E.g.,energy is lost through friction and the initial conditions
cannot be reached along the same path
?However,some processes are almost reversible
?If they occur slowly enough (so that system is almost in
equilibrium)
Physics 121,Lecture 25,Pg 29
The Carnot Engine
? No real engine operating between two energy reservoirs
can be more efficient than a Carnot engine operating
between the same two reservoirs.
A,A?B,the gas expands isothermally
while in contact with a reservoir at Th
B,B?C,the gas expands adiabatically
(Q=0)
C,C?D,the gas is compressed
isothermally while in contact with a
reservoir at Tc
D,D?A,the gas compressed
adiabatically (Q=0) V
P
A
B
CD
Weng
Physics 121,Lecture 25,Pg 30
The Carnot Engine
? All real engines are less efficient than the Carnot
engine because they operate irreversibly due to
friction as they complete a cycle in a brief time period.
? Carnot showed that the thermal efficiency of a Carnot
engine is:
h
c
c T
Te ?? 1
Physics 121,Lecture 25,Pg 31
Entropy and the 2nd Law
? Consider a reversible process between two equilibrium states
The change in entropy ?S between the two states is given by
the energy Qr transferred along the reversible path divided by
the absolute temperature T of the system in this interval.
? The Second Law of Thermodynamics
“There is a quantity known as entropy that in a closed system
always remains the same (reversible) or increases
(irreversible).”
? Entropy is a measure of disorder in a system.
T
QS r??
Physics 121,Lecture 25,Pg 32
Entropy and the 2nd Law
? What about the following situation
?Atoms all located in half the room
?Although possible,it is quite
improbable
? Disorderly arrangements are much
more probable than orderly ones
? Isolated systems tend toward greater
disorder
?Entropy is a measure of that
disorder
?Entropy increases in all natural
processes
no atomsall atoms
Physics 121,Lecture 25,Pg 33
Recap for today:
? Todays’ topics
?No more homeworks !
?About midterm 2,
?Final,Monday Dec,12 @ 3:30PM Room P-38.
? Today’s topics
?Heat and energy
?Laws of thermodynamics
?Work in thermodynamics
?First Law of thermodynamics and applications
?Heat engines and Second Law of thermodynamics
?Reversible/irreversible processes and Entropy
Physics 121,Lecture 25
Today’s Agenda
? Announcements
?No more homeworks !
?About midterm 2,
?Final,Monday Dec,12 @ 3:30PM Room P-38.
? Today’s topics
?Heat and energy
?Laws of thermodynamics
?Work in thermodynamics
?First Law of thermodynamics and applications
?Heat engines and Second Law of thermodynamics
?Reversible/irreversible processes and Entropy
Physics 121,Lecture 25,Pg 2
Chap,14,Energy in Thermal Processes
? Solids,liquids or gases have internal energy
?Kinetic energy from random motion of molecules
? translation,rotation,vibration
?At equilibrium,it is related to temperature
? Heat,transfer of energy from one object to another as a
result of their different temperatures
? Thermal contact,energy can flow between objects
T1 T2
U1 U2
>
Physics 121,Lecture 25,Pg 3
Heat
? Heat,Q = C ? T
? Q = amount of heat that must be supplied to raise
the temperature by an amount ? T,
? [Q] = Joules or calories.
? energy to raise 1 g of water from 14.5 to 15.5 oC
? James Prescott Joule found mechanical equivalent
of heat.
? C, Heat capacity
1 Cal = 4.186 J
1 kcal = 1 Cal = 4186 J
? Q = c m ? T
? c,specific heat (heat capacity per units of mass)
? amount of heat to raise T of 1 kg by 1oC
? [c] = J/(kg oC)
? Sign convention,+Q, heat gained- Q, heat lost
Physics 121,Lecture 25,Pg 4
Specific Heat, examples
? You have equal masses of aluminum and copper at the
same initial temperature,You add 1000 J of heat to each of
them,Which one ends up at the higher final temperature?
a) aluminum
b) copper
c) the same
Substance c in J/(kg-C)
aluminum 900
copper 387
iron 452
lead 128
human body 3500
water 4186
ice 2000
Physics 121,Lecture 25,Pg 5
Latent Heat
? L = Q / m
? Heat per unit mass
[L] = J/kg
? Q = ? m L
+ if heat needed (boiling)
- if heat given (freezing)
? Lf, Latent heat of fusion
solid ? liquid
? Lv, Latent heat of vaporization
liquid ? gas
? Latent heat,amount of energy needed to add or to remove
from a substance to change the state of that substance.
?Phase change,T remains constant but internal energy changes
?heat does not result in change in T (latent =,hidden”)
? e.g., solid ? liquid or liquid ? gas
heat goes to breaking chemical bonds
Lf (J/kg) Lv (J/kg)
water 33.5 x 104 22.6 x 105
Physics 121,Lecture 25,Pg 6
Latent Heats of Fusion and Vaporization
Energy added (J)
T (oC)
120
100
80
60
40
20
0
-20
-40 Water
Water
+
Ice
Water + Steam Steam
62.7 396 815 3080
Physics 121,Lecture 25,Pg 7
Energy transfer mechanisms
? Thermal conduction (or conduction):
?Energy transferred by direct contact.
?E.g.,energy enters the water through
the bottom of the pan by thermal
conduction,
?Important,home insulation,etc.
? Rate of energy transfer
?through a slab of area A and
thickness ?x,with opposite faces at
different temperatures,Tc and Th
?k, thermal conductivity
?x
Th TcA
Energy
flow
? =Q/?t = k A (Th - Tc ) / ?x
Physics 121,Lecture 25,Pg 8
Thermal Conductivities
Aluminum 238 Air 0.0234 Asbestos 0.25
Copper 397 Helium 0.138 Concrete 1.3
Gold 314 Hydrogen 0.172 Glass 0.84
Iron 79.5 Nitrogen 0.0234 Ice 1.6
Lead 34.7 Oxygen 0.0238 Water 0.60
Silver 427 Rubber 0.2 Wood 0.10
J/s m 0C J/s m 0C J/s m 0C
Physics 121,Lecture 25,Pg 9
Energy transfer mechanisms
? Convection:
?Energy is transferred by flow of substance
?E.g., heating a room (air convection)
?E.g., warming of North Altantic by warm waters
from the equatorial regions
?Natural convection,from differences in density
?Forced convection,from pump of fan
? Radiation:
?Energy is transferred by photons
?E.g., infrared lamps
?Stephan’s law
?s = 5.7?10-8 W/m2 K4,T is in Kelvin,and A is the surface area
?e is a constant called the emissivity
? = sAe T4, Power
Physics 121,Lecture 25,Pg 10
Resisting Energy Transfer
? The Thermos bottle,also called a
Dewar flask is designed to minimize
energy transfer by conduction,
convection,and radiation,The
standard flask is a double-walled
Pyrex glass with silvered walls and
the space between the walls is
evacuated.
Vacuum
Silvered
surfaces
Hot or
cold
liquid
Physics 121,Lecture 25,Pg 11
Chap,15,the Laws of Thermodynamics
0) If two objects are in thermal equilibrium with a third,they are in
equilibrium with each other.
1) There is a quantity known as internal energy that in an isolated
system always remains the same.
2) There is a quantity known as entropy that in a closed system
always remains the same (reversible) or increases (irreversible).
Physics 121,Lecture 25,Pg 12
Zeroth Law of Thermodynamics
? Thermal equilibrium,
when objects in
thermal contact cease
heat transfer
?same temperature
T1 T2
U1 U2
=
If objects A and B are separately in thermal equilibrium
with a third object C,then objects A and B are in
thermal equilibrium with each other.
A
C B
Physics 121,Lecture 25,Pg 13
1st Law,Work & Heat
? Two types of variables
?State variables,describe the system
(e.g,T,P,V,U).
?Transfer variables,describe the process
(e.g,Q,W).
=0 unless a process occurs
? change in state variables.
? Work done on gas
?W = F d cos? = -F ?y
= - PA ?y = - P ?V
?valid only for isobaric processes
(P constant)
?If not,use average force or calculus,
W = area under PV curve
PV diagram
Physics 121,Lecture 25,Pg 14
1st Law,Work & Heat
? Work:
?Depends on the path taken in the PV-diagram
?Same for Q (heat)
Physics 121,Lecture 25,Pg 15
b) = |W1|a) > |W1| c) < |W1|
i
f
p
V
2
1
? Consider the two paths,1 and 2,
connecting points i and f on the pV
diagram.
?The magnitude of the work,|W2|,
done by the system in going from i
to f along path 2 is
Lecture 25,ACT 1
Work
Physics 121,Lecture 25,Pg 16
First Law of Thermodynamics
? Isolated system
?No interaction with surroundings
?Q = W = 0 ??U = 0.
?Uf = Ui, internal energy remains constant.
? First Law of Thermodynamics
?U = Q + W
variation of internal energy
heat flow,in” (+) or,out” (-)
work done,on” the system
?Independent of path in PV-diagram
?Depends only on state of the system (P,V,T,…)
?Energy conservation statement ?only U changes
Physics 121,Lecture 25,Pg 17
Other Applications
? Cyclic process:
?Process that starts and ends at the state
?Must have ?U = 0 ? Q = -W,
? Adiabatic process:
?No energy transferred through heat ? Q = 0.
?So,?U = W,
?Important for
?expansion of gas in combustion engines
?Liquifaction of gases in cooling systems,etc.
? Isobaric process,(P is constant)
?Work is simply
?
W ? ? P d V ? ? P (V 2V
1
V 2? ? V
1 )
Physics 121,Lecture 25,Pg 18
Other Applications (continued)
? Isovolumetric process:
?Constant volume ? W =0.
?So ?U = Q ? all heat is transferred into internal energy
?e.g,heating a,can” (no work done).
? Isothermal process:
?T is constant
?Using PV=nRT,we find P= nRT/V.
?Work becomes,
?
W ? ? P d V ? ?V
1
V 2? n R T dV
V ?V 1
V 2? n R T ln V 1
V 2
?
? ?
?
? ?
?PV is constant.
?PV-diagram,isotherm.
Physics 121,Lecture 25,Pg 19
p
V
? Identify the nature of paths A,B,C,and D
?Isobaric,isothermal,isovolumetric,and adiabatic
Lecture 25,ACT 2
Processes
A
B
C
D
T1
T2
T3T4
Physics 121,Lecture 25,Pg 20
Heat Engines
? We now try to do more than just raise the temperature of an
object by adding heat,We want to add heat to get some work
done!
? Heat engines:
?Purpose,Convert heat into work using a cyclic process
?Example,Cycle a piston of gas between hot and cold
reservoirs* (Stirling cycle)
1) hold volume fixed,raise temperature by adding heat
2) hold temperature fixed,do work by expansion
3) hold volume fixed,lower temperature by draining heat
4) hold temperature fixed,compress back to original V
Physics 121,Lecture 25,Pg 21
Heat Engines
? Example,the Stirling cycle
Gas
T=TH
Gas
T=TH
Gas
T=TC
Gas
T=TC
V
P
TC
TH
Va Vb
1 2
3
4
We can represent this
cycle on a P-V diagram:
1 1 2
34
*reservoir,large body whose
temperature does not change
when it absorbs or gives up heat
Physics 121,Lecture 25,Pg 22
? Identify whether
?Heat is ADDED or REMOVED from the
gas
?Work is done BY or ON the gas for each
step of the Stirling cycle:
V
P
TC
TH
Va Vb
1 2
3
4
ADDED
REMOVED
BY
ON
1
HEAT
WORK
step
ADDED
REMOVED
BY
ON
2
ADDED
REMOVED
BY
ON
3
ADDED
REMOVED
BY
ON
4
QU0W ???? QW0U ???? QU0W ???? QW0U ???
Heat Engines
Physics 121,Lecture 25,Pg 23
Heat Engines and the 2nd Law of
Thermodynamics
? A schematic representation of a
heat engine,The engine receives
energy Qh from the hot reservoir,
expels energy Qc to the cold
reservoir,and does work W.
? If working substance is a gas
Hot reservoir
Cold reservoir
Engine
Qh
Qc
Weng
V
P
Area = Weng
Physics 121,Lecture 25,Pg 24
Heat Engines and the 2nd Law of
Thermodynamics
? A heat engine goes through a
cycle
?1st Law gives
?U = Q + W =0
? So Qnet=|Qh| - |Qc| = -W = Weng
Hot reservoir
Cold reservoir
Engine
Qh
Qc
Weng
Physics 121,Lecture 25,Pg 25
Efficiency of a Heat Engine
? How can we define a,figure of merit” for a heat engine?
? Define the efficiency e as:
It is impossible to construct a heat engine that,operating
in a cycle,produces no other effect than the absorption
of energy from a reservoir and the performance of an
equal amount of work
?
e ? W e n gQ
h
? Q h ? Q cQ
h
? 1 ? Q cQ
h
Physics 121,Lecture 25,Pg 26
Heat Engines and the Second law of
Thermodynamics
Reservoir
Engine
Qh
Weng
It is impossible to
construct a heat engine
that,operating in a cycle,
produces no other effect
than the absorption of
energy from a reservoir
and the performance of
an equal amount of work
Physics 121,Lecture 25,Pg 27
? Consider two heat engines:
?Engine I,
?Requires Qin = 100 J of heat added to system to get
W=10 J of work
?Engine II:
?To get W=10 J of work,Qout = 100 J of heat is exhausted
to the environment
? Compare eI,the efficiency of engine I,to eII,the efficiency of
engine II.
A) eI < eII B) eI > eII C) Not enough data to determine
Lecture 25,ACT 3
Efficiency
Physics 121,Lecture 25,Pg 28
Reversible/irreversible processes
? Reversible process:
?Every state along some path is an equilibrium state
?The system can be returned to its initial conditions along the
same path
? Irreversible process;
?Process which is not reversible !
? Most real physical processes are irreversible
?E.g.,energy is lost through friction and the initial conditions
cannot be reached along the same path
?However,some processes are almost reversible
?If they occur slowly enough (so that system is almost in
equilibrium)
Physics 121,Lecture 25,Pg 29
The Carnot Engine
? No real engine operating between two energy reservoirs
can be more efficient than a Carnot engine operating
between the same two reservoirs.
A,A?B,the gas expands isothermally
while in contact with a reservoir at Th
B,B?C,the gas expands adiabatically
(Q=0)
C,C?D,the gas is compressed
isothermally while in contact with a
reservoir at Tc
D,D?A,the gas compressed
adiabatically (Q=0) V
P
A
B
CD
Weng
Physics 121,Lecture 25,Pg 30
The Carnot Engine
? All real engines are less efficient than the Carnot
engine because they operate irreversibly due to
friction as they complete a cycle in a brief time period.
? Carnot showed that the thermal efficiency of a Carnot
engine is:
h
c
c T
Te ?? 1
Physics 121,Lecture 25,Pg 31
Entropy and the 2nd Law
? Consider a reversible process between two equilibrium states
The change in entropy ?S between the two states is given by
the energy Qr transferred along the reversible path divided by
the absolute temperature T of the system in this interval.
? The Second Law of Thermodynamics
“There is a quantity known as entropy that in a closed system
always remains the same (reversible) or increases
(irreversible).”
? Entropy is a measure of disorder in a system.
T
QS r??
Physics 121,Lecture 25,Pg 32
Entropy and the 2nd Law
? What about the following situation
?Atoms all located in half the room
?Although possible,it is quite
improbable
? Disorderly arrangements are much
more probable than orderly ones
? Isolated systems tend toward greater
disorder
?Entropy is a measure of that
disorder
?Entropy increases in all natural
processes
no atomsall atoms
Physics 121,Lecture 25,Pg 33
Recap for today:
? Todays’ topics
?No more homeworks !
?About midterm 2,
?Final,Monday Dec,12 @ 3:30PM Room P-38.
? Today’s topics
?Heat and energy
?Laws of thermodynamics
?Work in thermodynamics
?First Law of thermodynamics and applications
?Heat engines and Second Law of thermodynamics
?Reversible/irreversible processes and Entropy
