Chapter 18
The Second Law of Thermodynamics
Irreversible Processes
• Irreversible Processes:
always found to proceed in one direction
•
•
•
•
Examples:
free expansion of ideal gas
heat flow (for finite DT): from hot to cold
friction: converts mechanical energy to heat
Reversible Processes
• Reverisble processes:
just an idealization, but we can come close
• put system in equilibrium
(with itself and its surroundings)
• make slow, infinitesimal changes
in order to reverse a process
Heat Engine
• Heat engine:
transforms heat into
mechanical energy, W
• Reservoirs:
• hot reservoir at TH
• cold reservoir at TC
Heat Engine
• engine absorbs heat
QH from hot reservoir
• engine discards heat
QC to cold reservoir
• engine does work W
Heat Engine
• Working substance:
material in engine
undergoing the heat
transfer
• Cyclic process:
reuse same substance,
return it to initial state at
end of each cycle
Q = net heat
input
• For each cycle:
• heat QH enters engine
• heat QC leaves engine
• net heat in per cycle:
Q = QH + Q C
= QH – |QC|
Heat Engine
• cyclic process:
DU = 0 for engine
during one cycle
• 1st Law:
Q = W + DU
• Thus for each cycle:
Q=W
Heat Engine
• For each cycle:
Q = QH – |QC|
• So by 1st Law:
W=Q
= QH – |QC|
Engine
Efficiency
W= QH – |QC|
Perfect engine: |QC| = 0
All experiments: |QC| > 0
Efficiency: e
• measures how close an
engine gets to |QC| = 0
Engine
Efficiency
W
e
QH
QH  | QC |

QH
| QC |
 1
QH
Ideal Engine
= limiting behavior
of real engines
• engine: converts heat to
mechanical energy
• recall: converting
mechanical energy to
heat is ‘irreversible’
Ideal Engine
• avoid irreversibility
• carry out engine cycle
slowly and with
reversible heat flow
• the reversibility of
heat flow motivates
the allowed processes
Ideal Engine
• For reversible heat
flow, there can be no
finite DT between
engine and reservoir
• so during heat transfer
we need an
isothermal process
Ideal Engine
• If there is a finite DT
between engine and
reservoir, any heat
flow would be
irreversible
• So when there is a
finite DT we need an
adiabatic process
Ideal Engine
• so we are motivated to
consider a cycle with
only two processes:
• isothermal
• adiabatic
• called a ‘Carnot cycle’
Ideal Engine
• efficiency:
e=?
• working substance = ?
• We’ll calculate e for
substance = ideal gas
• Actually, e is
independent of the
working substance
Carnot Cycle
Example engine:
internal combustion engine
• TH is provided by combustion of air-fuel mixture
• TC is provided by exhaust gases venting to outside air
• fuel = gasoline
• working substance = mixture of air and burned fuel
• this engine doesn’t actually recycle the same mixture in
each cycle, but we can idealize with simple models
Problem 18-44
Refrigerator
• Refrigerator:
like a heat engine, but
operating in reverse
• Reservoirs:
• hot reservoir at TH
• cold reservoir at TC
Refrigerator
• refrigerator absorbs
heat QC
• refrigerator discards
heat QH
• work W done on
refrigerator
Refrigerator
• cyclic process:
DU = 0
during one cycle
• 1st Law:
Q = W + DU
• Thus for each cycle:
Q=W
Refrigerator
• So by 1st Law:
W= Q
= QH + QC
– |W| = – |QH| + QC
|QH| = QC + |W|
Announcements
• Today:
• Wednesday:
• Thursday:
finish Chapter 18
review
no class
(office hours 12:30-2:30)
• HW 6 (Ch. 15):
returned at front
• Midterms:
returned at front
(midterm scores: classweb, solutions: E-Res)
The Second Law of
Thermodynamics
2nd Law of Thermodynamics
Four equivalent approaches, in terms of:
•
•
•
•
Engines
Refrigerators
Carnot Cycle
Entropy
Heat Engine
• engine absorbs heat QH
• engine discards heat QC
• engine does work W
• engine efficiency :
| QC |
W
e
 1
QH
QH
2nd Law:
Engines
• There are no perfect
engines (real engines
have e < 1)
• It is impossible for a
system to change heat
completely into work,
with no other change to
the system taking place
2nd Law:
Engine Version
If 2nd Law weren’t true:
• ships could move by
cooling the ocean
• cars could move by
cooling surrounding air
Refrigerator
• refrigerator absorbs
heat QC
• refrigerator discards
heat QH
• work W done on
refrigerator
2nd Law:
Refrigerator
Version
• There are no perfect
refrigerators (for real
refrigerators |W|> 0).
• It is impossible for heat
to flow from hot to cold
with no other change to
the system taking place
‘Engine’ and ‘Refrigerator’
versions of 2nd Law are equivalent
See notes
(a)
If you can build a perfect refrigerator,
then you can build a perfect engine
(perfect refrigerator) + (real engine) = perfect engine
See notes
(b)
If you can build a perfect engine,
then you can build a perfect refrigerator
(perfect engine) + (real refrigerator) = perfect refrigerator
Engine Efficiency Revisited
2nd Law:
Real engines have efficiency e < 1.
• How close can we get to e = 1?
• Is there a maximum possible efficiency?
• Yes!
Carnot Cycle
2nd Law: Carnot Cycle Version
e < eCarnot
The efficiency e of a real engine operating
between temperatures TH and TC can never
exceed that of a Carnot engine operating
between the same TH and TC:
eCarnot
Proof
TC
 1
TH
Problem 18-44, Revisited
See notes
Defining Entropy
• For Carnot cycle:
QC
TC

QH
TH
QH QC

0
TH TC
Q
T 0
• Any reversible cycle = sum of many Carnot cycles
• For cycle, in limit of infinitesimally close isotherms:
dQ
 T 0
Entropy (S)
• Result for a reversible cycle:
• This defines a new
state variable, entropy (S):
dQ
 T 0
 dS  0
dQ
dS 
T
dQ
dS 
T
(for a reversible process)
• Entropy measures the disorder of a system
• The more heat dQ you add, the higher the
entropy increase dS that results
• The smaller the temperature T, the larger the
entropy increase dS that results
Entropy (S)
• For any reversible path from state 1 to 2:
2
2
dQ
DS   dS  
T
1
1
• This must also equal DS for an irreversible
path from 1 to 2, since S is a state variable
Exercises 18-20, 18-26
Example 18-8, Problem 18-50
Example 18-8:
Free Expansion of Ideal Gas
dQ
 T 0
 dS  0
dQ
dS 
T
Entropy (S)
• For any reversible path from state 1 to 2:
2
2
dQ
DS   dS  
T
1
1
• This must also equal DS for an irreversible
path from 1 to 2, since S is a state variable
Exercises 18-20, 18-26
Example 18-8, Problem 18-50
2nd Law: Entropy Version
If we consider all systems taking part in a process,
For a reversible process:
DStotal = 0
For an irreversible process:
DStotal > 0
Laws of Thermodynamics
Law
state variable
Zeroth Law
temperature: T
1st Law
DU = Q – W
internal energy: U
2nd Law
DS  0
entropy: S
Announcements
• Today:
• Wednesday:
• Thursday:
finish Chapter 18
review
no class
(office hours 12:30-2:30)
• HW 6 (Ch. 15):
returned at front
• Midterms:
returned at front
(midterm scores: classweb, solutions: E-Res)