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Thermal Physics
Second Law of Thermodynamics
• Heat Engines
• Statements of the Second Law (Kelvin, Clausius)
• Carnot Cycle
• Efficiency of a Carnot engine
• Carnot’s theorem
• Introduction to idea of entropy
• Entropy as a function of state
• Entropy form of second law
• Example calculations
Heat Engines
•Heat engines operate in a
cycle, converting heat to work
then returning to original state
at end of cycle.
•A gun (for example) converts
heat to work but isn’t a heat
engine because it doesn’t
operate in a cycle.
•In each cycle the engine takes
in heat Q1 from a “hot
reservoir”, converts some of it
into work W, then dumps the
remaining heat (Q2) into a
“cold reservoir”
HOT
Q1
Engine
Q2
COLD
W
Efficiency of a heat engine
Definition:
efficiency  
work done per cycle
heat input per cycle
W

Q1
•Because engine returns to original
state at the end of each cycle,
U(cycle) = 0, so W = Q1 - Q2
HOT
Q1
Engine
Q2
•Thus:
Q1  Q 2
Q2

1
Q1
Q1
COLD
W
Efficiency of a heat engine
• According to the first law of thermodynamics (energy
conservation) you can (in principle) make a 100% efficient
heat engine.
BUT………….
• The second law of thermodynamics says you can’t:
Kelvin Statement of Second Law:
“No process is possible whose
SOLE RESULT is the complete
conversion of heat into work”
William Thomson, Lord Kelvin (1824-1907)
Heat flow
Both processes
opposite are perfectly
OK according to First
Law (energy
conservation)
But we know only one
of them would really
happen – Second Law
COLD
Q
WARM
COLD
COLDER
HOT
WARM
Q
HOT
HOTTER
Clausius Statement of Second Law:
“No process is possible whose SOLE RESULT is the net transfer
of heat from an object at temperature T1 to another object at
temperature T2, if T2 > T1”
Rudolf Clausius (1822-1888)
How to design a “perfect” heat engine
1) Don’t waste any work
So make sure engine operates reversibly (always equilibrium
conditions, and no friction).
2) Don’t waste any heat
So make sure no heat is used up changing the temperature
of the engine or working substance, ie ensure heat
input/output takes place isothermally
Sadi Carnot (1796-1832)
The Carnot Cycle (I): isothermal expansion
Working substance (gas)
expands isothermally at
temperature T1, absorbing
heat Q1 from hot source.
Hot
Source
T1
a
Q1
P
a
b
b
T1
V
Gas
Q1
T1
Piston
The Carnot Cycle (II): adiabatic expansion
Gas isolated from hot source,
Gas isolated from hot source,
expands adiabatically and
expands adiabatically, and
temperature falls from T to T .
temperature falls from T11 to T22
Gas
P
a
b
b
T1
c
T2
V
c
Piston
The Carnot Cycle (III): isothermal compression
3) Gas is compressed
isothermally
Gas is compressed
T2 expelling heat
at temperatureisothermally
at temperature
2 expelling heat
Q2 to coldTsink.
Q2 to cold sink.
Gas
T2
P
a
d
b
T1
c
d
T2
V
Q2
c
Piston
Q2
Cold
Sink
T2
The Carnot Cycle (IV): adiabatic compression
Gas
Gasisiscompressed
compressedadiabatically
adiabatically,,
temperature
temperaturerises
risesfrom
fromTT22to
toTT11
and
andthe
thepiston
pistonisisreturned
returnedto
toits
its
original
donedone
is
originalposition.
position.Work
The work
the
pershaded
cycle isarea.
the shaded area.
P
a
d
a
b
T1
W
d
Gas
c
T2
V
Piston
Efficiency of ideal gas Carnot engine
Q1
P
a
Q1  Q 2
Q2

1
Q1
Q1
b
T1
W
d
c
T2
V
Q2
•We can calculate the efficiency using our knowledge of the
properties of ideal gases
Isothermal expansion (ideal gas)
Q1
P
a
b
T1
Va
Q1  Wab
Vb
V
Vb
 nRT1 ln
Va
Isothermal compression (ideal gas)
P
a
b
T1
c
d
T2
V
Q2
Q 2  Wcd
Vc
 nRT 2 ln
Vd
Efficiency of ideal gas Carnot engine
Q1
P
a
b
T1
W
d
c
V
nRT 2 ln c
Vd
Q
 1 2 1
Vb
Q1
nRT1 ln
Va
T2
V
Q2
T1V b  1  T 2Vc  1 (1)
T1Va
 1
 T 2Vd
 1
Adiabatic processes
(2)
(1)  (2)
V b Vc

V a Vd
T2
 1
T1
Q1 Q 2

T1 T 2
Can you do better than a Carnot?
HOT
Q3
Q1
Carnot
W
“Super
Carnot”
Q4
Q2
COLD
c 
W
Q1
W
sc 
W
Q3
sc  c
Q 1  Q 3
A Carnot engine is reversible……..
HOT
HOT
Q1
Q1
Engine
Q2
COLD
W
Engine
Q2
COLD
…..so you can drive it backwards
W
•Drive Carnot backwards with
work output from “Super
Carnot”
HOT
Q3
Q1
Carnot
W
“Super
Carnot”
Q4
Q2
COLD
•Heat leaving hot reservoir
=Q3 – Q1, which is negative
•So, net heat enters the hot
reservoir
•Since the composite engine
is an isolated system, this
heat can only have come
from the cold reservoir
•Net result, transfer of heat
from cold body to hot body,
FORBIDDEN BY CLAUSIUS
STATEMENT OF SECOND
LAW
•Carnot’s Theorem:
HOT
Q3
Q1
Carnot
W
“Super
Carnot”
Q4
Q2
COLD
“No heat engine operating
between a hot (Th) and a
cold (Tc) reservoir can be
more efficient than a Carnot
engine operating between
reservoirs at the same
temperatures”
•It follows, by exactly the
same argument, that ALL
Carnot engines operating
between reservoirs at same
Tc, Th, are equally efficient (ie
independent of “working
substance”) – our ideal gas
result holds for all Carnot
Cycles
So……..
For all Carnot Cycles, the following results hold:
Tc
 1
Th
Qc Q h
Qc Q h



0
Tc T h
Tc T h
Conservation of
“Q/T”
What about more general cases?????
The expression
Qc Q h
Qc Q h



0
Tc T h
Tc T h
Was derived from expressions for efficiency, where only
the magnitude of the heat input/output matters. If we now
adopt the convention that heat input is positive, and heat
output is negative we have:
Qc Q h

0
Tc T h
Arbitrary reversible cycle
Arbitrary reversible cycle can be built
up from tiny Carnot cycles (CCs)
For 2 CCs shown:
Q1
T1

Q 2
T2

Q 3
T3

Q 4
T4
Q1
Q3
0
For whole reversible cycle:
Qn
T
n
Q2
0
n
Q4
In infinitesimal limit, Q→dQ, → ,fit to cycle becomes exact:

dQ
0
T
cycle
For any reversible cycle:
Entropy
To emphasise the fact that the relationship we have just
derived is true for reversible processes only, we write:

dQrev
0
T
cycle
We now introduce a new quantity, called ENTROPY (S)
dQrev
dS 
T
Entropy is conserved for a reversible cycle
Is entropy a function of state?
For whole cycle:
S 

Reversible cycle
dQrev
0
T
B
P
cycle
B
 dQrev

T
A

B

A

 dQrev



T
path 1  B


 dQrev 

T 

A
path 1
B
path1


0

path 2

 dQrev 

T 


A
path 2
S BA (path 1)  S BA (path 2)  S B  S A
A
path2
V
Entropy change is
path independent →
entropy is a
thermodynamic
function of state
Example calculation:
Calculate the entropy change of a 10g ice cube at an initial temperature of
-10°C, when it is reversibly heated to completely form liquid water at
0°C……..
Specific heat capacity of ice = 2090 J kg-1K-1
Specific latent heat of fusion for water = 3.3105 J kg-1
Irreversible processes
Carnot Engine
 Qc
c  1  
 Qh

Tc

1
Th
rev
Irreversible Engine
irrev
 Qc
 1  
 Qh

Tc

1
Th
irrev
For irreversible case:
 Qc
 
 Qh

 Qc 
Tc
Tc



 

Th
Th
irrev
 Qh irrev
 Qc 
 Qh 
  
 0


 T h irrev  Tc irrev
 Qc

 Tc

Qh


Th
irrev
Irreversible processes
Following similar argument to that for arbitrary reversible cycle:

dQirrev
0
T
For irreversible cycle
cycle
Irreversible cycle
P
Path 1
(irreversible)
B ( irrev )
B
Path 2
(reversible)

A( irrev )
B ( irrev )

A
V
A( irrev )
dQirrev

T
dQirrev

T
A( rev )

B ( rev )
B ( rev )

A( rev )
dQrev
0
T
dQrev
T
Irreversible processes
B ( irrev )

A( irrev )
dQirrev

T
B ( rev )

A( rev )
B ( irrev )
dQrev
T

A( irrev )
 B (rev )



 dS  S B  S A  S BA 
A

 (rev )


dQirrev

T
B ( rev )
 dS
A( rev )
dQirrev
dS 
T
General Case
dQ
dS 
T
Equality holds for reversible change, inequality holds for irreversible change
“Entropy statement” of Second Law
We have shown that:
dQ
dS 
T
For a thermally isolated (or completely isolated) system, dQ = 0
dS  0
“The entropy of an isolated system cannot decrease”
What is an “isolated system”
The Universe itself is the ultimate “isolated system”, so you
sometimes see the second law written:
“The entropy of the Universe cannot decrease”
(but it can, in principle, stay the same (for a reversible process))
It’s usually a sufficiently good enough approximation to
assume that a given system, together with its immediate
surroundings constitute our “isolated system” (or
universe)………
Entropy changes: a summary
For a reversible cycle:
S(system) = S(surroundings) = 0
S(universe) = S(system) + S(surroundings) = 0
For a reversible change of state (A→B):
S(system) = -S(surroundings) = not necessarily 0
S(universe) = S(system) + S(surroundings) = 0
For an irreversible cycle
S(system) = 0; S(surroundings) > 0
For a irreversible change of state (A→B):
S(system) ≠ - S(surroundings)
S(universe) = S(system) + S(surroundings) > 0
Example: entropy changes in a Carnot Cycle
Entropy calculations for irreversible processes
•Suppose we add or remove heat in an irreversible way to our system,
changing state from A to B
•At first sight we might think it’s a problem to use the relation:
B ( rev )
S BA 

A( rev )
dQrev
T
•However, because S is a function of state, S for the change of state
must be the same for all paths, reversible or not.
•So, we can “pretend” the heat required to produce the given change of
state was added or removed reversibly and use the formula anyway!
•BUT… the entropy change of the surroundings must be different for the
reversible (S(universe)=0) and irreversible (S(universe)>0) cases.
Entropy calculations for adiabatic processes
•Here, we have no “dQ” term at all, so where do we start with the
calculation………..
•To calculate the entropy change of the system we can “invent” a nonadiabatic process that takes the system between the same 2 states as
our actual, adiabatic process and use that “dQ” to do the calculation.
•Again we rely on the path independence of S
Example: Joule expansion of an ideal gas: (irreversible, adiabatic
change of state)
Rigid,
adiabatic
wall
GAS
Vi
Before
GAS
Vf
After
Joule expansion of an ideal gas
Rigid,
adiabatic
wall
GAS
Vi
Before
GAS
Vf
After
•For this process, U(gas)=0
•For ideal gas, U is a function of T only, so T = 0
•So, our “model” process is a reversible, isothermal expansion from Vi to
Vf (S(gas) = nRln(Vf/Vi), see Carnot cycle calculation)
•But, for Joule expansion, S(universe) = S(gas): gas is an isolated
system
•For reversible isothermal expansion, S(surroundings) =-nRln(Vf/Vi),
S(universe) = 0
Some past (part) exam questions….
1) A copper block of heat capacity 150J/K, at an initial
temperature of 60°C is placed in a lake at a temperature of
10°C. Calculate the entropy change of the universe as a result
of this process
2) A thermally insulated 20 resistor carries a current of 10A for
1 second. The initial temperature of the resistor is 10C, its
mass is 5x10-3kg, and its heat capacity is 850 Jkg-1K-1.
Calculate the entropy change in a) the resistor and b) the
universe.