1. Second Law of Thermodynamics
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

2. Heat Engines HOT COLD Q1 Engine W Q2
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 W Q2
COLD

3. Efficiency of a heat engine
Definition:
HOT
COLD
Engine Q1 Q2 W
Because engine returns to original state at the end of each cycle, U(cycle) = 0, so W = Q1 - Q2
Thus:

4. 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:

5. Kelvin Statement of Second Law:
“No process is possible whose SOLE RESULT is the complete conversion of heat into work”
 William Thomson, Lord Kelvin  ( )

6. Heat flow
COLD
HOT Q
Both processes opposite are perfectly OK according to First Law (energy conservation)
But we know only one of them would really happen – Second Law
WARM
COLD
HOT Q
HOTTER
COLDER

7. 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 ( )

8. How to design a “perfect” heat engine
Don’t waste any work
So make sure engine operates reversibly (always equilibrium conditions, and no friction).
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 ( )

9. The Carnot Cycle (I): isothermal expansion
Working substance (gas) expands isothermally at temperature T1, absorbing heat Q1 from hot source.
Gas
Hot
Source T 1 T1 Q 1 a
Piston b

10. The Carnot Cycle (II): adiabatic expansion
Gas isolated from hot source, expands adiabatically and temperature falls from T1 to T2.
Gas
Gas isolated from hot source, expands adiabatically, and temperature falls from T1 to T2 b c
Piston

11. The Carnot Cycle (III): isothermal compression3)
Gas is compressed
isothermally
at temperature T 2
expelling heat Q
to cold sink.
Gas is compressed isothermally at temperature T2 expelling heat Q2 to cold sink.
Gas
Cold
Sink T 2 T2 Q 2 d c
Piston

12. The Carnot Cycle (IV): adiabatic compression
Gas is compressed adiabatically, temperature rises from T2 to T1 and the piston is returned to its original position. Work done is the shaded area.
Gas is compressed adiabatically, temperature rises from T2 to T1 and the piston is returned to its original position. The work done per cycle is the shaded area.
Gas a d
Piston

13. Efficiency of ideal gas Carnot engine
We can calculate the efficiency using our knowledge of the properties of ideal gases

14. Isothermal expansion (ideal gas)

15. Isothermal compression (ideal gas)

16. Efficiency of ideal gas Carnot engine
Adiabatic processes

17. Can you do better than a Carnot?
HOT Q1 Q3
“Super
Carnot”
Carnot W W Q2 Q4
COLD

18. A Carnot engine is reversible……..
HOT
HOT Q1 Q1
Engine
Engine W W Q2 Q2
COLD
COLD
…..so you can drive it backwards

19. HOT COLD Q1 Q3 W “Super Carnot” Carnot Q2 Q4
Drive Carnot backwards with work output from “Super Carnot”
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
HOT Q1 Q3 W
“Super
Carnot”
Carnot Q2 Q4
COLD

20. HOT COLD Q1 Q3 W “Super Carnot” Carnot Q2 Q4 Carnot’s Theorem:
“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
HOT Q1 Q3 W
“Super
Carnot”
Carnot Q2 Q4
COLD

21. So…….. For all Carnot Cycles, the following results hold:
Conservation of “Q/T”
What about more general cases?????

22. The expression
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:

23. Arbitrary reversible cycle
Arbitrary reversible cycle can be built up from tiny Carnot cycles (CCs)
Q1
Q3
Q2
Q4
For 2 CCs shown:
For whole reversible cycle:
In infinitesimal limit, Q→dQ, → ,fit to cycle becomes exact:
For any reversible cycle:

24. Entropy is conserved for a reversible cycle
To emphasise the fact that the relationship we have just derived is true for reversible processes only, we write:
We now introduce a new quantity, called ENTROPY (S)
Entropy is conserved for a reversible cycle

25. Is entropy a function of state?
For whole cycle:
Reversible cycle B P
path1
path2 A V
Entropy change is path independent → entropy is a thermodynamic function of state

26. 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

27. Irreversible processes
Carnot Engine
Irreversible Engine
For irreversible case:

28. Irreversible processes
Following similar argument to that for arbitrary reversible cycle:
For irreversible cycle
Irreversible cycle B P
Path 1
(irreversible)
Path 2
(reversible) A V

29. Irreversible processes
General Case
Equality holds for reversible change, inequality holds for irreversible change

30. “Entropy statement” of Second Law
We have shown that:
For a thermally isolated (or completely isolated) system, dQ = 0
“The entropy of an isolated system cannot decrease”

31. 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)………

32. 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
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 r

33. Example: entropy changes in a Carnot Cycle

34. 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:
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.

35. 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 non-adiabatic 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)
Before
After
Rigid, adiabatic wall
GAS Vi Vf

36. Joule expansion of an ideal gas
Before
After
Rigid, adiabatic wall
GAS Vi Vf
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

37. Some past (part) exam questions….
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
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.