1. HEAT ENGINES AND REFRIGERATORS
THERMODYNAMICS
Department of Physics, Faculty of Science
Jazan University, KSA
Part-7
222 PHYS

2. 2
Example-8 Consider two engines, the details of which are given in the following
diagrams. For both engines, calculate the heat flow to the cold reservoir and the changes
in entropy of the hot reservoir, cold reservoir and engine. Which engine violates the
Second Law? What is the efficiency of the working engine?

3. 3
Solution
First Law: U = Q – W
Engine: cyclic process U = 0
 Q = W  |Qh| - |QC|  |QC | = |Qh| - W
Engine 1: |QC| = 1000 - 200 = 800 J
Engine 2: |QC| = 1000 - 300 = 700 J

4. 4
S(total) = - |Qh|/ Th + |QC|/ TC
Engine 1: S = (- 2.5 + 2.7) J.K-1 = + 0.2 J.K-1 > 0
 Second Law validated
Engine 2: S = (- 2.5 + 2.3) J.K-1 = - 0.2 J.K-1 < 0
 Second Law not validated
Engine 1 is the working engine
Efficiency (η) = (work out / energy input)  100
= (200 / 1000)(100) = 20 %

5. 5
Example-9: (not required)
A hybrid petrol-engine car has a higher efficiency than a petrol-only car
because it recovers some of the energy that would normally be lost as
heat to the surrounding environment during breaking.
(a)
If the efficiency of a typical petrol-only car engine is 20%, what efficiency
could be achieved if the amount of heat loss during breaking is halved?
(b)
Is it possible to recover all the energy lost as heat during braking and
convert it into mechanical energy? Explain your answer.

6. Solution
Identify / Setup
efficiency 1h C C
h h h
W Q Q Q
Q Q Q


   
Second Law of Thermodynamics
100% of heat can not be transformed into mechanical energy η < 1
(a) 1 0.2 0.80C C
h h
Q Q
Q Q
     
Reduce heat loss by half 0.40 0.6C
h
Q
Q
  
(b) Would require |QC| = 0, this would be a violation of the Second Law
1 1 0 0C C
C
h h
Q Q
Q
Q Q
       

7. Stirling heat engine
The gasses used inside a Stirling engine never leave the engine.
There are no exhaust valves that vent high-pressure gasses, as
in a gasoline or diesel engine, and there are no explosions taking
place. Because of this, Stirling engines are very quiet.
The Stirling cycle uses an external heat source, which could be
anything from gasoline to solar energy to the heat produced by
decaying plants. Today, Stirling engines are used in some very
specialized applications, like in submarines or auxiliary power
generators, where quiet operation is important.
It is a simple, practical heat engine using a gas as a working
substance. It’s more practical than Carnot, though its efficiency
is pretty close to the Carnot maximum efficiency. The Stirling
engine contains a fixed amount of gas which is transferred
back and forth between a "cold" and a "hot" end of a long
cylinder. The "displacer piston" moves the gas between the two
ends and the "power piston" changes the internal volume as the
gas expands and contracts.
T
4
1
2 3
V2V1
T1
T2

8. Efficiency of Stirling Engine
T
4
1
2 3
V2V1
T1
T2
1-2     00
2
3
12121212  WTTnRTTCQ V
2-3 2323
1
2
2223 0ln
2
1
2
1
WQ
V
V
nRT
V
dV
nRTPdVQ
V
V
V
V






 
3-4     00
2
3
34212134  WTTnRTTCQ V
4-1
1 1
2 2
1
41 1 1 41 41
2
ln 0
V V
V V
VdV
Q PdV nRT nRT Q W
V V
 
 
      
 
   
     
2 1 2 2 1
12 23 2
23 41 2 1 2 1 2 1 2 1 max
3
ln /
1 3 12
ln / 2ln /
h
h C
T T T V V
Q QQ T
Q Q Q Q T T V V T T V V 
 

     
   
In the Stirling heat engine, a working substance, is an ideal monatomic gas, at initial
volume V1 and temperature T1 takes in “heat” at constant volume until its temperature
is T2, and then expands isothermally until its volume is V2. It gives out “heat” at
constant volume until its temperature is again T1 and then returns to its initial state by
isothermal contraction. The efficiency compared to a Carnot engine operating
between the same two temperatures.

9. Internal Combustion Engines (Otto cycle)
Otto cycle. Working substance – a mixture of air and
vaporized gasoline. No hot reservoir – thermal energy is
produced by burning fuel.
0 – 3 intake (fuel+air is pulled into the cylinder
by the retreating piston)
3 – 4 isentropic compression
4 – 1 isochoric heating
1 – 2 isentropic expansion
2 – 3 – 0 exhaust
P
V
3
4
1
2
ignition
exhaust
V2V1
Patm
intake/exhaust
0
- Engines where the fuel is burned inside the engine cylinder as opposed to that where the
fuel is burned outside the cylinder (e.g., the Stirling engine). More economical than ideal-
gas engines because a small engine can generate a considerable power.

10. 1
2 1
1 2
1 1
V T
V T



 
    
 
1
2
V
V
- the compression ratio
The efficiency:
 = 1+2/f - the adiabatic exponent
For typical numbers V1/V2 ~8 ,  ~ 7/5  η = 0.56, (in reality, η = 0.2 – 0.3)
(even an “ideal” efficiency is smaller than the second law limit 1-T3/T1)
2V
- minimum cylinder volume
- maximum cylinder volume
1V

11. Otto cycle (cont.)
P
V
3
4
1
2
ignition
exhaust
V2V1
Patm
intake/exhaust
0
S
V
3
4
1 2
V2V1
S2
S1
1 C
h
Q
Q
  
CQHQ
0Q
0Q
 1 4h VQ C T T   3 2 0C VQ C T T  
 
 
3
1
2 32 3 2 2 2 2 2 2 2 1
41 4 1 1 4 1 1 1 1 1 2
1
1
1 1 1 1 1 1
1
P
P PT T V V P P V P T V
PT T V P P V P V P T V
P



  
             
   
VC
R
1
/
1
2
1
VR C
V
V

 
   
 

12. The Third Law:
The entropy of a perfect
crystal at 0 K is zero.
Perfect crystal means:
Everything in its place
No molecular motion
The Third Law of Thermodynamics

13. • The entropy of a perfect crystal at 0 K is zero.
• It is impossible to reach a temperature of absolute zero
• It is impossible to have a (Carnot) efficiency equal to 100% (this would
imply Tc = 0).
(a) T=0 K, S=0
(b) T>0 K, S>0

14. Entropy Curve Solid GasLiquid
S
(qrev/T)
(J/K)
Temperature (K)
0
0
 fusion
 vaporization
S° (absolute entropy)
can be calculated for
any substance
ln ln1 0S K K   For macrostate ensamples

15.  For substances in internal equilibrium, undergoing an isothermal process, the entropy
change goes to zero as T (in K) goes to zero.
0
lim 0
T
S

 
 The law is valid for pure substances and mixtures.
 Close to Zero Kelvin, the molecular motions have to be treated using quantum mechanics
→ still it is found that quantum ideal gases obey the third law.

16. S = entropy, 𝛺 = the number of ways of arranging the components of a system having
equivalent energy, and K = the Boltzmann constant = R/NA (R = universal gas
constant, NA = Avogadro’s number) = 1.38 x 10-23 J/K.
A system with relatively few equivalent ways to arrange its components (smaller 𝛺)
has relatively less disorder and low entropy.
A system with many equivalent ways to arrange its components (larger 𝛺) has
relatively more disorder and high entropy.
𝑆 = 𝐾ln𝛺

17. Statistical view of entropy
Entropy and multiplicity
lnS K 
Ω is the multiplicity associated with the configuration whose entropy we wish
to calculate.
is called Boltzmann’s entropy equation

18. For one particle
S1=kBlnΩ1
S2=kBlnΩ2
∆S=S2-S1= kBlnΩ2-kBlnΩ1=kBln(Ω2/Ω1)
∆S= kBln(2Ω1/Ω1)=kBln2= …….
Ex.

19. 1 2
1 2
2 2
1 1
2
1
2
1
1 ln( )
ln( )
V V
V V
V
V
V
For mole of gas S R
V
V
For n mole of gas S nR
V





 
 
Definition of entropy in term of probability

20. Measuring Entropy Even if we cannot calculate S,
we can still measure it:
For V=const and N=const :
T
dTTC
T
Q
T
dU
dS V )(


 


T
V
T
TdTC
STS
0
)(
)0()(
By heating a cup of water (200g, CV =840 J/K) from 200C to 1000C, we
increase its entropy by 373
293
(840 J/K) 202 J/K
dT
S
T

  

At the same time, the multiplicity of the system is increased by
25
105.1 
e
holds for all reversible (quasi-static) processes (even if V is
changed in the process).T
Q
dS


This is the “thermodynamic” definition of entropy, which Clausius introduced in
1854, long before Boltzmann gave his “statistical” definition S  kln .
but
Ex.

21. But C  0 as T  0
 


T
V
T
TdTC
STS
0
)(
)0()( 0as0)(  TTC
For example, let’s fix P :
dTTCQ P )(
)(
00
 



T
P
T
T
TdTC
T
Q
- finite,  CP(0) = 0
Similar, considering V = const ----- CV(0) = 0 .
Thus, the specific heat must be a function of T. We know that at high T, CV
approaches a universal temperature-independent limit that depends on N
and # of degrees of freedom. This high - T behavior cannot persist down to
T = 0 –---- quantum effects will come into play.

22. Example
For a mole of aluminum, CV = aT + bT3 at T < 50 K (a = 1.35x10-3 J/K2,
b = 2.48x10–5 J/K4). The linear term – due to mobile electrons, the cubic term –
due to the crystal lattice vibrations. Find S(T) and evaluate the entropy at
T = 1K, 10 K.
  3
0
3
0
3
)(
)( T
b
aT
T
TdTbTa
T
TdTC
TS
TT
V






 
3 2 5 4 3 31
(1 ) 1.35 10 J/K 1K 2.48 10 J/K 1K 1.36 10 J/K
3
S K   
       T = 1K
- at low T, nearly all the entropy comes from the mobile electrons
T = 10K 3 2 5 4 3 3 21
(10 ) 1.35 10 J/K 10K 2.48 10 J/K 10 K 2.18 10 J/K
3
S K   
       
- most of the entropy comes from lattice vibrations
3 2
20 21
23 23
(1 ) 1.35 10 J/K (10 ) 2.18 10 J/K
10 1.6 10
1.38 10 J/K 1.38 10 J/KB B
S K S K
K K
 
 
 
    
 
- much less than the # of particles, most degrees of freedom are still frozen out.

23. Example:
Imagine that one macropartition of a combined system of two Einstein solids has
an entropy of 1 J/K, while another (where the energy is more evenly divided) has
an entropy of 1.001 J/K. How many times more likely are you to find the system in
the second macropartition compared to the first?
23
2
19
23
1
/ 0.72464*10
7.2*102
/ 0.72536*10
1
Prob(mp2)
Prob(mp1)
B
B
S k
S k
e e
e
e e

   
