SECOND LAW OF THERMODYNAMICS

Noida Institute of Engineering and Technology, Greater Noida
2nd Law of thermodynamics & Entropy
Nisha Yadav
Assistant Professor
[Department of Mechanical
Engineering]
5/6/2020
1
Unit: 2
Nisha Yadav Thermodynamics[TD] Unit - 2
Subject Name
Thermodynamics
Course Details: B.Tech. (3rd Sem)

• Course Objectives
• Course Outcomes
• CO-PO Mapping
• CO-PSO Mapping
• Prerequisite and recap
• Limitations of first law of thermodynamics
• 2nd law of thermodynamics
• Direction of change
• Heat Engines
• Heat Pumps & Refrigerators
• Carnot Cycle
• 2nd law statements
• Equivalence of both statements
• Entropy of ideal gas
• The Carnot Principle
• P-V diagram
2
Content
Nisha Yadav Thermodynamics[TD] Unit : 25/6/2020

• To learn about work and heat interactions, and balance of energy
between system and its surroundings.
• To learn about application of First law to various energy conversion
devices.
• To evaluate the changes in properties of substances in various
processes.
• To understand the difference between high grade and low-grade
energies and second law limitations on energy conversion.
• To get conversant with pure substance , psychrometric charts , human
comfort conditions , vapour processes and thermodynamics vapour
cycles , Refrigeration cycles and performance estimation.
3
Course Objective
Nisha Yadav Thermodynamics[TD] Unit : 2
5/6/2020

• CO1- Identify and use units and notations in Thermodynamics, State
and illustrate Zeroth and First law of Thermodynamics.
• CO2- To understand the concepts of Second law of Thermodynamics
and Entropy, Apply the first and second laws to various gas process
and cycle.
• CO3- Understand the role of Thermodynamic cycles, Availability and
Irreversibility, second law of efficiency and different thermodynamic
relations.
• CO4- Ability to recognize and understand the different forms of pure
substances and to get conversant with properties of steam, dryness
fraction measurement. Also get conversant with Psychrometric Charts,
Psychrometric processes and human comfort conditions.
• CO5- Study and analyze different Refrigeration Cycles. Understand
the effect of Supercooling and sub-cooling and change in evaporator
and condenser pressure on performance of Refrigeration Cycle. 4
Course Outcome

CO PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12
KME301.1 3 3 2 2 1 2 2 1 1 1 1 2
KME301.2 3 3 2 3 1 2 2 1 1 1 1 2
KME301.3 3 3 2 3 1 2 2 1 1 1 1 2
KME301.4 3 3 3 2 1 2 2 1 3 1 1 2
KME301.5 3 3 2 2 1 2 2 1 2 1 1 2
5
CO-PO Mapping
5/6/2020

6
CO-PSO Mapping
CO PSO1 PSO2 PSO3 PSO4
KME301.1 3 2 3 2
KME301.2 3 3 3 3
KME301.3 3 3 3 3
KME301.4 3 2 3 3
KME301.5 3 2 1 2
5/6/2020

• Student must aware with the basics of heat , temperature and
work etc.
• Student should have basic mathematical equations.
7
Prerequisite and Recap
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8
Why 2nd Law of thermodynamics?
• The 1st law of thermodynamics states that a certain energy flow
takes place when a system under goes a process or change of
state.
• But , it does not give any information on whether that process or
change of state is possible or not.
• This change of state is possible or not is find out with the help of
2nd law of thermodynamics.
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9
According to 1st Law of thermodynamics?
• Work is completely converted in to heat or heat is completely
converted into work.
(δW = δQ and δQ=δW)
• Potential energy can be transformed into kinetic energy or kinetic
energy can be transformed into Potential energy.
(PE→KE and KE→PE)
• Heat flows from hot to cold or from cold to hot.
(Th→ Tl and Tl → Th)
• Gas expands from high pressure to low pressure or from low
pressure to high pressure
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10
Limitation of 1st Law of termodynamics
• Heat is not completely converted into work.(Q ˃ W)
• kinetic energy can not be transformed into Potential energy.
(KE ≠> PE)
• Heat flow from cold to hot is not possible.
(Tl ≠> Th)
• Gas expands from low pressure to high pressure is not
possible.
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11
Limitation of 1st Law of termodynamics
• 1st law does not help to predict whether the certain process
is possible or not.
• A process can be proceed in particular direction only , but 1st
law does not gives information about direction.
• 1st law not provides sufficient condition for certain process to
take place.
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12
2nd Law of Thermodynamics
Heat can not flow itself from cold body to hot
body.
• It is a spontaneous process.
• The 2nd law of thermodynamics is also used to determine the
theoretical limits for the performance of mostly used in
engineering systems like heat engines and refrigerators.
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13
Direction of Change
• The second law of thermodynamics asserts that processes
occur in a certain direction and that the energy has quality as
well as quantity.
Hot
Container
Possible
Impossible
Cold
Surroundings
• In above example, two bodies are different temperature are
brought into contact, heat energy flows from the body at
high temperature to that at low temperature. Heat energy
never flow from lower temperature level to higher
temperature without applying external work.
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14
Heat Engines
• A heat engine cycle is a thermodynamic cycle in which there is
a net heat transfer to the system and a net work transfer from
the system.
• The system which executes a heat engine cycle is called a heat
engine.
• The function of the eat engine cycle is to produce the work
continuously at the expense of heat input to the system.
• So the efficiency of the heat engine and heat engine cycle is
defined as follows:
 𝑡ℎ
=
Net work output of the cycle
Total heat input to te cycle
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15
Heat Engines
The thermal efficiency is always less than 1 or less than
100percent.
 𝑡ℎ
=
𝑊𝑛𝑒𝑡 ,𝑜𝑢𝑡
𝑄𝑖𝑛
where
Wnet ,out = Wout − Win
Qin ≠ Qout
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16
Heat Engines
Applying the first law to the cyclic heat engine
Qnet , in Wnet , out  U
Wnet, out Qnet , in
Wnet, out  Qout
The cycle thermal efficiency may be written as

th 
Wnet , out
Q

Qin  Qout
Qin
 1
Qout
in
Qin
5/8/2020

17
Heat Engines
A thermodynamic temperature scale related to the heat transfers
between a reversible device and the high and low-temperature
reservoirsby
QL
QH
=
TL
TH
The heat engine that operates on the reversible Carnot cycle is
called the Carnot Heat Engine in which its efficiencyis
1
𝑇 𝐻
𝑇 𝐿
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18
Heat Engines
The work-producing device that
best fit into the definition of a
heat engine is the steam power
plant, which is an external
combustion engine.
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19
Heat Pumps and Refrigerators
• A device that transfers heat from a low temperature medium to a high
temperature one is the heat pump.
• Refrigerator operates exactly like heat pump except that the desired
output is the amount of heat removed out of the system
• The index of performance of a heat pumps or refrigerators are
expressed in terms of the coefficient of performance.
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20
COPR =
TL
TH − TL
=
1
TH
TL
−1
COPHP =
TH
TH − TL
=
1
TL
TH
−1
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21
COPHP =
QH
QH − QL
=
QH
W net,in
COPR =
QL
QH − QL
=
QL
W net,in
5/8/2020

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Carnot Cycle

23
Carnot Cycle
Execution of Carnot cycle in a piston cylinder device

24
Carnot Cycle
Process Description
a-b (1-2)( Isothermal Expansion) Reversible isothermal heat addition at
high temperature
b-c (2-3)( Adiabatic Expansion) Reversible adiabatic expansion from
high temperature to low temperature
c-d(3-4)(Isothermal Compression) Reversible isothermal heat rejection at
low temperature
d-a(4-1)( Adiabatic Compression) Reversible adiabatic compression from
low temperature to high temperature
5/8/2020

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Carnot Cycle

26
Carnot Cycle

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
th

Wnet , out
in
Q
 Qin QoutWnet , out
 Qin Qout
Q
in

th = 1 -
Q
out
in
Q
Qout
Q
in
= m Cp L
T
= m Cp TH
Carnot Cycle

28
Carnot Cycle
The thermal efficiencies of actual and reversible heat engines
operating between the same temperature limits compare as follows:
The coefficients of performance of actual and reversible refrigerators
operating between the same temperature limits compare as follows
5/8/2020

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Carnot Cycle

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Carnot Cycle

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Carnot Cycle

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Carnot Cycle

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Carnot Cycle

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Carnot Cycle

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Carnot Cycle

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Carnot Cycle

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Class Test (Numericals)
Q.1 A Carnot engine operates between two reservoirs whose difference in
temperature is 200◦C. If the work output of the engine is 0.5 times the eat rejected ,
make calculations for the temperature of source and sink and the thermal efficiency of
the engine.
Q.2 An engine operation on a Carnot cycle works with in temperature limits of 600
K and 300 K . If the engine receives 2000 KJ of heat , evaluate the work done and
thermal efficiency of the engine.
Q.3 A perfect as flows through a nozzle where it expands in a reversible adiabatic
manner. The inlet conditions are 22 bar, 500 ◦C , 38 m/s. At exit the pressure is 2
bar. Determine the exit velocity and exit area if the flow rate is 4 Kg/s. Take R
190 J/Kg K . And ϒ = 1.35.

38
Carnot Cycle
Example1:
A steam power plant
produces 50 MW of net
work while burning fuel
to produce 150 MW of
heat energy at the high
temperature. Determine
the cycle thermal
efficiency and the heat
rejected by the cycle to
the surroundings.
Solution:

th 
Wnet , out
QH

50 MW
150 MW
 0.333 or 33.3%
Wnet , out  QH QL
QL  QH  Wnet, out
 150 MW  50 MW
 100 MW
5/8/2020 Nisha Yadav Thermodynamics[TD] Unit : 2

39
KELVIN PLANK STATEMENT
• It is impossible to construct a heat engine that operates on cycle
to receive heat from single reservoir and produce equivalent
amount of work.
• It implies that it is impossible to build a heat engine that has
100%thermal efficiency.
TH
QH
Heat engine
Wnet
Impossible

40
CLAUSIUS STATEMENT
• It is impossible to construct a device as heat pump that
operates in a cycle and produces no effect other then the
transfer of heat from a lower temp body to higher temp body.
• Heat can not itself flow from colder body to hater body.
Heat pump
QL
TH
TL
Impossible
5/9/2020

41
EQUIVALENCE OF THE TWO
STATEMENTS
• It can be shown that the violation of one statement leads to a
violation of the other statement, i.e. they are equivalent.
1. Violation of Kelvin-plank statement leading to violation of
Clausius statement.
2. Violation of Clausius statement leading to violation of
Kelvin-plank statement .
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42
EQUIVALENCE OF THE TWO
STATEMENTS
Wnet =QH
TL
QL
QH+QL
TH
QH
Heat engine Heat pump
Heat pump
TH
QH
QL
TL
Heat transfer from low-temp body to high-
temp body without work; A violation of the
Clausius statement
A 100% efficient heat engine; violates K-P
Statement
5/9/2020

43
VIOLATION OF KELVIN-PLANK STATEMENT
LEADING TO VIOLATION OF CLAUSIUS
STATEMENT

44
VIOLATION OF KELVIN-PLANK
STATEMENT LEADING TO VIOLATION
OF CLAUSIUS STATEMENT

45
VIOLATION OF CLAUSIUS STATEMENT
LEADING TO VIOLATION OF KELVIN-PLANK
STATEMENT

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PMM II

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Carnot Theorem and its Corollaries

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Carnot Theorem

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Efficiency of Reversible Engine

12/19/2020 50Nisha Yadav Thermodynamics[TD] Unit : 2
Numericals on Carnot Theorem

12/19/2020 54
Thermodynamic Temperature Scale
“A temperature scale which is independent of the property of
thermometric substance is defined as thermodynamic
temperature scale”.

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Thermodynamic Temperature Scale

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Clausius Inequality

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Clausius Inequality

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Faculty Name Subject code and
abbreviation Unit Number
59
Clausius Inequality

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Clausius Inequality

Clausius Inequality

Numerical on Clausius Inequality

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Solution of Numerical on Clausius Inequality

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68
ENTROPY
• Entropy is a thermodynamic function of the state of the system can
be interpreted as the amount of order or disorder of a system
• Change in Entropy:-
The change in entropy  S of a system when an amount of thermal
energy Q is added to a system by a reversible process at constant
absolute temperature T is given by:
 S = dQ / T
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69
Example
A heat engine removes 100 J each cycle from a heat reservoir at
400 K and exhausts 85 J of thermal energy to a reservoir at 300 K.
Compute the change in entropy for each reservoir.
• Since the hot reservoir loses heat, we have that:
 S = Q / T = -100 J / 400 K = -0.25 JK-1
For the cold reservoir we have:
 S = Q / T = 85 J / 300 K = 0.283 JK-1
Therefore:
The increase in entropy of the cold reservoir is greater than the
decrease for the hot reservoir.
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70
ENTROPY OF IDEAL GAS
• An expression for the entropy change of an ideal gas can be
obtained from Eq.
ds =
𝑑𝑢 𝑃 𝑑𝑣
𝑇 + ----- Eq. 1
𝑇
Or
ds = 𝑑ℎ 𝑣 𝑑𝑃
𝑇 𝑇
- ----- Eq. 2
• By employing the property relations forideal gases. By
substituting du=𝑪 𝒗dTand P =
𝑹𝑻
𝑽
into
Eq. 1, the differential entropy change of an ideal gas becomes
ds = 𝐶𝑣
𝑑𝑇 𝑑𝑣
+ R
𝑇 𝑣
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71
The entropy change for a process is obtained by integrating this
relation between the endstates.
∆𝑆 = 𝐶𝑣
𝑑𝑇
𝑇
+ 𝑅 ln
𝑣2
𝑣1
…………Eq.3
 A second relation for the entropy change of an ideal gas is
obtained in a similar manner by substitutingdh = 𝐶 𝑝dT and V =
𝑅𝑇
𝑇
in Eq. 2 and integrating. The result is
∆𝑆= 𝐶 𝑝
𝑑𝑇
𝑇
+ 𝑅 ln
𝑝1
𝑝2
………….Eq.4
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72
From Eq.(3) and (4), we see that for constant volume process,
∆𝑆 = 𝐶𝑣
𝑑𝑇
𝑇
= 𝐶𝑣 ln
𝑇2
𝑇1
For constant pressure process, we have
∆𝑆 = 𝐶 𝑝
𝑑𝑇
𝑇
= 𝐶 𝑝 ln
𝑇2
𝑇1
For isothermal process,
∆𝑆 = 𝑅 ln
𝑣2
𝑣1
= 𝑅 ln
𝑝2
𝑝1
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73
THE CARNOT PRINCIPLE
The Carnot cycle is a thermodynamic process that describes how a
fluid is used to convert thermal energy into work.
• Characteristics:
• High Efficiency
• Multi-Source Engine
• Better reliability and easier maintenance
• Reversible
• Safe, discrete and oxygen-free
• Modularity and flexibility
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74
P-V DIAGRAM
• A pressure volume diagram is used to describe corresponding
changes in volume and pressure in a system.
• The PV diagram, called an indicator diagram, was developed
by James Watt and his employee John Southern (1758–1815)
to improve the efficiency of engines.
isothermal segments (AB and CD) occur when there is perfect
thermal contact between the working fluid and one of the
reservoirs, so that whatever heat is needed to maintain constant
temperature will flow into or out of the working fluid, from or
to the reservoir.
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75
P-V DIAGRAM
This is the PV diagram…..
5/9/2020

76
P-V DIAGRAM
 adiabatic segments (BC and DA)
occur when there is perfect thermal
insulation between the working
fluid and the rest of the universe,
including both reservoirs, thereby
preventing the flow of any heat into
or out of the working fluid.
5/9/2020

Youtube/other Video Links
https://www.youtube.com/watch?v=77EIce2Nu8Y
https://www.youtube.com/watch?v=zD2XDeBu8sI
https://www.youtube.com/watch?v=0Cn--CqXals
https://www.youtube.com/watch?v=tJAH_cgXc3Q
https://www.youtube.com/watch?v=KOEJ4UvlnAc
https://www.youtube.com/watch?v=XyQ9P4eKn6E
https://www.youtube.com/watch?v=ePTc1mtBmcM
77
Faculty Video Links, Youtube & NPTEL Video Links and Online
Courses Details

78
Youtube & NPTEL Video Links and Online Courses Details
NPTEL/Swayam online course: Fundamental of thermodynamics
https://www.youtube.com/watch?v=tyr6vNII3Bk
https://www.youtube.com/watch?v=uEErWdtkRUM
https://www.youtube.com/watch?v=kNNwrp7tHPU
https://www.youtube.com/watch?v=9GMBpZZtjXM&list=PLseVjnVx50AN97Kav4
UYk3JCuyRQDkL6T

79
Daily Quiz
Q.1.Statement of II law of thermodynamics and equivalence of both the statement.
Q.2.Explain throttling process.
Q.3.What do you mean by “Perpetual motion machine of second kind-PMM 2”?
Q.4.Define principle of entropy increase.
Q.5.What are the limitation of first law of thermodynamics?
Q.6.Explain Carnot engine and Carnot theorem.
Q.7.Statement of third law of thermodynamics.
Q.8.Explain Clausius Inequality.
Q.9. What is the working of Heat engine?
Q.10. What is the difference between Heat pump and Refrigerator?

80
Weekly Assignment
Q.1. A Carnot refrigerator exhaust 250 KJ of heat from a reservoir at 150K and
rejects heat to a reservoir at 450 K this heat serves as the energy input to a
second Carnot refrigerator which separates between 450K and 1500 K.
Determine the COP of
(i) The cold refrigerator
(ii) The Hot refrigerator
(iii) The composite system
Check the answer of composite system with the relation established above.
Q.2. Find the entropy change of 5 Kg of a perfect gas whose temperature
varies from 150°C to 200°C. During the constant volume process the specific
heat varies linearly with absolute temperature and is presented by the relation
(V = 10.45+ 0.009T ) KJ/Kg K.
Q.3. Three Carnot Engines E1, E2, E3 operate between temperatures of 1000K
and 300K. Make calculations for the intermediate temperatures if the works
produced by the engines are in the ratio of 4:3:2.
Q.4. What are the statement of II law of thermodynamics and equivalence of
both the statement?

81
Weekly Assignment
Q.5.Calculate the change in entropy at 368.3 g of ice at 32.0oF32.0oF melts.
The latent heat of fusion of water is 333000 J/kg
Q.6. What is the relationship between Entropy and The Second Law of
Thermodynamics?
N2(g) + 3H2(g) 2NH3(g)
So
298 (J/mol
K)
191.5 130.6 192.3
Q.7.What is the standard entropy change of the reaction below at 298 K with
each compound at the standard pressure?
Q.8.A heat pump is used to meet the heating requirements of a house and
maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C,
the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump
under these conditions has a COP of 2.5, determine the rate at which heat is
absorbed from the cold

82
Q.9.During a cyclic process, a heat engine absorbs 500 J of heat from a hot
reservoir, does work and ejects an amount of heat 300 J into the
surroundings (cold reservoir). Calculate the efficiency of the heat engine?
Q.11.There are two Carnot engines A and B operating in two different
temperature regions. For Engine A the temperatures of the two reservoirs are
150°C and 100°C. For engine B the temperatures of the reservoirs are 350°C
and 300°C. Which engine has lesser efficiency?
Q.10.A cyclic machine, as shown below, receives 325 kJ from a 1000 K
energy reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle
produces 200kJ of work as output. Is this cycle reversible, irreversible, or
impossible?
Weekly Assignment

83
MCQ s
1. The entropy of an isolated system can never ____
a) increase
b) decrease
c) be zero
d) none of the mentioned
2. This set of Thermodynamics Multiple Choice Questions & Answers
(MCQs) focuses on “Entropy Principle and its Applications”.
Which of the following is true?
a) for an isolated system, dS>=0
b) for a reversible process, dS=0
c) for an irreversible process, dS>0
d) all of the mentioned
3. According to entropy principle, the entropy of an isolated system can never
decrease and remains constant only when the process is reversible.
a) true
b) false

84
MCQ s
4. Entropy may decrease locally at some region within the isolated system.
How can this statement be justified?
a) this cannot be possible
b) this is possible because entropy of an isolated system can decrease.
c) it must be compensated by a greater increase of entropy somewhere
within the system.
5. Clausius summarized the first and second laws of thermodynamics as
a) the energy of the world is constant
b) the entropy of the world tends towards a maximum
c) both of the mentioned
6. The entropy of an isolated system always ____ and becomes a ____ at the
state of equilibrium.
a) decreases, minimum
b) increases, maximum
c) increases, minimum
d) decreases, maximum
5/11/2020

85
MCQ s
7. Entropy principle is the quantitative statement of the second law of
thermodynamics.
a) true
b) false
8. Which of the following can be considered as an application of entropy
principle?
a) transfer of heat through a finite temperature difference
b) mixing of two fluids
c) maximum temperature obtainable from two finite bodies
d) all of the mentioned
9. The final temperatures of two bodies, initially at T1 and T2 can range from
a) (T1-T2)/2 to sqrt(T1*T2)
b) (T1+T2)/2 to sqrt(T1*T2)
c) (T1+T2)/2 to (T1*T2)
d) (T1-T2)/2 to (T1*T2)
5/11/2020

86
MCQ s
10. Which of the following processes exhibit external mechanical
irreversibility?
a) isothermal dissipation of work
b) adiabatic dissipation of work
c) both of the mentioned

87
Old Question Papers
5/11/2020

Q.1.A steam engine boiler is maintained at 250°C and water is converted into steam.
This steam is used to do work and heat is ejected to the surrounding air at
temperature 300K. Calculate the maximum efficiency it can have?
Q.2.Three Carnot Engines E1, E2, E3 operate between temperatures of 1000K and
300K. Make calculations for the intermediate temperatures if the works produced by
the engines are in the ratio of 4:3:2.
Q.3.Find the entropy change of 5 Kg of a perfect gas whose temperature varies from
150°C to 200°C. During the constant volume process the specific heat varies linearly
with absolute temperature and is presented by the relation (V = 10.45+ 0.009T )
KJ/Kg K.
Q.4.A Carnot refrigerator exhaust 250 KJ of heat from a reservoir at 150K and
rejects heat to a reservoir at 450 K this heat serves as the energy input to a second
Carnot refrigerator which separates between 450K and 1500 K. Determine the COP
of
(i) The cold refrigerator
(ii) The Hot refrigerator
(iii) The composite system
88
Expected Questions for University Exam

89
Q.5.There are two Carnot engines A and B operating in two different
temperature regions. For Engine A the temperatures of the two reservoirs are
150°C and 100°C. For engine B the temperatures of the reservoirs are 350°C
and 300°C. Which engine has lesser efficiency?
Q.6.During a cyclic process, a heat engine absorbs 500 J of heat from a hot
reservoir, does work and ejects an amount of heat 300 J into the surroundings
(cold reservoir). Calculate the efficiency of the heat engine.
maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the
house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under
these conditions has a COP of 2.5, determine the power consumed by the heat
pump.
Q.8.A car engine operates with a thermal efficiency of 35%. Assume the air-
conditioner has a coefficient of performance of 3 working as a refrigerator
cooling the inside using engine shaft work to drive it. How much fuel energy
should be spend extra to remove 1 kJ from the inside?

90
Q.9.A cyclic machine, as shown below, receives 325 kJ from a 1000 K
energy reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle
produces 200kJ of work as output. Is this cycle reversible, irreversible, or
impossible?
maintain it at 20°C. On a day when the outdoor air temperature drops to
2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat
pump under these conditions has a COP of 2.5, determine the rate at
which heat is absorbed from the cold

SECOND LAW OF THERMODYNAMICS

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