Download Module - 1: Thermodynamics

Module - 1: Thermodynamics
Thermodynamics:
Thermodynamics (Greek: thermos = heat and dynamic = change) is the study of the conversion of energy
between heat and other forms, mechanical in particular. All those problems that are related to the interconversion of heat energy and work done are studied in thermodynamics. In thermodynamics, we discuss
different cycles such as Carnot cycle, Rankine cycle, Otto cycle, diesel cycle, refrigerator, compressors,
turbines and air conditioners.
Thermal equilibrium and Temperature:
The central concept of thermodynamics is temperature. Temperature is familiar to us all as the measure of
the hotness or coldness of objects. We shall learn afterwards that temperature is a measure of the average
internal molecular kinetic energy of an object.
It is observed that a higher temperature object which is in contact with a lower temperature object will
transfer heat to the lower temperature object. The objects will approach the same temperature, and in the
absence of loss to other objects, they will then maintain a constant temperature. They are then said to be in
thermal equilibrium. Thermal equilibrium is the subject of the Zeroth Law of Thermodynamics.
The Zeroth Law of Thermodynamics:
The "zeroth law" states that if two systems are in thermal equilibrium with a third system, then they must be
in thermal equilibrium with each other.
This law states that if object A is in thermal equilibrium with object B, and
object B is in thermal equilibrium with object C, then object C is also in
thermal equilibrium with object A.
The message of the zeroth law is: “Every body has a property called
temperature”. Two objects are defined to have the same temperature if they
are in thermal equilibrium with each other. This law allows us to build
thermometers. For example the length of a mercury column (object B) may be
used as a measure to compare the temperatures of the two other objects.
The zeroth law, which has been called a logical afterthought, came to light
only in 1930s, long after the first and second laws of thermodynamics had
been discovered and numbered. Because the concept of temperature is
fundamental to those two laws, the law that establishes temperature as a
valid concept should have the lowest number-hence the zero.
The zeroth law has a fairly straightforward statistical interpretation and this will allow us to begin to establish
the equivalence between the statistical definitions and the conventional thermodynamic ones.
Temperature and Heat:
If you take a can of cola from the refrigerator and leave it on the kitchen table, its temperature will riserapidly at first but then more slowly – until the temperature of the cola equals that of the room(the two are
then in thermal equilibrium).
In generalizing this situation, we describe the cola or coffee as a system (with temperature TS) and the
relevant part of the kitchen as the environment (with temperature TE) of that system. Our observation is that if
TS is not equal to TE, then TS will change until the two temperatures are equal and thus thermal equilibrium is
reached.
Such a change in temperature is due to the transfer of energy between the thermal energy of the system and
the system‟s environment. It may be mentioned that thermal energy is an internal energy that consists of the
kinetic and potential energies associated with the random motions of the atoms, molecules and other
microscopic bodies within an object. The transferred energy is called heat and is symbolized Q. Heat is
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positive when energy is transferred to a system‟s thermal energy from its environment (we say that heat is
absorbed). Heat is negative when energy is transferred from a system‟s thermal energy to to its environment
( we say that heat is released or lost).
We are then led to this definition of heat:
“Heat is the energy that is transferred between a system and its environment because of a temperature
difference that exists between them.”
Recall that energy can also transferred between a system and its environment as work W via a force acting
on a system. Heat and work, unlike temperature, pressure, and volume, are not intrinsic properties of a
system. They have meaning only as they describe the transfer of energy into or out of a system.
Let us now look into the Molecular Theory of Matter for an explanation of heat and temperature. Molecular
Theory of Matter states that matter is made up of tiny particles called molecules. These particles are in
constant motion within the bounds of the material. Since the relationship between kinetic energy of an object
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and its velocity is: KE = ½ mv , which means that the more energy an object has, the faster it is traveling (or
vice versa).
Thus, when you provide extra energy to an object, you cause its molecules to speed up. Those molecules, in
turn, can cause other molecules to speed up. The sum effect of the speed or energy of these molecules is
the phenomenon we call heat. Molecules can go into high-energy motion, causing heat, from various energy
sources such as Light, Chemical reactions, Electrical resistance, Friction and nuclear reactions.
Heat 
n
 (K .E )i
i 1
Heat is defined as "The total kinetic energy of all the molecules of a body" and temperature is a measure
of “the average internal molecular kinetic energy of an object”.
Units of Heat:
Before scientists realized that heat is transferred energy, heat was measured in terms of its ability to raise
the temperature of water. Thus, the calorie (cal) was defined as the amount of heat that would raise the
temperature of 1 g of water from 14.50C to 15.50C.
In 1948, the scientific community decided that since heat (like work) is transferred energy, the SI unit for heat
should be the one we use for energy, namely, the joule. The calorie is now defined to be 4.1860 J (exactly).
The “calorie” used in nutrition is really is kilocalorie.
1.
2.
3.
Calorie (cal): It is the amount of heat required to increase the temperature of 1 g of water from
14.5oC to 15.5oC. (1 cal=4.186 J)
Kilocalorie (kcal): It is the amount of heat required to raise the temperature of 1 kg of pure
water through 1oC.
British Thermal Units (BTU): It is defined as the quantity of heat required to raise the
o
temperature of 1 pound (lb) of pure water through 1 F. It is also referred to as pound-degree
Fahrenheit unit.
Conversion:
1 B.T.U = 251.996 Cal ( or 252 Cal)
1 calorie = 4.186 Joule ( or 4.2 joule)
Flow of Heat: Natural flow of heat always takes place from a region of high temperature to a region of
lower temperature.
Scales of Temperature:
There are three main scales of temperature:
2
(1) Centigrade or Celsius Scale
(2) Kelvin Scale
(3) Fahrenheit Scale
C F  32 K  273
Conversion:


5
9
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Heat Capacity (The Storage of Thermal Energy):
The heat capacity C, of a substance is defined as “the amount of heat required to increase the temperature
of that substance by one Celsius degree”.
For such a system C  Q
T

o
The unit is cal/ C or J/ oC.
Specific Heat:
The specific heat of a material is the “amount of heat energy required to raise the temperature of unit
mass of a substance by one Kelvin or 1 °C is known as specific heat of the substance."
Mathematical Expression:
If Q is the amount of heat required to raise the temperature of "m" kg of the substance through „T Kelvin,
Then
Q
m, -------------------------------(i)
 Q   T , ------------------------------(ii)
or
Q=mST
where „S‟ is the constant called specific heat of substance.
Q
C
Or,
S

mT m
Unit of Specific Heat: Unit of specific heat is J/kg K in S.I. system.
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C which is higher than any other common
substance. As a result, water plays a very important role in temperature regulation. The specific heat per
gram for water is much higher than that for a metal, hence it takes a lot of heat to change the temperature of
water.
Water Equivalent:
The water equivalent of a body is defined as the amount of water, which will absorb same quantity of heat as
the substance for the same rise of temperature. The symbol representing water equivalent is „W‟ and unit
gm.
Mechanical Equivalent of Heat:
The mechanical equivalent of heat (J) is defined as the amount of work that is necessary to produce one unit
(calorie) of heat.
J = 4.186
joules/cal
7
J = 4.186 x 10 ergs/cal
J = 778
ft-lbs/ B.Th.U
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Latent Heat:
Latent heat describes the amount of energy in the form of heat that is required for a material to undergo a
change of phase.
A solid consists of molecules that are tightly bound to each other by forces acting between them. Energy
must be supplied in the form of heat to overcome these forces. As the bonding forces weaken, groups of
molecules break free, and as more heat is supplied the solid changes into a liquid, in which small groups of
molecules constantly break apart, and reform and groups slide freely past each other. While this is
happening, the temperature of the substance remains unchanged. All of the heat energy is used in loosening
the bonds between molecules.
If still more energy is applied, it has the effect of making the groups of molecules move faster. When they
move faster they strike harder against any object with which they come into contact. It is the speed of motion
of molecules that a thermometer measures as temperature. Once a solid has melted to become a liquid, the
application of additional heat raises the temperature of the liquid.
When a liquid is cooled, its molecules lose energy and move more slowly, and the temperature of the liquid
falls. When their energy falls to a certain level the molecules start bonding together. This requires less
energy than moving about, and so heat energy is released as the liquid solidifies. The temperature of the
substance remains unchanged, but the surrounding medium is warmed by the release of energy.
For a phase change, the heat liberated or absorbed is given by
Q  ml                  (1)
where l is the latent heat of fusion (latent heat of melting) or latent heat of vaporization.
How much heat does it take to
80 cal / gram absorbed
540 cal / gram absorbed
get water to change state? If
the water is at a temperature of
100 degrees C (that is, the
Gas
boiling point, or 212 degrees F) it
Solid
Water
vapor
Liquid
takes an additional 540 calories
Ice
Water
of heat to convert one gram of
water from the liquid state to the
vapor state. When the vapor
80 cal / gram released
540 cal / gram released
converts to the liquid state, 540
calories of energy will be
released per gram of water. If you are converting solid water (ice) to liquid water at 0 degrees C, it will
require about 80 calories of heat to melt one gram of ice, and the 80 calories will be released when the liquid
water is frozen to the solid state.
Latent Heat of Fusion:
Fusion is the change of state from solid to liquid. In the process of fusion, the molecule absorbs energy. This
energy is latent heat.
When a solid substance changes from the solid phase to the liquid phase, energy must be supplied in order
to overcome the molecular attractions between the constituent particles of the solid. This energy must be
supplied externally, normally as heat, and does not bring about a change in temperature.
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The units of heat of fusion are usually expressed as joules per mole (the SI units) or calories per gram or Btu
per pound-mole
l fusion  80 cal/gm or 334 kJ/kg
The specific latent heat of fusion is defined as
"The specific latent heat of fusion of a substance is the
amount of heat required to convert unit mass of the
solid into the liquid without a change in temperature."
The specific latent heat of fusion of ice at 0 ºC, for example, is 334 kJ.kg-1. This means that to convert 1 kg of
ice at 0 ºC to 1 kg of water at 0 ºC, 334 kJ of heat must be absorbed by the ice. Conversely, when 1 kg of
water at 0 ºC freezes to give 1 kg of ice at 0 ºC, 334 kJ of heat will be released to the surroundings.
Latent Heat of Vaporization:
Evaporation is the change of state from liquid to vapor. In the process of evaporation, the molecule absorbs
energy. This energy is latent heat.
How did you make the water evaporate? Probably you added heat. You might have set it out in the sun, or
possibly put it over a fire. To make water evaporate, you put energy into it. The individual molecules in the
water absorb that energy, and get so energetic that they break the hydrogen bonds connecting them to other
water molecules. They become molecules of water vapor.
lvaporization  540 cal/gm or 2260kJ/kg
The definition of the specific latent heat of vaporization is
'The specific latent heat of vaporization is the amount of
heat required to convert unit mass of a liquid into the
vapour without a change in temperature."
For water at its normal boiling point of 100 ºC, the latent
specific latent heat of vaporization is 2260 kJ.kg-1. This means that to convert 1 kg of water at 100 ºC to 1 kg
of steam at 100 ºC, 2260 kJ of heat must be absorbed by the water. Conversely, when 1 kg of steam at 100
ºC condenses to give 1 kg of water at 100 ºC, 2260 kJ of heat will be released to the surroundings.
Heating / Cooling Curves:
The diagram on the left shows the uptake of heat by 1 kg of
water, as it passes from ice at -50 ºC to steam at temperatures
above 100 ºC, affects the temperature of the sample.
A: Rise in temperature as ice absorbs heat.
B: Absorption of latent heat of fusion.
C: Rise in temperature as liquid water absorbs heat.
D: Water boils and absorbs latent heat of vaporization.
E: Steam absorbs heat and thus increases its temperature.
The above is an example of a heating curve. One could reverse the process, and obtain a cooling curve. The
flat portions of such curves indicate the phase changes.
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First Law of Thermodynamics:
“The energy can be neither created nor destroyed; it can only change forms”. This principle is based on
experimental observations and is known as the first law of thermodynamics or the conservation of energy
principle.
Mathematical Representation: Let a system absorbs Q amount of heat energy. Addition of heat
energy increases the internal energy of system from U1 to U2 and some useful work is also performed by the
system.
Increase in internal energy is given by:
U = U1 – U2 and work done is W
According to the first law of thermodynamics:
Q = U + W ------------------------------ (1)
(+)
Sign Convention:
Q = positive if heat is added to a system
Q = negative if heat is released from a system
W = positive if work is done by the system
W = negative if work is done on the system
(-)
SYSTEM
(-)
(+)
For a gas, the internal energy, U, is directly proportional to its temperature measured in Kelvin. This means
that U only reflects a change in the kinetic energy of the gas molecules. Remember that the potential
energy can not change except when there is a phase change: liquid to solid or liquid to vapor.
Work Done by Gas (Pressure-Volume work):
Consider a cylinder fitted with a frictionless and weightless non-conducting piston of area of cross-section "A"
as shown in Fig. An ideal gas is enclosed in the cylinder. Let the volume of gas at initial state is "V 1". An
external pressure "P" is exerted on the piston.
If we supply "Q" amount of heat to the system then it will increase its internal energy by U. "After a certain
limit gas exerts pressure on the piston. If piston is free to move, it will be displaced by "h" & volume of
system increases from V to V2.
We know that pressure is the force per unit area i.e.
P = F/A
Or
F = PA ........ (i)
We also know that the work done by the gas on the
W = F d
Where d = displacement of piston = h
Putting the value of F and d, we get
W = (PA) h
Or
W = P (Ah)
But Ah = change or increase in volume = V
Hence

V
Application of the First Law to Gas Processes, Four Special Cases:
The variables that give the “state” of the gas P, V, and T are seen to be important, along with the
thermodynamic variables Q, U, and W. In principle, all six of these could change, but there are some
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Q
W
interrelations that reduce the freedom of independent adjustment. With one or more variables being kept
constant, we select four to illustrate the versatility of thermodynamics:
Quantity held constant
Type of process
V
Constant volume, Isochoric
P
Constant pressure, Isobaric
T
Constant temperature, Isothermal
Q
No heat flow, Adiabatic
In discussing each of these cases, we shall make reference to the trends of pressure and volume on the P-V
diagram.
1. Constant volume, V = 0, Isochoric: Heat supplied at
constant volume is also known as "ISOCHORIC SYSTEM". An
isochoric process is one in which the volume of system during the
supply of heat does not change. This is achieved only when the piston
of cylinder is fixed.
Consider a cylinder fitted with a frictionless piston. An ideal gas is
enclosed in the cylinder. The piston is fixed at a particular position so
that the volume of cylinder remains constant during the supply of heat.
Let Q amount of heat is added to the system. Addition of heat causes
the following changes in the system:
Internal energy increases from U1 to U2.
Volume of the system remains unchanged.
Temperature increases from T1 K to T2 K.
Pressure increases from P1 to P2.
No work is performed.
Applying First Law of Thermodynamics
Q  U  W
But
Thus ,
As ,
 W  P V
Q   U  P V
V  0
Q  U  P (0)
Q  U
This expression indicates that the heat supplied under isochoric process is consumed in increasing the
internal energy of the system but no work is performed.
2. Constant Pressure P = 0, Isobaric Process: A
thermodynamic process in which pressure of the system remains
constant during the supply of heat is called an ISOBARIC
PROCESS.
Explanation:
Consider a cylinder fitted with a frictionless piston. The piston is
free to move in the cylinder. An ideal gas is enclosed in the
cylinder.
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Let the initial volume of the system is V1 and initial internal energy is U1. Let QP the gas is heated from T1 K
to T2 K. Addition of heat causes the following changes in the system:
Internal energy increases from U1 to U2.
Volume of the system increases from V1 to V2.
Temperature increases from T1 K to T2 K.
Work (W) is done by the gas on the piston.
According to the first law of thermodynamics:
Q   U  W
But
 W  P V
Thus , Q  U  PV
As ,
V  (V 2  V1 )
3. Constant Temperature T = 0, Isothermal Process:
A thermodynamic process in which the temperature of the system remains
constant during the supply of heat is called an ISOTHERMAL PROCESS.
Isothermal Compression: Consider a cylinder of non-conducting walls
and good heat conducting base. The cylinder is fitted with a frictionless piston.
An ideal gas is enclosed in the cylinder. In the first stage pressure on the piston
is increased and the cylinder is placed on a cold body. Due to compression, the
temperature of the system increases but at the same time Q amount of heat is removed from the system
and the temperature of the system is maintained.
According to the first law of thermodynamics:
Q  U  W
Since temperature is constant, therefore, there is no change in internal energy of the system. i.e. ΔU = 0
As the work is done on the system, therefore, W is negative,
 Q  0  W
But
W  PV
Thus,
Q   W
Isothermal Expansion:
In another situation the cylinder is placed over a hot body and the pressure on the system is decreased. Due
to expansion, the temperature of the system is decreased but at the same time Q amount of heat is
According to the first law of thermodynamics:
Q  U  W
Since temperature is constant, th
As the work is done by the system, therefore, ΔW is positive,
Q  0  W
Thus,
Q   W
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ΔU = 0
4. No Heat Flow Q = 0, Adiabatic Process:
Insulation
A thermodynamic process in which no heat flows to or from the system
and hence
QAB  0
Sometimes we say that the system is thermally isolated but this
condition does not mean that temperature is constant. In fact all the
thermodynamic variables P, V, and T change in an adiabatic process.
Adiabatic
System
Applying First Law of Thermodynamics:
ΔQ = ΔU+ ΔW
U A B  W AB
This tells us that if work is done by the system (that is, if W is
positive), the internal energy of the system decreases by the amount
of work. Conversely, if work is done on the system (that is, if W is
negative), the internal energy of the system increases by that
amount.
In an idealized adiabatic process, heat cannot enter or leave the
system because of the insulation. Thus, the only way energy can be transferred between the system and its
environment is by work. If we remove shot from the position and allow the gas to expand, the work done by
the system (the gas) is positive and the internal energy of the gas decreases. If, instead, we add shot and
compress the gas, the work done by the system is negative and the internal energy of the gas increases.
Reversible and Irreversible Process:
Reversible Process: A process that, once having taken place, can be reversed and in doing so leaves
no change in either the system or the surroundings. Net work and net heat transfer must be zero. A
reversible process is an ideal that we never achieve – all processes are irreversible.
Why do we study reversible processes?
They are ideals that we can shoot for. They are standards to which we can compare different processes.
Irreversible Process: An irreversible process is such that it cannot be traced in the opposite direction
by reversing the controlling factors. Causes of irreversibility:
- Friction
- Unrestrained expansion
- Heat transfer through a finite temperature difference
- Mixing of different substances
- Others (chemical reactions, inelastic deformation of solids, electric resistance)
During an irreversible process heat energy is always used to overcome friction. Energy is also dissipated in
the form of conduction and radiation. This loss takes place whether the engine works on one direction or the
reverse direction. Such energy cannot be regained. In actual practice all engine is irreversible.
If electric current is passed through a wire, heat is produced. If the direction of current is reversed heat is
again produced. This is also an example of an irreversible process.
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Q=0
Carnot Reversible Engine
Heat engine are used to convert heat into mechanical work. Sadi Carnot (French) conceived a theoretical
engine which is free from all defects of practical engines. Its efficiency is maximum and it is an ideal heat
engine.
For any engine, there are three essential requisites:
(i) Source (ii) Sink and (iii) Working substance
Piston
Conducting
Conducting
Cylinder
Working
Substance
Source
Insulating
Conducting
Sink
Stand
At T1
At T2
Fig: Carnot‟s Reversible Engine
(i) Source: The source should be at a fixed high temperature T1 from which the heat engine can draw heat.
It has infinite thermal capacity and any amount of heat can be drawn from it at constant temperature T 1.
(ii) Sink: The sink should be at a fixed lower temperature T2 to which any amount of heat can be rejected. It
has infinite thermal capacity and its temperature remains constant at T 2.
(iii) Working Substance: A cylinder with non-conducting sides and conducting bottom contains the perfect
gas as the working substance.
A perfect non-conducting and frictionless piston is fitted into the cylinder. The working substance undergoes
a complete cyclic operation.
A perfectly non-conducting stand is also provided so that the working substance can undergo adiabatic
operation.
P
A
T1
Working Principle of a Carnot Engine
An ideal cycle of reversible engine operations in which a
substance maintain the following four processes:
isothermal
Q in
adiabatic
Q=0
D
-
Reversible isothermal expansion
Reversible adiabatic expansion
Reversible isothermal compression
Reversible adiabatic compression
Net work
done
B
adiabatic
Q=0
C
T2
isothermal
Q out
E
F
G
H
Step-1: Place the engine containing the working substance over
Fig: Carnon Cycle
the source at temperature T1. The working substance is also at
temperature T1. Its pressure is P1 and volume is V1 as shown by
the point A in Fig. Decrease the pressure. The volume of the
working substance increases. Work is done by the working substance. As the bottom is perfectly conducting
10
V
to the source at temperature T1, it absorbs heat. The process is completely isothermal. The temperature
remains constant. Let the amount of heat absorbed by the working substance be Q1 at the temperature T1.
The point B is obtained.
Consider one gram molecule of the working substance work done from A to B (Isothermal Process)
W1 
V2
V
P .dV          (1)
PV  RT1
1
P
RT1
V
RT1
dV
V
dV
 RT1 VV2
1 V
V
 RT1 ln 2          (2)
V1
 W1  VV2
1
Step-2: Place the engine on the stand having an insulated top. Decrease the pressure on the working
substance. The volume increases. The process is completely adiabatic. Work is done by the working
substance at the cost of its internal energy. The temperature falls. The working substance undergoes
adiabatic change from B to C. At C the temperature is T2.
Work done from B to C (adiabatic process)
PV   Cons tant
W2  V 3 P .dV          (3)
2
V
P 
 W2 
V3 Cons tan t
V
V
2
 cons tan t 
V3 dV
V2
V
Cons tant
V
dV
V3
V  1
 cons tan t
   1V
PV  RT
2
 1
V 3 1
 V2
 P3V 3
 P2V 2
 1
 1
PV
P V  P2V 2 R (T2  T1 )
PV
 3 3  2 2  3 3

 1  1
 1
 1

PV  P1V1  P2V2  cons tan t
Here  
Cp
Cv
R(T1  T2 )
           ( 4)
 1
 W2 
(iii) Step 3: Place the engine on the sink at temperature T2. Increase the pressure. The work is done on
the working substance. As the base is conducting to the sink, the process is isothermal. A quantity of heat Q 2
is rejected to the sink at temperature T2. Finally the point D is reached.
Work done from C to D (isothermal process)
W3 
V4
V
P .dV          (5)
3
PV  RT2 P 
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RT2
V
RT2
dV
V
dV
 RT2 VV4
3 V
V
 RT2 ln 4
V3
 W3  VV4
3
 RT2 ln
V3
         (6)
V4
The –ve sign indicates that work is done on the working substance.
(iii) Step 4: Place the engine on the insulating stand. Increase the pressure. The volume decreases. The
process is completely adiabatic. The temperature rises and finally the point A is reached.
Work done from D to A (adiabatic process)
W4 
V1
V
4
P .dV  
R(T1  T2 )
         (7)
 1
W 2 and W 4 are equal and opposite and cancel each other.
Net work done = W 1 + W 3
V
V
 RT1 ln 2  RT2 ln 4
V1
V3
V
V
 RT1 ln 2  RT2 ln 3                (8)
V1
V4
Efficiency of a Heat Engine
In common-sense language, efficiency is
output
input
For a heat engine, we are giving heat energy as input and we are getting work as output. So the efficiency of
a heat engine can be written as
Efficiency of a heat engine  
From First Law of Thermodynamics we know that
W net
------------------------ (9)
Q in
Q = U+ W
But over the course of a cycle, the gas returns to its original state, so there can be no net change in the
internal energy. Thus U = 0, so we have Q = 0+ W
But the net heat added Q = Qin - Qout.
W net = Qin - Qout.
So, the net work done by the gas "W net". So we have
Qin  Qout
Q
 1  out ---------------------- (10)
Therefore, the efficiency of a heat engine  
Qin
Qin
Efficiency in terms of temperature:
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Qin  RT1 ln
V2
V1
Q out  RT2 ln
V4
V3
V
ln 4
Q out T2
V3

V
Qin
T1 ln 2
V1
Since ln
  1
V2
V
= ln 4 , then we can write,
V1
V3
Q out
T
 1  2          (11)
Q in
T1
Solved Problems
Problem 1: Given that the specific heat capacity of water is 11 times that of copper, calculate the
mass of copper at a temperature of 100 °C required to raise the temperature of 150 g of water from
30.0 °C to 34.0 °C, assuming no energy is lost to the surroundings.
Solution:
Let specific heat capacity of copper, Sc cal/g°C. Then specific heat capacity of water, Sw = 11Sc cal/g°C.
Since no energy is lost to the surroundings, we can write
Heat lost by the hot body = Heat gained by the cold body
mcSc(100 – 34) = mwSw(34 – 30)
or mcSc× 66 = 150 × 11Sc× 4
or mc = 100 g
(Ans.)
Problem 2: A series of thermodynamic processes is shown in the pV-diagram of adjacent figure.
In process ab, 150 J of heat are added to the system, and in process bd, 600 J of heat are added.
Find [1] the internal energy change in process ab; [2] the internal energy change in process abd;
and [3] the total heat added in process acd.
P
Solution:
b
4
d
8.0 x 10 Pa
[1] Since, no volume change occurs during process ab, so W ab
= 0 and  Uab = Qab= 150 J.
[2] Process bd occurs at constant pressure, so the work done
by the system during this expansion is
4
-3
3
-3
3
W ab=p[V2 – V1]=[8.0x10 Pa][5.0x10 m – 2.0x10 m ] = 240 J
4
3.0 x 10 Pa
The total work for process abd is
O
W abd = W ab + W bd =0 + 240 J = 240 J
and the total heat is
Qabd = Qab + Qbd = 150 J+ 600 J = 750 J
Applying First law of thermodynamics, we get
 Uabd = Qabd –W abd = 750 J – 240 J = 510 J
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a
c
-3
2.0 x 10 m
3
-3
5.0 x 10 m
3
V
[3] Since,  U is independent of path, the internal energy change is the same for path acd as for path
abd ; that is,
 Uacd=  Uabd = 510 J.
The total work for the path acd is
4
-3
3
-3
3
W acd = W ac+ W cd = p[V2 – V1] + 0 = [3.0x10 Pa][5.0x10 m – 2.0x10 m ]= 90 J
Therefore, total heat added in the process acd :
Qacd =  Uacd + W acd = 510 J + 90 J = 600 J
3
3
Problem 3: One gram of water [1 cm ] becomes 1671 cm of steam when boiled at a constant
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6
pressure of 1 atm [1.013x10 Pa]. The heat of vaporization at this pressure is L V = 2.256x10 J/kg.
Compute [a] the work done by the water when it vaporizes; [b] its increase in internal energy.
Solution: [a] For a constant pressure process, work done by the vaporizing water:
W = p[V2 – V1] = [1.013x105 Pa][1671x10-6 m3 – 1.0x10-6 m3 ] = 169 J
[b] The heat added to water is the heat of vaporization:
Q = mLv = [10-3 kg] [ 2.256x106 J/kg] = 2256 J.
Therefore, change in internal energy:
 U = Q – W = 2256 J – 169 J = 2087 J.
So, to vaporize one gram of water, we have to add 2256 J of heat. Most of this added energy [2087 J]
remains in the system as an increase in internal energy. The remaining 169 J leaves the system again
as it does work against the surroundings while expanding from liquid to vapor.
Problem 4: (i) How much heat is needed to take ice of mass m = 720 g at -10 oC to a liquid state at
15oC? (ii) If we supply the ice with a total heat only 210 kJ, what then is the final state of the
water?
Solution:
Heat needed to reach the temperature 0 oC for ice, Q1 = mice×Sice×ΔT = 720×0.5×( 0 – ( -10)) cal
= 3600 cal = 15120 J
= 15.12 KJ
Heat needed to melt ice, Q2 = mice×Lice = 720×80 cal = 57600 cal = 241920 J = 241.92 KJ
Heat needed to raise the temperature of water to 15oC, Q3 = mw×Sw×ΔT = 720×1×(15 – 0) cal = 10800 cal
Now the total heat required, Q = Q1 + Q2 + Q3 = 3600 + 57600 + 10800 = 72000 cal = 302400 J
= 302.4 KJ
But the supplied heat is 210 KJ. Now we see that total ice will not melt because of supplying heat is
smaller than required to melt total ice.
Now the remaining heat = 210 – 15.12 = 194.88 KJ = 194880 J = 46400 cal
Now we consider that m g ice will melt. So we can write
m×Lice = 46400 cal
m ×80 cal/g = 46400 cal
or m = 580 g
So the remaining mass of the ice = 720 – 580 = 140 g
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(Ans.)
Problem Sheet – 01
o
o
o
1. The temperature of equal masses of three different liquids A, B and C are 12 C, 19 C and 28 C
o
o
respectively. The temperature when A and B are mixed is 16 C and when B and C are mixed is 23 C.
What would be the temperature when A and C are mixed?
2. Determine the change in the internal energy of a system that (i) absorbs 580 cal of thermal energy and
620 J of work is done by the system. (ii) releases 720 cal of thermal energy while 680 J of external work
is done on the system. (iii) maintains at a constant volume while 1430 cal of heat is removed from the
system.
0
3. Consider 1.0 kg of liquid water at 100 C by boiling at standard atmosphere pressure (which is 1.00 atm
or 1.01×105 Pa).The volume of that water changes from an initial value of 1.00 × 10-3 m3 as a liquid to
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1.765 m as steam. (i) How much work is done by the system during this process? (ii) How much energy
is transferred as heat during the process? (iii) What is the change in the system‟s internal energy during
the process?
4. A Carnot engine operates between 240 0C and 100 0C, absorbing 3.35x105 J per cycle at the higher
temperature. (i) What is the efficiency of the engine? (ii) How much work per cycle is this engine capable
of performing?
5. A Carnot engine has a power output of 120 kW. The engine operates between two reservoirs at 42oC and
550oC. (i) How much thermal energy is absorbed per minutes? (ii) How much thermal energy is lost per
minutes?
6. A total of 0.8 kg of water at 20oC is placed in a 1 kW electric kettle. How long a time is needed to raise the
temperature of the water to 100oC?
7. In an industrial process 20 kg of water per hour is to be heated from 20 0C to 60 0C.To do this, steam at
140 0C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and
is returned to the boiler as water at 80 0C. How many kg of steam is required per hour? Specific heat of
steam = 1996 J/kg-K, Latent heat of steam =2.268x106 J/Kg.
8. What would be the final temperature of the mixture when 5 gm of ice at -10 0C are mixed with 30 gm of
water at 20 0C? (Sp. heat of ice = 0.5 cal/gm-0C; latent heat of fusion of ice = 80cal/gm.)
0
9. A 625 g iron block is heated to 352 C is placed in an insulated container (of negligible heat capacity)
containing 40.0 g of water at 15.0 0C. What is the equilibrium temperature of this system? If your answer
is 100 0C, determine the amount of water that has vaporized. The average specific heat of iron over this
temperature range is 560 J/(kg K), and the specific heat of liquid water is 4186 J/(kg K).
10.An aluminium container of mass 150 gm contains 250 gm of ice at -120C.Heat is added to the system at
the rate of 420 joules per seconds. What is the temperature of the system after 3.5 minutes? Specific
o
heat of aluminium = 840 J/(kg C).
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