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Chapter 8
ALL ABOUT HEAT
Hot water and cold water ”contain” the same amount of heat: none
at all.
−J.P. Holman (Thermodynamics)
A beginner in thermodynamics may think that a hot object contains
more heat than what a cold object contains. The truth is none of them
contains any heat at all since heat does not reside in an object. The
beginner may also think that the temperature of an object cannot be raised
unless heat is provided to the object, which is a wrong notion. Even one with
a fair knowledge of thermodynamics, often confuses heat with enthalpy.
This chapter is presented to give a clear idea about what heat is, and what
heat is not.
162 Chapter 8
8.1 What is Heat?
We know heat is energy. In the past, even scientists and engineers did
not know that. They thought heat was a fluid. Almost 50 years after
the construction of the first successful steam engine in 1712 by Thomas
Newcomen, Professor Joseph Black founded a theory on heat. This theory
is known as caloric theory, and it said heat was a colourless, weightless fluid
known as caloric, and it was conserved. That means caloric, or heat, could
not be created or destroyed, but it could transfer itself from one object to
another. If a metal was heated using fire, it was explained, then the fluid
caloric was transferred from the fire to the metal. James Watt, associated
with Professor Black’s laboratory, modified Thomas Newcomen’s engine in
1765 and made the first efficient steam engine. In a steam engine, heat
generated from burning the coal is converted into work required to do a
job, such as rotating the wheels of a train.
In 1824, a French engineer named Sadi Carnot, a believer in the caloric
theory, presented on paper an ideal engine that could provide the maximum
amount of work for a specific amount of heat given to the engine. Carnot
theorized that the work was obtained from an engine because of heat,
which he believed as fluid, falling from a high temperature source to a low
temperature source. He showed that the thermal efficiency of his ideal
engine depended only on these two temperatures. No real engine can be
more efficient than the Carnot engine, and this result, very interestingly, is
still valid, even though Carnot believed that heat was fluid.
James Prescott Joule, in 1840s, performed a series of experiments where
falling wights stirred a liquid and heated it up. He showed that the heat
produced had always the same quantitative relationship to the energy lost
by the falling weights, and concluded that heat was just another form of
energy. Joule’s ideas and those of Carnot were reconciled, simply and
effectively, by Rudolf Julius Emanuel Clausius in 1850 who wrote down
the First law of Thermodynamics as “the total energy of the system is a
constant”. William John Macquorn Rankine, a Scottish engineer, defined
thermodynamic efficiency of a heat engine in 1853 when applying the theory
of thermodynamics to heat engines, and wrote the first thermodynamics
textbook in 1859.
All about Heat 163
Our world is full of heat engines where heat is converted into useful
work. Car engines and jet engines, for example, are powered by the heat
generated from burning a fuel. The source of electricity generation in coal,
thermal and nuclear power stations is heat. Heat released from the burning
of coal, petroleum and natural gas accounts for about 85% of the global
energy consumption, which is in the order of hundreds of exajoules.
Without the recognition of the exact nature of heat and its relationship
to work, the world would not have gone so far in its utilization of heat to
provide for the gigantic amount of energy consumed by the human race to-
day. This colossal amount of energy consumption has resulted in pollution
of all kinds. One of the consequences of which is global warming, and the
resulting life-threatening climate change.
8.2 Heat Supply and Common Sense
We know that heat flows from a hot object to a cold object when the
two touch each other. Thus, we have a tendency to believe that a hot
object contains more heat than a cold object does. The truth is heat does
not reside in an object, and there is no such thing as the heat content of
an object. The thermal energy that resides in an object is not heat, but
internal energy. When a hot object comes into contact with a cold object,
the internal energy content of the hot object decreases and the internal
energy content of the cold object increases, until the temperatures of both
become the same. The energy that is transferred between the two objects
during such a process, driven by the temperature gradient between the two
objects, is called heat. This means, we can refer to a form of energy as
heat only when it is being transferred from one object to the other because
of the temperature difference existing between the two objects.
When heat is supplied to water, the temperature of water increases.
Such familiar everyday experiences have made some of us to conclude that
‘when heat is provided to a substance, its temperature should necessarily
increase’. Is that really true? If so, how could we explain the following
observations?
164 Chapter 8
•Heat provided to ice at its melting point turns ice to water at the
temperature of the melting point. That is, the temperature remains
constant but there is change of phase.
•When heat is added to water at its boiling temperature, the temper-
ature does not change until all water is turned into steam. That is,
there is phase change at constant temperature.
Now, some may conclude that ‘when a substance is heated either the
temperature of the substance should increase or the phase of the substance
should change’. Is that true? No. The following sections are designed to
give us an insight into what else may happen when heat is supplied to a
substance.
8.3 Heat Supplied to
Increase the Temperature
Take air in a cylindrical container with enough force applied to the pis-
ton to keep the air volume constant, as in Figure 8.1. Insulate the walls
of the cylinder and the piston so that heat is not lost to the surroundings.
Supply heat to air using a heating coil.
ppppp
p
p
pp
p
p
p
pppppp
ppppp
p
p
ppppp
p
p
pppppp
pppppp
pppppp
initial state
ppppp
p
p
ppppp
p
p
pppppp
ppppp
p
p
ppppp
p
p
pppppp
pppppp
pppppp
final state
Figure 8.1 Heat supplied at constant volume.
6
pp
Since no work transfer occurs, the heat supplied to air goes to increase
the internal energy of air in accordance with the first law of thermodynamics
Qin =ΔU
All about Heat 165
Assuming ideal gas behaviour, the increase in the specific internal energy
of air may be accounted for by
Δu=Tf
To
CvdT
Combining the two equations above, we get
Qin =mTf
To
CvdT
which says that the heat supplied to a substance increases its temperature.
It is a very familiar experience for us, and therefore we have no problem
recognizing it, and accepting it.
8.4 Heat Supplied to Do Work
Consider air in the set up shown in Figure 8.2, and let us carry out the
following thought experiment. Allow the piston to move away by a tiny dis-
tance such that the volume occupied by the air in the cylinder is increased
by a tiny amount. Do not supply heat to the air during this step. The tiny
increase in the volume of air would cause a tiny decrease in the air pressure
and hence the temperature of air would drop by a tiny amount. Let us now
hold the volume of the air in the cylinder constant at its new value, and
supply heat to the air such that the temperature of air is brought back to
its original value.
ppppp
p
p
ppppp
p
p
pppppp
ppppp
p
p
ppppp
p
p
pppppp
pppppp
pppppp
p
pp
pp
p
p
pp
pp
pp
p
pp
pp
p
p
pp
pp
pp
p
pp
pp
p
p
pp
pp
pp
p
p
p
p
ppppp
p
p
ppppp
p
p
pppppp
ppppp
p
p
ppppp
p
p
pppppp
pppppp
pppppp
initial state final state
Figure 8.2 Heat supplied maintaining temperature constant.
By continuing this procedure, it is possible to supply heat to the air
while maintaining its temperature about a constant value. Here we have
166 Chapter 8
a process in which the heat supplied neither increases the temperature nor
changes the phase of a substance. We may ask then what happened to the
heat supplied to air. The answer to the question shall be found as follows.
First law of thermodynamics applied to the air assumed to behave as
an ideal gas, gives
Qin =−Win +ΔU=Wout +mTf
To
CvdT
Since Tis maintained a constant, the above becomes Qin =Wout.
That is, all the heat supplied to the ideal gas maintained at a constant
temperature is used by the ideal gas to do work in pushing the piston away.
8.5 Temperature Increase without
Heat Supply
Consider air in the set up shown in Figure 8.3. Insulate the system so
that heat does not cross the system boundary. Push the piston slowly to
compress the air. Since no heat is transferred across the system boundary,
the first law of thermodynamics applied to air yields
Win =ΔU=mTf
To
CvdT
if ideal gas behaviour is assumed. The above expression clearly shows that
the work done on the air is responsible for increasing the temperature. That
is, the temperature of air increases with no heat supplied to it.
ppppp
p
p
ppppp
p
p
pppppp
ppppp
p
p
ppppp
p
p
pppppp
pppppp
pppppp
p
pp
pp
p
p
pp
pp
pp
p
pp
pp
p
p
pp
pp
pp
p
pp
pp
p
p
pp
pp
pp
p
p
p
p
ppppp
p
p
ppppp
p
p
pppppp
ppppp
p
p
ppppp
p
p
pppppp
pppppp
pppppp
initial state final state
Figure 8.3 No heat is supplied, but force is applied on the piston.
xsr r
q
q
pp
prxsr r
q
q
pp
pr
All about Heat 167
Summarizing what we have learnt so far in this chapter, we could say
that changes in the properties, such as the temperature, pressure or volume
of a system, may alter the internal energy of the system. Remember that
temperature, pressure, volume and internal energy are all properties of a
system, and are related to each other. Any change in the internal energy of
a closed system, in turn, is related to heat and/or work transfers between
the system and its surroundings by the first law of thermodynamics applied
to a closed system, which is
Qin +Win =ΔU
Let us once again state that heat is defined as the form in which energy
crosses the boundary of a system owing to a temperature difference between
the system and its surroundings. Any other form of energy transfer across
the boundary of a system is in the form of work, such as the boundary
work.
8.6 Direct Evaluation of Qin
In the examples that we have so far worked out, the system was either
taken as an adiabatic system, for which Qin = 0, or the numerical value of
Qin or Qout was given. Whenever we were asked to determine how much
heat was transferred to or from the system, we calculated the value of Qin
by use of the first law of thermodynamics. In this section, we will explore
how else Qin could be evaluated.
Suppose a system is heated by a heat source, such as an electric heater.
The amount of heat added to the system from the heat source can be
evaluated using
Qin =tf
to
˙
λδt
where ˙
λis the rate at which heat is added to the system from the heat
source, and toand tfare the initial and final times, respectively.
If ˙
λis a constant then we have
Qin =˙
λ(tf−to)
168 Chapter 8
where, if ˙
λis in kW and tfand toareininseconds,Qin will be in kJ.
Consider a system heated by heat transferred to it from its surround-
ings. We know that heat flows into or out of a system, only if there is a
temperature difference between the system and surroundings. Let us say
that the system is at temperature Tsys , and the surroundings is at temper-
ature Tsurr, which is higher than Tsys . The heat transfer dQin during a
very short time dt to the system from the surroundings is then given by
dQin =κA(Tsurr −Tsys )dt
where κis the overall heat transfer coefficient in kJ/m2·K·s, and A
is the surface area in m2across which heat enters the system from the
surroundings.
If the surroundings are at a lower temperature than the system, then
Tsurr is less than Tsys . Consequently, dQin of the above expression takes
a negative value. It means that the heat flows from the system to its
surroundings.
To determine the total heat transfer, the above equation is integrated
to obtain
Qin =tf
to
κA(Tsurr −Tsys )dt
To evaluate the above integral we need to know how the temperature
of the system and the temperature of its surroundings vary with time.
Owing to the complex nature of the evaluation of Qin using the above
expression, beginners in thermodynamics are seldom expected to evaluate
Qin using the above described method.
8.7 Zeroth Law of Thermodynamics
Let us now look at something that is seemingly obvious to us. We know
that heat flows from a hot object to a cold object when they are brought
into contact. If no heat flows between them when they are brought into
contact, then the two objects should be at the same temperature. These
two objects are then in thermal equilibrium with each other.
The zeroth law of thermodynamics, formulated in 1931, states that if
a system A is in thermal equilibrium with a system C and another system
All about Heat 169
B is also in thermal equilibrium with the system C, then the systems A and
B will be in thermal equilibrium with each other.
Let us consider systems A and B as liquids in two different containers
which are not in contact, and system C as a thermometer. Let us suppose
that the two systems A and B are in thermal equilibrium with the ther-
mometer C, which means that systems A and B give the same temperature
reading. According to the zeroth law, the two systems A and B are in ther-
mal equilibrium even if they are not in contact since they have the same
temperature reading.
The zeroth law, like the first law of thermodynamics, is not provable,
even though it seems obvious and trivial.
8.8 Heat and Enthalpy
Even one with a fair knowledge of thermodynamics often confuses heat
with enthalpy. This section would help us to see clearly the relationship
between heat and enthalpy.
Enthalpy defined by (4.1) as H=U+PV, when differentiated, takes
the form
dH =dU +PdV +VdP (8.1)
Substituting dU from (3.5), which is the differential form of the first
law applied to closed simple compressible systems, in (8.1), we get
dH =dQin +dWin +PdV +VdP
which could be rearranged to yield
dQin =dH −dWin −PdV −VdP (8.2)
On integration of (8.2), we get
Qin =ΔH−Win −Vf
Vo
PdV −Pf
Po
VdP (8.3)
which relates the heat provided to a closed simple compressible system to
the enthalpy increase of the system. Remembering (8.3) would keep us
away from confusing heat with enthalpy.
170 Chapter 8
If the closed system undergoes a quasistatic process then dWin given
by (7.5) will be used to reduce (8.2) to
dQin =dH −VdP (8.4)
If the given process is a constant-pressure process then dP = 0. Thus,
(8.4) becomes
dQin =dH (8.5)
which upon integration yields
Qin =ΔH(8.6)
Equation (8.6) is applicable to a a closed system undergoing
a quasistatic constant-pressure process involving no forms of
work transfer other than boundary work.
It is important to note that enthalpy is a property of a system, and
therefore the enthalpy change ΔHis the difference between the enthalpies
at the initial and the final states. The amount of heat entering the system
Qin depends on the path that the system takes between the initial and the
final states of the system. And, these two very different entities equal each
other only under special circumstances such as the one above.
8.9 Heat and Internal Energy
The circumstances under which heat exchanged with a system could be
related to the internal energy change of a system is explored in this section.
Let us consider a closed system undergoing a constant-volume process. The
first law, given by (3.4), applied to this process becomes
Qin =ΔU(8.7)
provided there is no other forms of work transfer associated with the closed
system.
It is important to note that the internal energy change ΔUis the dif-
ference between the internal energies at the initial and the final states.
The amount of heat entering the system Qin depends on the path that
the system takes between the initial and the final states of the system.
All about Heat 171
And, these two very different entities equal each other only under special
circumstances such as the one above.
8.10 Heat and Specific Heat
We have learned about the specific heat at constant volume (Cv)in
Section 5.4, and about the specific heat at constant pressure (Cp) in Section
5.5. Nowhere in these sections, however, we find any reference to the
quantity heat. Then, why do we call Cvand Cpspecific heats? It is a
question asked by many beginners in thermodynamics. In this section, we
shall see how to relate the specific heats to the quantity heat.
First, let us deal with specific heat at constant pressure. From the
expression for Cpgiven by (5.9), we get
Cp=dh
dT
P
=1
m
dH
dT
P
(8.8)
When using (8.5) applicable to a quasistatic constant-pressure process
involving only boundary work, (8.8) becomes
Cp≡1
m
dQin
dT
P
(8.9)
Therefore, Cpis the heat required to raise the temperature of a unit
mass of a substance by one degree in a quasistatic constant-pressure process
involving no forms of work transfer other than boundary work.
From the expression for Cvgiven by (5.5), we get
Cv=du
dT
v
=1
m
dU
dT
v
(8.10)
For a constant-volume process, the boundary work is zero, and therefore
the first law yields dQin =dU provided no other forms of work transfer is
involved. Thus, (8.10) becomes
Cv≡1
m
dQin
dT
v
(8.11)
172 Chapter 8
Therefore, Cvis the heat required to raise the temperature of a unit
mass of a substance by one degree in a constant-volume process involving
no forms of work transfer.
Note that some textbooks introduce Cpand Cvin terms of (8.9) and
(8.11), respectively. Upon integration of (8.9) and (8.11), we get the
following two very useful expressions for the direct evaluation of heat in the
following processes:
For a quasistatic constant-pressure process involving no forms of work
transfer other than boundary work:
Qin =mTf
To
CpdT (8.12)
For a constant-volume process involving no forms of work transfer:
Qin =mTf
To
CvdT (8.13)
Note that (8.12) and (8.13) are applicable for any simple compressible
substance.
8.11 Worked Examples
Example 8.1
An ideal gas (Cv= 0.744 kJ/kg ·K) in a piston-
cylinder arrangement is compressed such that 93 kJ/kg of work is done on
the gas. Assuming adiabatic condition prevails, find out what happens to
the work provided to the system.
Solution to Example 8.1
The first law of thermodynamics applied to the given adiabatic system gives
Win =ΔU=mC
vΔT. Therefore,
ΔT=Win
mC
v
=93
0.744 = 125 K
All about Heat 173
That is, the work provided to the system has increased the temperature of
the system. Note that the temperature of the system has increased without
receiving any heat.
Example 8.2
The piston-cylinder device in Figure 8.4 con-
tains 0.2 kg of air with a molecular weight of 29, Cvof 0.718 kJ/kg·Kand
Cpof 1.005 kJ/kg ·K. The initial pressure of air is 1 MPa, and it is just
enough to balance the weight of the piston
and the atmospheric pressure acting on the
piston. The initial temperature is 127◦C.
Heat is transferred to air until the piston,
assumed to be frictionless, reaches the stops.
Further heat is transferred to air until the
air reaches 2 MPa and 927◦C. Sketch the
path of the process on a P-Vdiagram, and
determine the amount of heat supplied to
the air.
air
piston
= ZZ~
stops
Figure 8.4
+
Solution to Example 8.2
At the initial state A, PA=1MPaandTA= 400 K. Heat is transferred to
air so that it expands until the piston reaches the stops shown in Figure 8.4. This
intermediate state is denoted by B. During process A→B, the expansion of air
is at a constant pressure of 1 MPa, since this pressure just balances the weight
of the piston and the atmospheric pressure acting on the piston. Thus PB=
PA= 1 MPa. Assume this constant-pressure process to be quasistatic. After
the piston reaches the stops, further heat is added to the air until the process
reaches its final state, denoted by C, at which PC=2MPaandTC= 1200 K.
During process B→C, the volume of air is a constant. The path A→B→Con
the P-Vdiagram of Figure 8.5 describes the entire process.
To evaluate the total amount of heat transferred to air during the entire
process, let us determine the heat transfers separately for the processes A→B
and B→C. Since A→B is a quasistatic constant-pressure process, we can use
(8.12) to determine the heat transfer as
(Qin)A→B=mC
p(TB−TA)(8.14)
174 Chapter 8
Since B→C is a constant-volume process of a simple compressible closed
system, we can use (8.13) to determine the heat transfer as
(Qin)B→C=mC
v(TC−TB)(8.15)
P(MPa)
V(m3)
A
C
VAVB=VC
2
1
Figure 8.5 The path of the process given in Example 8.2.
B
rr
r
Adding (8.14) and (8.15), we get
(Qin)A→B→C=mC
p(TB−TA)+mC
v(TC−TB)
Substituting all the numerical values that we already know, we get
(Qin)A→B→C=0.2×1.005 ×(TB−400) + 0.2×0.718 ×(1200 −TB)(8.16)
where the unknown TBcan be found using
TB=PBTC
PC
=1×1200 K
2= 600 K,
at states B and C along the constant-volume path B→C. Using the numerical
value of TBin (8.16), we get (Qin)A→B→C= 126.4 kJ.
Example 8.3
An ideal gas of 0.01 kmol is taken through a
cyclic process consisting of the following four processes: Process A to B
is an isothermal expansion at 800 K from 8 bar to 6 bar; Process B to C
is an adiabatic expansion to 3 bar; Process C to D is a constant-pressure
cooling; Process D to A is a constant-volume heating. Sketch the cyclic
All about Heat 175
process on a P-Vdiagram. Assuming that all processes are quasistatic and
taking γto be 1.38, determine the temperatures at states C and D, and
calculate the heat and work transfers for the entire cyclic process.
Solution to Example 8.3
The cyclic process sketched on a P-Vdiagram is shown in Figure 8.6.
P
V
D
Figure 8.6 The path of the cyclic process of Example 8.3.
A
C
B
r
r
rr
9
) isothermal expansion at 800 K
adiabatic expansion
(a) Determination of the temperatures at states C and D:
Table 8.1 shows the data at states A, B, C and D. Since B →C is a quasistatic
adiabatic expansion of an ideal gas, (7.31) can be used to find TCas
TC=PC
PB(γ−1)/γ
TB=3
6(1.38−1)/1.38
×800 K= 661 K
Since D →A is a constant-volume process of an ideal gas, the ideal gas equation
of state can be used to find TDas
TD=PD
PATA=3
8×800 K= 300 K
APA=8bar TA= 800 K VA=?
BPB=6bar TB= 800 K VB=?
CPC=3bar TC=? VC=?
DPD=3bar TD=? VD=VA=?
Table 8.1 Data at states A, B, C and D of the cyclic process.
176 Chapter 8
(b) Calculation of the heat and work transfers:
Process A →B is a quasistatic isothermal expansion of an ideal gas at 800
K, and therefore the work transfer is calculated using
(Win)A→B=−VB
VA
PdV =−nRT
Aln PA
PB
=−0.01 ×8.314 ×800 ×ln 8
6kJ =−19.1kJ
and the heat transfer is calculated using
(Qin)A→B=−(Win)A→B=19.1kJ,
since there is no internal energy change for an isothermal process of an ideal
gas.
Process B →C is a quasistatic adiabatic expansion of an ideal gas, and
therefore the heat transfer becomes (Qin)B→C= 0, and the work transfer is
calculated using
(Win)B→C=(ΔU)B→C=nC
v(TC−TB)=nR
γ−1(TC−TB)
=0.01 ×8.314
1.38 −1×(661 −800) kJ =−30.4kJ
Process C →D is a quasistatic constant-pressure cooling of an ideal gas,
and therefore the heat transfer is calculated using (8.12) as
(Qin)C→D=nC
p(TD−TC)=nγR
γ−1(TD−TC)
=0.01 ×1.38 ×8.314
1.38 −1×(300 −661) kJ =−109.0kJ
and the work transfer is calculated using
(Win)C→D=(ΔU)C→D−(Qin)C→D
==nC
v(TD−TC)−nC
p(TD−TC)=−nR(TD−TC)
=−0.01 ×8.314 ×(300 −661) kJ =30.0kJ
Process D →A is a constant-volume heating of an ideal gas, and therefore
the work transfer becomes (Win)D→A=0,andtheheattransferiscalculated
using (8.13) as
(Qin)D→A==nC
v(TA−TD)=nR
γ−1(TA−TD)
=0.01 ×8.314
1.38 −1×(800 −300) kJ = 109.4kJ
All about Heat 177
The net heat transfer for the entire cyclic process therefore becomes
(Qin)net =(19.1+0−109.0 + 109.4) kJ =19.5kJ
and the net work transfer for the entire cyclic process becomes
(Win)net =(−19.1−30.4+30.0+0) kJ =−19.5kJ
Comment: Note that we have (Qin)net +(Win)net =0. Itisbecauseina
cyclic process the net internal energy change is zero owing to the initial and the
final states being the same.
Example 8.4
An ideal gas enclosed in a piston-cylinder as-
sembly is compressed adiabatically to increase its temperature from TLK
to THK. Heat is then supplied to the ideal gas such that it expands isother-
mally at THK. The heat supply is cut off and the ideal gas is allowed to
continue expanding adiabatically until its temperature drops to TLK. Fi-
nally, the ideal gas is compressed isothermally until it returns to its initial
state during which heat is rejected to the surroundings. Sketch the cyclic
process on a P-Vdiagram. Assuming that all processes are quasistatic,
obtain an expression, in terms of TLand TH, for the thermal efficiency of
the cycle, ηth, defined as the net work output of the cycle per unit of heat
added to the cycle.
If TL= 300 K and TH= 1500 K, determine the amount of heat added
to the ideal gas to produce 100 kJ of net work output. Also, determine the
amount of heat rejected to the surroundings by the ideal gas.
Solution to Example 8.4
The cyclic process sketched on a P-Vdiagram is shown in Figure 8.7, where
A→B is the quasistatic adiabatic compression, B →C is the quasistatic isother-
mal heating at the temperature TH,C→D is the quasistatic adiabatic expansion,
and D →A is the quasistatic isothermal cooling at the temperature TL.
178 Chapter 8
P
V
A
Figure 8.7 The path of the cyclic process of Example 8.4.
B
D
C
s
s
s
s
adiabatic
expansion
isothermal
cooling at TL
adiabatic
compression
isothermal
heating at TH
The thermal efficiency of the cycle, ηth, is defined as the net work output of
the cycle, Wnet, per unit of heat added to the cycle, Qin, and therefore we have
ηth ≡Wnet
Qin
(8.17)
In a cyclic process, the net change in the internal energy is zero, and therefore
the first law applied to the cyclic process gives
Wnet =Qin −Qout (8.18)
where Qin is the heat added to the ideal gas during the isothermal heating B →
CandQout is the heat removed from the ideal gas during the isothermal cooling
D→A. No heat is added or removed during the adiabatic processes A →Band
C→D.
Combining (8.17) and (8.18), we get
ηth =Qin −Qout
Qin
=1−Qout
Qin
(8.19)
Since the internal energy remains constant for an isothermal process of an
ideal gas, isothermal heat addition during B →C may be expressed using the
first law as
Qin =−Win =nVC
VB
PdV =nRT
Hln VC
VB(8.20)
where THin the temperature of the ideal gas during the isothermal heating
process. Isothermal heat rejection during D →A may be expressed using the
All about Heat 179
first law as
Qout =Win =−nVA
VD
PdV =−nRT
Lln VA
VD=nRT
Lln VD
VA
(8.21)
where TLin the temperature of the ideal gas during the isothermal cooling
process.
Combining (8.19), (8.20) and (8.21), we have
ηth =1−TLln (VD/VA)
THln (VC/VB)(8.22)
Since A →B is a quasistatic adiabatic expansion of an ideal gas, (7.30) can
be used to get
TA
TB
=VB
VAγ−1
(8.23)
Since C →D is a quasistatic adiabatic expansion of an ideal gas, (7.30) can
be used to get
TC
TD
=VD
VCγ−1
(8.24)
Combining (8.23) and (8.24) using TB=TC=THand TD=TA=TL,we
have VB
VA
=VC
VD
which can be rearranged to give
VD
VA
=VC
VB
(8.25)
Using (8.25), we can reduce (8.22) to
ηth =1−TL
TH
The above expression for thermal efficiency is known as the Carnot efficiency,
and therefore we write it as follows:
ηCarnot ≡1−TL
TH
(8.26)
If TL= 300 K and TH= 1500 K, then (8.26) gives
ηCarnot =1−300
1500 =0.80 = 80%
180 Chapter 8
Since the thermal efficiency of the cycle is 80%, using (8.17), we can calculate
the amount of heat to be added to the cycle to produce 100 kJ of net work output
from the cycle as,
Qin =Wnet
ηCarnot
=100 kJ
0.8= 125 kJ
The amount of heat rejected to the surroundings from the cycle can be
calculated using (8.18) as
Qout =Qin −Wnet = 125 kJ −100 kJ =25kJ
Comment: Carnot efficiency is a very important concept in thermodynamics.
It is because the thermal efficiency of no engine, that converts heat to work
operating in a cyclic process between the maximum temperature THKandthe
minimum temperature TLK, can be higher than the Carnot efficiency (the proof
of which is given in Chapter 13), which Sadi Carnot presented in 1824, while
not knowing for sure that heat is energy.
Example 8.5
In a cyclic process, air initially at 1 bar and
300 K is compressed adiabatically to reduce its volume to one eighth of
the initial value. The compressed air is heated at constant volume to 1500
K. The air in then expanded adiabatically and finally cooled at constant
volume to its initial state. Sketch the cyclic process on a P-Vdiagram.
Assume that all processes are quasistatic and that air behaves as as ideal
gas with γ=1.4. Determine the thermal efficiency of the cycle and the
amount of heat added to the air to produce 100 kJ of net work output.
Determine the numerical values of the pressures at the end of each
process in the cycle.
Solution to Example 8.5
The cyclic process sketched on a P-Vdiagram is shown in Figure 8.8, where
A→B is the quasistatic adiabatic compression, B →C is the constant-volume
heating, C →D is the quasistatic adiabatic expansion, and D →Aisthe
constant-volume cooling.
All about Heat 181
P
V
Figure 8.8 The path of the cyclic process of Example 8.5.
A
B
C
D
Let us first determine the thermal efficiency of the cycle, ηth, the definition
of which is given by (8.17). For the given cyclic process, we can calculate ηth
using (8.19), where Qin is the amount of heat added during the constant-volume
heating B →C, and Qout is the amount of heat rejected during the constant-
volume cooling D →A. No heat is added or removed during the adiabatic
processes A →BandC→D.
Constant-volume heat addition can be calculated using (8.13) as
Qin =mC
v(TC−TB)(8.27)
Constant-volume heat rejection can be calculated using (8.13) as
Qout =−mC
v(TA−TD)=mC
v(TD−TA)(8.28)
Combining (8.19), (8.27) and (8.28), we have
ηth =1−mC
v(TD−TA)
mC
v(TC−TB)=1−TD−TA
TC−TB
(8.29)
where TA= 300 K, TC= 1500 K and TBand TDare unknown. To determine
TBand TD, let us tabulate all the known data at the states A, B, C and D as
in Table 8.2.
APA=1bar TA= 300 K VA=?
BPB=? TB=? VB=VA/8
CPC=? TC= 1500 K VC=VB=VA/8
DPD=? TD=? VD=VA=?
Table 8.2 Data at states A, B, C and D of the cyclic process.
182 Chapter 8
Since A →B is a quasistatic adiabatic compression of an ideal gas, (7.30)
can be used to find TBas
TB=VA
VB(γ−1)
TA=8
0.4×300 K= 689 K
Since C →D is a quasistatic adiabatic expansion of an ideal gas, (7.30) can
be used to find TDas
TD=VC
VD(γ−1)
TC=1
80.4
×1500 K= 653 K
Substituting the numerical values of the temperatures in (8.29), we get
ηth =1−653 −300
1500 −689 =0.565 = 56.5%
Since the thermal efficiency of the cycle is 56.5%, using (8.17), we can
calculate the amount of heat to be added to the cycle to produce 100 kJ of net
work output from the cycle as,
Qin =Wnet
ηth
=100 kJ
0.565 = 177 kJ
The amount of heat rejected to the surroundings from the cycle can be
calculated using (8.18) as
Qout =Qin −Wnet = 177 kJ −100 kJ =77kJ
We know PA= 1 bar, and we are to find PB,PCand PD.SinceA→Bis
a quasistatic adiabatic compression of an ideal gas, (7.29) can be used to find
PBas
PB=VA
VBγ
PA=8
1.4×1bar =18.4bar
Since B →C is a constant-volume process of an ideal gas, ideal gas equation
of state can be used to find PBas
PC=TC
TBPB=1500
689 ×18.4bar =40.1bar
Since C →D is a quasistatic adiabatic expansion of an ideal gas, (7.29) can
be used to find PDas
PD=VC
VDγ
PC=1
81.4
×40.1bar =2.2bar
All about Heat 183
Comment: The cycle studied in this problem is known as the ideal Otto cycle.
It is an idealized cycle used to understand the working of an engine in which the
fuel-air mixture contained in a piston-cylinder arrangement is first compressed
and then ignited using a spark, as in a 4-stroke car engine fueled by petrol. Also,
notice that the efficiency of the ideal Otto cycle operating between 1500 K and
300 K is only 56.5%, where as the Carnot efficiency for the same temperature
extremes is 80% (see Example 8.4).
Example 8.6
In a cyclic process, air initially at 1 bar and 300
K is compressed adiabatically to reduce its volume to one twentieth of the
initial value. The compressed air is heated at constant pressure to 1500
K. The air in then expanded adiabatically and finally cooled at constant
volume to its initial state. Sketch the cyclic process on a P-Vdiagram.
Assuming that all processes are quasistatic and that air behaves as as ideal
gas with γ=1.4, determine the thermal efficiency of the cycle, ηth.
Determine the quantity of heat added and the net work output per kg
of air, taking the molar mass of the air as 29 kg/kmol.
Solution to Example 8.6
The cyclic process sketched on a P-Vdiagram is shown in Figure 8.9.
P
V
Figure 8.9 The path of the cyclic process of Example 8.6.
B
A
C
D
184 Chapter 8
A→B is the quasistatic adiabatic compression, B →C is the constant-
pressure heating, C →D is the quasistatic adiabatic expansion, and D →Ais
the constant-volume cooling.
The thermal efficiency of the cycle, ηth, is defined by (8.17). For the given
cyclic process, we can calculate ηth using (8.19), where Qin is the amount of
heat added during the constant-pressure heating B →C, and Qout is the amount
of heat rejected during the constant-pressure cooling D →A. No heat is added
or removed during the adiabatic processes A →BandC→D. Constant-pressure
heat addition can be calculated using (8.12) as
Qin =mC
p(TC−TB)(8.30)
Constant-volume heat removal can be calculated using (8.13) as
Qout =−mC
v(TA−TD)=mC
v(TD−TA)(8.31)
Combining (8.19), (8.30) and (8.31), we have
ηth =1−1
γTD−TA
TC−TB(8.32)
where TA= 300 K, TC= 1500 K and TBand TDare unknown. To determine
TBand TD, let us tabulate all the known data at the states A, B, C and D as
in Table 8.3.
APA=1bar TA= 300 K VA=?
BPB=? TB=? VB=VA/20
CPC=PB=? TC= 1500 K VC
DPD=? TD=? VD=VA=?
Table 8.3 Data at states A, B, C and D of the cyclic process.
Since A →B is a quasistatic adiabatic compression of an ideal gas, (7.30)
can be used to find TBas
TB=VA
VB(γ−1)
TA=20
0.4×300 K= 994 K(8.33)
Since C →D is a quasistatic adiabatic expansion of an ideal gas, (7.30) can
be used to find TDas
TD=VC
VD(γ−1)
TC=VC
VA0.4
×1500 K(8.34)
All about Heat 185
where the unknown VC/VAcan be found, using the fact that B →Cisa
constant-pressure expansion of an ideal gas, as follows:
VC
TC
=VB
TB
VC
VB
=TC
TB
VC
VA/20 =TC
TB
VC
VA
=TC
20 ×TB
=1500
20 ×994 =0.076 (8.35)
Combining (8.34) and (8.35), we get
TD=0.0760.4(1500 K) = 535 K
Substituting the numerical values of the temperatures and γin (8.32), we
get
ηth =1−1
1.4535 −300
1500 −994=0.668 = 66.8%
Quantity of heat added per kg of air can be calculated using (8.30) as
Qin
m=Cp(TC−TB)= γR
γ−1(TC−TB)
=1.4×8.314
29 ×0.4(1500 −994) kJ/kg = 507.7kJ/kg
The net work output per kg of air can be calculated as
Wnet
m=ηth ×Qin
m=0.668 ×507.7kJ/kg = 339.2kJ/kg
Comment: The cycle studied in this problem is known as the ideal Diesel
cycle. It is an idealized cycle used to understand the working of an engine in
which air contained in a piston-cylinder arrangement is compressed to a high
temperature and ignited by injecting the fuel into the hot air, as in a 4-stroke
car engine fueled by diesel. Also, notice that the efficiency of the ideal Diesel
cycle operating between 1500 K and 300 K is only 66.8%, where as the Carnot
efficiency for the same temperature extremes is 80% (see Example 8.4).
186 Chapter 8
Example 8.7
In a cyclic process, air initially at 1 bar and
300 K is compressed adiabatically to reduce its volume to one tenth of
the initial value. The compressed air is then heated at constant volume
until its pressure increased by 50% of the value at the end of adiabatic
compression. The air in then expanded adiabatically and finally cooled at
constant pressure to its initial state. Sketch the cyclic process on a P-V
diagram. Assuming that all processes are quasistatic and that air behaves
as as ideal gas with γ=1.4, determine the thermal efficiency of the cycle.
Solution to Example 8.7
The cyclic process sketched on a P-Vdiagram is shown in Figure 8.10,
where A →B is the quasistatic adiabatic compression, B →C is the constant-
volume heating, C →D is the quasistatic adiabatic expansion, and D →Ais
the constant-pressure cooling.
P
V
Figure 8.10 The path of the cyclic process of Example 8.7.
A
B
C
D
We can calculate ηth using (8.19), where Qin is the amount of heat added
during the constant-volume heating B →C, and Qout is the amount of heat re-
jected during the constant-pressure cooling D →A. No heat is added or removed
during the adiabatic processes A →BandC→D.
Constant-volume heat addition can be calculated using (8.13) as
Qin =mC
v(TC−TB)(8.36)
All about Heat 187
Constant-pressure heat removal can be calculated using (8.12) as
Qout =−mC
p(TA−TD)=mC
p(TD−TA)(8.37)
Combining (8.19), (8.36) and (8.37), we have
ηth =1−γTD−TA
TC−TB(8.38)
where TA= 300 K, and TB,TCand TDare unknown.
To determine TB,TCand TD, let us tabulate all the known data at the
states A, B, C and D as in Table 8.4.
APA=1bar TA= 300 K VA=?
BPB=? TB=? VB=VA/10
CPC=1.5×PB=? TC=? VC=VB=VA/10
DPD=PA=1bar TD=? VD=?
Table 8.4 Data at states A, B, C and D of the cyclic process.
Since A →B is a quasistatic adiabatic compression of an ideal gas, (7.30)
canbeusedtofindTBas
TB=VA
VB(γ−1)
TA=10
0.4×300 K= 753.6K
Since B →C is a constant-volume process of an ideal gas, ideal gas equation
canbeusedtofindTCas
TC=PC
PBTB=1.5×753.6K= 1130.4K
Since C →D is a quasistatic adiabatic expansion of an ideal gas, (7.30) can
be used to find TDas
TD=PD
PC(γ−1)/γ
TC=PA
1.5×PB(γ−1)/γ
TC
=1bar
1.5×PB(1.4−1)/1.4
×1130.4K
where the unknown PBcould be found as
PB=VA
VBγ
PA=10
1.4×1bar =25.1bar
188 Chapter 8
since A →B is a quasistatic adiabatic process of an ideal gas.
Therefore, we have
TD=1
1.5×25.1(1.4−1)/1.4
×1130.4K= 400.9K
Substituting the numerical values of the temperatures in (8.38), we get
ηth =1−1.4400.9−300
1130.4−753.6=0.625 = 62.5%
8.12 Summary
•Heat does not reside in an object. Heat is simply the energy that transfers
from one object to the other, driven by a temperature difference.
•It is not always necessary to provide heat to increase the temperature of a
system. Work done on a system could also result in temperature increase
of the system (see Example 8.1).
•Heat provided to a simple compressible closed system is related to the
enthalpy increase of the system by the following:
Qin =ΔH−Pf
Po
VdP−Vf
Vo
PdV −Win (8.3)
•For a simple compressible closed system undergoing a constant-pressure
quasistatic process:
Qin =ΔH=mTf
To
CpdT
•For a simple compressible closed system undergoing a constant-volume
process:
Qin =ΔU=mTf
To
CvdT