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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