chapter 7 thermodynamics - universiti tunku abdul …staff.utar.edu.my/limsk/engineering...
TRANSCRIPT
Thermodynamics 101
Chapter 7
Thermodynamics
_____________________________________________
7.0 Introduction
Thermodynamics is a branch of physics, which deals with the energy and work
of a system, was first studied in the 19th
century as scientists were first
discovering how to build and operate steam engines. Thermodynamics deals
with the large-scale response of a system in microscopic change, which can be
observed and measured in experiments. Small-scale gas interactions are
described by the kinetic theory of gasses, which is a compliment to
thermodynamics.
In this chapter, the laws of thermodynamic shall be discussed in great detail.
7.2 Review the Basis of Thermodynamic
Let’s review some basis of thermodynamic. They are covered in the following
paragraphs.
Heat is the mechanism by which energy is transferred between system and
its environment because there is temperature difference between them.
Internal energy U is the energy associated with the microscopic
component of the system such as atom and molecule, when view from reference
frame of rest with respect to the system. It includes kinetic energy and potential
energy associated with the random translational, rotational, and vibrational
motion of the atoms or molecules and intermolecular forces.
For an ideal monoatomic gas, the internal energy is just the translational
kinetic energy of the linear motion of the atom. For polyatomic gases, there is
rotational and vibrational kinetic energy as well. In liquids and solids there is
potential energy associated with the intermolecular attractive forces. A
simplified visualization of the contributions to internal energy is shown in Fig.
7.1.
Thermodynamics 102
Figure 7.1: Internal energy contribution of gas, liquid, and solid
If the water is placed in a pot and heated by flame, the internal energy of water
increases due to the heat transferred from the flame to the water. Comparison of
internal energy for copper and water is shown in Fig. 7.2.
Figure 7.2: Comparison of internal energy for copper and water
To set-up a temperature scale, one has to pick a reproducible thermal
phenomenon and arbitrary assigns a certain Kelvin temperature to its
environment. This standard fixed-point temperature is select at the triple point
of water. The triple point temperature T3 of water is T3 = 273.16 K. The triple
point temperature is the temperature where liquid water, solid ice, and water
vapor can coexist in thermal equilibrium at a fixed set value of pressure and
temperature. The triple point temperature can be measured using a constant –
volume gas thermometer shown in Fig. 7.3, filled with different type of gas such
as hydrogen, helium or nitrogen. At constant volume, the pressure P of the in
thermometer is dependent on the temperature of the liquid where it is immersed.
Thermodynamics 103
Figure 7.3: Constant-volume gas thermometer
Thus,
T = Cp (7.1)
where C is a constant and P the pressure is is equal to P = Po - h g. Po is the
atmospheric pressure which is 1.01x105 pascal Pa (Newton/m
2) or 760 torr or
14.7 lb/in2. At triple temperature the pressure shall be P3, and then an equation
relating it is
!!"
#$$%
&'
3P
PK15.273T (7.2)
However, it was found that at boiling point of water, different gas gives
different result. Thus, it is necessary to measure the triple point temperature
when the volume in the thermometer approaches zero, where is now
independent of the gas type. Equation (7.2) shall then be changed to
!!"
#$$%
&'
(3
0V P
PlimK15.273T . Figure 7.4 shows the result of the temperature-pressure plot
of constant-volume thermometer filled with helium, nitrogen, and hydrogen.
Thermodynamics 104
Figure 7.4: Result of temperature-pressure plot of constant-volume thermometer filled with
three-gas type
There are two ways that the unit of heat is being defined. In British system, it is
the British thermal unit BTU, in which one BTU is defined as the heat required
to raise the temperature of 1.0 lb of water from 630 F to 64
0 F. Alternatively,
one calorie is the energy required to raise the temperature of 1.0 g water from
14.50 C and 15.5
0 C.
In 1948, scientist linked the heat like work defined as the energy transfer
and defined one calorie c as equals to 4.186 J, which is equal to 3.969x10-3
BTU.
This shall mean 1.0 BTU = 251.9 calorie = 1,954.67 J.
The specific heat cm of a material is defined as the amount of energy
necessary to raise the temperature of one kilogram of that material by one
degree Celsius. In mathematical expression it is equal to )TT(m
Qcm
if )' , where
m is the mass, Q is the heat or energy.
Figure 7.5 shows the specific heats of some materials at room temperature.
Specific Heat Substance
cal/g-K J/kg-K
Lead 0.0305 128
Tungsten 0.0321 134
Silver 0.0564 236
Copper 0.0923 386
Aluminum 0.215 900
Brass 0.092 380
Thermodynamics 105
Granite 0.19 790
Glass 0.20 840
Ice (-100C) 0.530 2220
Mercury 0.033 140
Ethyl alcohol 0.58 2430
Sea water 0.93 3900
Water 1.00 4190
Figure 7.5: specific heats of some materials at room temperature
7.3 Thermal Expansion
When the temperature of material increases, its internal energy increases. This
result its dimensions increase due to increase in inter-atomic or intermolecular
distance that can be caused by vibration, rotational moment, and etc.
If the temperature of metal rod of length L is raised by temperature *T, its
length is increased by an amount following equation (7.3).
*L = L+*T (7.3)
where + is the coefficient of linear expansion, which has unit per Kelvin or per
degree Celsius.
The volume of the material will also be increased with the increase of
temperature, whereby the increase of temperature follows equation (7.4).
*V = V,*T (7.4)
where , is the coefficient of volume expansion. The relationship between
coefficient of linear expansion and coefficient of volume expansion is , = 3+.
Example 1 On a hot day in Las Vegas, an oil tanker loaded 37,000 liters of diesel fuel. The
driver unloaded the entire load of diesel in Payson, Utah, where the temperature
is 25.0 K lower than in Las Vegas. How many liters did he deliver? What is the
percentage increase of the cost price per liter of diesel? Given that the
coefficient of volume expansion of diesel is 9.54x10-4
/0C.
Thermodynamics 106
Solution
The change in volume of diesel in Payson is 37,000x9.54x10-4
x(-25) = -882.45
liters. In Payson the driver delivers 37,000 – 882.45 = 36,117.55 liters of diesel.
The increase of the cost of diesel is (37,000/36,117.55 –1)100 % = 2.44 % per
liter.
7.4 Heat Transformation (Latent Heat)
When a matter absorbs energy, it is not necessary used to raise the temperature
of the material. The energy may be used to change from one phase or state to
another. All matters can exist in three common phases. i.e. solid, liquid, and gas
states. Take for an example, the temperature of solid ice and ice water is the
same, and the temperature of boiling water and steam is the same. The energy
absorbed by the solid ice is used to break the intermolecular force so that the
molecules can become loosely bonded. This energy is termed as heat of fusion.
It has unit J/kg. Boiling water absorbs energy to break the weak intermolecular
bond into free molecular gas. The energy required is termed as heat of
vaporization. Figure 7.6 shows the heat of fusion and heat of vaporization of the
some substances.
Melting Boiling
Substance Melting Point
K
Heat of Fusion
kJ/kg
Boling
Point K
Heat of
Vaporization kJ/kg
Hydrogen 14.0 58.0 20.3 455
Oxygen 54.8 13.9 90.2 213
Water 273 333 373 2,256
Mercury 234 11.4 630 296
Lead 601 23.2 2,017 858
Copper 1,356 207 2,868 4,730
Silver 1,235 105 2,323 2,336
Figure 7.6: Heat of fusion and heat of vaporization of some materials
From the table, it shows that the heat of vaporization is higher than heat of
fusion, which is logically true because it requires less energy to break the solid
bond into loose bond than to free the atom or molecule from inter-atomic or
intermolecular bonds.
Thermodynamics 107
Example 2
How much heat must be absorbed by ice of mass m = 720 g at – 100C to take it
to liquid state at 150C?
SolutionThree steps are to be considered which are shown in the diagram below.
The energy required to raise the temperature of solid ice from – 10
0C to 0
0C is
10x0.720x2,200J/kg-K = 15.84x103 J.
The energy required to break the intermolecular bond to loose bond is the heat
of fusion, which is equal to 0.72x333x103 = 239.8x10
3 J.
The energy required to heat ice water from 00C to 15
0C is 0.72x15x4,190 =
45.25x103 J.
The total energy required is 300.80x103 J.
7.5 Heat Transfer Mechanism
Heat can transfer between a system and its environment via three mechanisms,
which are conduction, convection, and radiation.
Conduction
Conduction is transfer of kinetic energy and vibrational energy of atoms to the
atom of lower kinetic energy and vibrational energy. The rate of conduction
Pcond, which is also equal to dt
dQ follows equation (7.5).
Thermodynamics 108
L
TTAP CH
cond
)' k (7.5)
where k is the thermal conductivity, L is the thickness of the transfer material,
and A is the face contact area. Figure 7.7 illustrates the conduction.
Figure 7.7: Thermal conduction between hot reservoir and cold reservoir via a slab
Thermal resistance R is defined as
R = k
L (7.6)
Equation (7.5) shall become R
TTAP CH
cond
)' after substituting equation (7.6). If
there are two insulating layer of difference thermal conductivity k, it can be
shown that the rate of conduction is 21
CHcond
RR
TTAP
-
)' , where R1 and R2 are
respectively the thermal resistance of material 1 and material 2.
If two materials are connected in parallel with the ends contact the hot and
cold reservoirs then the rate of conductor Pcond(total) shall be equal to sum of rate
of conduction Pcond(1) of material 1 and rate of conduction Pcond(2) of material 2.
Thus, equation (7.5), the rate of conduction Pcond(total) is
Pcond(total) = 1
CH1
R
TTA
) +
2
CH2
R
TTA
) (7.7)
Thermodynamics 109
Example 3
The wall of house is made of white pine, unknown materials, and a brick wall.
The thickness Ld of brick wall is two times the thickness of white pine La. The
thermal conductivity kd of the brick
is five times the thermal
conductivity ka of white pine. The
temperature of indoor T1 is = 250C.
The temperature T2 = 200C, T5 = -
100C. If steady state transfer of heat
is attained, what is the temperature
T4 of the interface between brick
and the unknown material?
Solution
Since steady state of transfer of heat is attained, the conduction rate of white
pine is same as the brick wall, which isa
21cond
L
TTAP
)' ak and
d
54
condL
TTAP
)' dk .
Thus,
a
21cond
L
TTAP
)' ak
a
54
d
54
L2
TTA5
L
TTA
)'
)' ad kk . The equation is now contains one
unknown T4 and T4 is found to be – 80C.
Convection
Convection is occurred when liquid or gases come in contact with object whose
temperature is higher than that of liquid or gas. The gas or liquid comes in
contact hot object. Its temperature increases and becomes less dense. It buoyant
force causes it to rise. The surround liquid or gas is then flowed in to take place
of rising warm liquid or gas. The process continues is termed convection. An
illustration is shown in Fig. 7.8.
Figure 7.8: An illustration of convection of thermal heat transfer
Thermodynamics 110
Radiation
The transfer of heat between a system and its environment via electromagnetic
waves is called thermal radiation. A very good example of thermal radiation is
visible light. The rate of thermal radiation Prad at which an object emits energy
via electromagnetic radiation depends on surface area A, temperature of that
area and emissivity . that has a value between 0 and 1, which surface dependent.
A surface that has maximum emissivity of 1 is said to be a blackbody radiator.
The rate of radiation Prad follows equation (7.8).
Prad = (7.8) 4AT/.
where / is the Stefan-Boltzmann constant that has value 5.67x10-8
J/s-m2-K
4.
Indeed Stefan-Boltzmann constant is a universal constant that applies to all
bodies, regardless of the nature of surface.
The rate of absorption Pabs of an object from its environment, which has
uniform temperature Tenv, is
Pabs = (7.9) 4
envAT/.
Owing to the fact that the object radiates energy to the environment and absorbs
from the environment, the net rate of radiation Pnet shall be
Pnet = 0 14
env
4 TTA )/. (7.10)
Example 4 The supergiant star Betelgeuse in constellation Orion has a surface temperature
about 2,900 K and emits a radiant power of approximately 4.0x1030
W. The
temperature is about half and the power is about 10,000 greater than that of the
sun. Assuming that Betelgeuse and sun have perfect emissivity and spherical
shape, calculate the radii of the supergiant and sun.
Solution
From Stenfan-Boltzmann equation, Prad = , the surface area A is A=4AT/.4
rad
T
P
/..
The radius R of supergiant shall be R = 4
rad
T4
P
2/.=
48
30
2900x10x67.5x4
10x0.4)2
=
2.82x1011
m, which is larger than the orbit of mar, which is 2.28x1011
m.
Thermodynamics 111
The radius of sun is R = 4
rad
T4
P
2/.=
48
26
5800x10x67.5x4
10x0.4)2
= 7.04x108 m.
Example 5 A wood-burning stove stands unused in a room where the temperature is 18
0C.
A fire is started inside the stove. Eventually the temperature of the stove surface
reaches a constant temperature of 1980C and the room warms to a constant
temperature of 290C. The stove has an emissivity of 0.9 and a surface area of
3.50 m2. Determine the net radiant power generated by the stove when the stove
(a) is unheated and has temperature equal to the room temperature (b) has
temperature of 1980C.
Solution
The power generated by unheated stove is Prad = = 5.67x104AT/. -
8x0.9x3.5x291
4 = 1,280.75 W.
The power absorbed from the surrounding by the stove is Prad = =
5.67x10
4AT/.-8
x0.9x3.5x2914 = 1280.75 W. The net power generated by the stove is
zero.
The net power generated by the stove when it is heated is Pnet =
0 14
env
4 TTA )/. = = 7304.1 W. )302471(10x67.5x5.3x9.0 448 ))
7.6 Thermodynamic Processes
A thermodynamic process is the way that a system changes from one state of
thermal equilibrium to another state such as the gas may be in thermodynamic
equilibrium for specified temperature, pressure, and volume. Heat transfer and
work done will change this equilibrium state to another.
Thermodynamic process can be divided into quasi-static (reversible)
processes and irreversible processes. If a thermodynamic system undergoes a
change from one state in thermal equilibrium to another state slowly enough
such that in any instant of time the system is in thermal equilibrium is said to be
reversible process. In reality this process does not exist. Any thermal process
that is not a quasi-static process is an irreversible process. In nature, all
thermodynamic processes are irreversible.
There are four special types of quasi-static processes that are useful in
equilibrium thermodynamics. The processes are characterized by the
Thermodynamics 112
thermodynamic parameter that is kept constant during process while other
parameters change. The types are: isothermal process where the temperature of
the system is kept constant, isobaric process where the pressure is kept constant,
isochoric process where the volume is kept constant and adiabatic process or
isentropic process where no heat is transfer in or out of the system from or to
the environment.
7.7 Laws of Thermodynamics
There are three principal laws of thermodynamics, which are zeroth law – the
thermodynamic equilibrium law, first law - internal energy law, second law -
entropy law. Each law leads to the definition of thermodynamic properties,
which help to understand and predict the operation of a physical system. There
is another law, which is third law state that it is impossible to reach zero Kelvin
temperature in finite number steps. This law shall not be discussed.
7.8 Zeroth Law of Thermodynamics
Zeroth law of thermodynamic involves some simple definitions of
thermodynamic equilibrium and temperature. It is observed that some properties
of an object, like the pressure in a volume of gas, the length of a metal rod, or
the electrical conductivity of a wire, can change when the object is heated or
cooled. Some of these phenomenons have been demonstrated in the earlier
section.
Zeroth law states that if two systems are at the same time in thermal
equilibrium with a third system, they are in thermal equilibrium with each other.
If the two objects are initially at different temperatures into physical contact,
they will eventually achieve thermal equilibrium. During the process of
reaching thermal equilibrium, heat is transferred between the objects and there
is a change in the property of both objects such as their internal energy.
Thermodynamic equilibrium leads to the large-scale definition of temperature.
When two objects are in thermal equilibrium they have the same temperature.
The details of the process of reaching thermal equilibrium are described in the
first and second laws of thermodynamics.
As an illustration in Fig. 7.5, object A and object B are in physical contact
and in thermal equilibrium. Object B is also in thermal equilibrium with object
C. There is initially no physical contact between object A and object C. But, if
object A and object C are brought into contact, it is observed that they are in
Thermodynamics 113
thermal equilibrium. This simple observation allows making of thermometers,
in which it can be calibrated to measure the change in a thermal property such
as the length of a column of mercury by putting the thermometer in thermal
equilibrium with a known physical system. If the thermometer is brought into
thermal equilibrium with any other system such as placing under tongue, the
temperature of the other system is known by noting the change in the thermal
property. Objects in thermodynamic equilibrium have the same temperature.
Figure 7.5: Illustration of Zeroth law of thermodynamic
7.9 First law of thermodynamics
First law of thermodynamics relates the various forms of energy, kinetic and
potential, in a system to the work, which a system can perform and to the
transfer of heat. It is used extensively in the discussion of heat engines.
The first law of thermodynamics defines the internal energy E is equal to
the difference of the heat transfer Q into a system and the work W done by the
system.
*U = U2 - U1 = Q – W (7.11)
Heat Q is absorbed by the system uses to increase the internal energy U2 – U1.
The internal energy is used to perform work done W, resulting decrease of
internal energy. This is the rationale how equation is formed.
Heat removed from a system would be assigned a negative sign in the
equation. Similarly work done on the system is assigned a negative sign. Figure
7.7 shows the interpretation of the law.
Thermodynamics 114
In chemistry textbook, first law is written as *U= Q + W. It is the same law,
of course - the thermodynamic expression of the conservation of energy
principle. It is just that W is defined as the work done on the system instead of
work done by the system.
In the context of physics, the common scenario is one adding heat to a
volume of gas and using the expansion of the gas to do work, as in case pushing
down of a piston in an internal combustion engine. In the context of chemical
reactions and process, it may be more common to deal with situations where
work is done on the system rather than by the system.
Figure 7.7: Interpretation of first law of thermodynamic
State 2 shows that there is decrease in volume, therefore work done is negative.
Work done W is also defined as . It is clearly shown that if V3'2
1
v
V
pdVW 2 < V1,
the work done is negative.
Example 6 The temperature of three moles of a monoatomic ideal gas is reduced from
temperature Ti = 540 K to Tf = 350 K by two different methods. In the first
method 5,500 J of heat flows into the gas, while in the second method, 1,500 J
of heat flows into it. In each case find the change of internal energy and (b)
work done by the gas.
SolutionSince the gas is monoatomic type, the internal energy of the gas is only the
translational kinetic energy, which follows equation nRT2
3. The change in
Thermodynamics 115
internal energy from temperature 540 K to 350 K shall be )TT(nR2
3i)f =
)540350(31.8x32
3) = -7,105 J.
First method shall yield W = Q - *U = 5,500 J +7,105 J = 12,605 J
Second method shall yield 1,500 J + 7,105 J = 8,605 J.
Let’s understand the relationship of specific heat capacity and first law of
thermodynamic and also look at some special process cases that applying first
Law of Thermodynamics.
7.9.1 Specific Heat Capacities and First Law of Thermodynamics
As mentioned in the earlier section the heat capacity is defined as Q = Cm*T.
The heat capacity for a gas shall then be Q = Cn*T. Instead of mass, it is now
replaced by number of mole.
The internal energy U of an ideal monoatomic gas is U = nRT2
3. The
change of internal energy *U from temperature Ti to Tf shall be *U =
)TT(nR2
3if ) .
For isobaric process, the work done is W = P(Vf – Vi). However, for ideal
gas PV = nRT. This shall mean that the work done is also W = nR(Tf – Ti). For
isochoric process, the work done is equal to zero since there is no volume
change.
Using the first law equation Q = *U + W, for isochoric process Q =
)TT(nR2
3if ) . This implies that the specific heat C is equal to R
2
3. We shall use
CV to denote specific heat for constant volume.
Similarly the heat supplied for isobaric process is equal to Q =
)TT(nR2
3if ) + nR(Tf – Ti). This shall mean that the specific heat capacity is
R2
5. We shall use CP to denote specific heat for constant pressure.
Thermodynamics 116
7.9.2 Thermodynamic Process
Let’s look at several thermodynamic processes in detail here.
Adiabatic process
The Adiabatic process is a process that occurs so rapidly that there is no transfer
of heat between the system and its environment. From first law of
thermodynamics with Q = 0 shows that the change in internal energy is in the
form of work done. Therefore, from equation (7.6) become U2 - U1 = – W.
Free expansion is an adiabatic process whereby there is no transfer of heat
between the system and its environment, no work done on and by the system.
Thus, from equation (7.6), Q = W = 0.
For adiabatic condition PV4 = constant K, where 4 is the ratio of Cp/Cv heat
capacity at constant pressure and constant volume. This implies that P = K/V4
and the work done shall be 3 4'
2
1
v
V
dVV
KW =
4)
) 4)4)
1
)VV(K 1
i
1
f . The pressure-volume
diagram is shown in Fig.7.8.
Figure 7.8: The work done graph of an adiabatic process
Note that for an ideal monatomic gas, 4 value is equal to 5/3.
Isochoric Process
Isochoric process is a constant volume process that there is no change in volume.
Thus, the work done is zero. Therefore, equation (7.6) becomes U2 - U1 = Q,
Thermodynamics 117
which is equal to CVn*T. From the ideal heat equation V*P = nR*T. This shall
mean Q = )PP(VR
CVif ) . The pressure-volume diagram is shown in Fig.7.9.
Figure 7.9: The work done graph of an isochoric process
Cyclical process
Cyclical process is a process after certain exchange of heat and work done, the
system restores back to its initial state. Thus, there is no change in its internal
energy. This implies that equation (7.6) becomes Q = W.
Isothermal Process
The isothermal process is a process whereby the temperature is kept constant
that is no change of internal energy. This implies that equation (7.6) becomes Q
= W. The work done is . Substituting P = 3'2
1
v
V
pdVWV
nRT into the equation
3'2
1
v
V
dVV
nRTW = nRT !!
"
#$$%
&
i
f
V
Vln . The pressure- volume diagram is shown in Fig.
7.10.
Figure 7.10: The work done graph of an isothermal process
Thermodynamics 118
Example 7
Two moles of the monatomic gas argon expand isothermally at temperature 298
K from an initial volume of Vi = 0.025 m3 to a final volume of Vf = 0.050 m
3.
Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the
change in the internal energy of the gas and (c) the heat supplied to the gas.
Solution
The work done by isothermal process is = nRTW !!"
#$$%
&
i
f
V
Vln = 2x8.31x298 !
"
#$%
&025.0
050.0ln
= 3,432.9 J.
The internal energy of U = nRT2
3. Since the temperature is constant, the change
of kinetic energy is zero.
Using first law equation, the heat supplied Q is 3,432.9 J.
Isobaric process
Isobaric process is process where the pressure is kept constant. The work done
W is W = P V = P(Vf – Vi). T = PV/(nR), thus the change in temperature T =
)VV(nR
Pi)f . The heat absorbed shall be Q = Cp )VV(
R
Pi)f . The PV diagram is
shown in Fig. 7.11.
Figure 7.11: The work done graph of an isobaric process
7.9.3 Thermodynamics Potentials
There are four thermodynamic potentials that are useful in the chemical
thermodynamics of reactions and non-cyclic processes. They are internal energy,
enthalpy, Helmholtz free energy, and Gibbs free energy.
Thermodynamics 119
Helmholtz free energy F states that F = U – TS, where T is the absolute
temperature and S is the final entropy. The system in an environment of
temperature T, energy can be obtained by spontaneous heat transfer to and fro
between environment and the system. Helmholtz free energy is a measure of the
amount of energy that put in to create a system once the spontaneous energy
transfer to the system from the environment is accounted.
Enthalpy is H defined as H = U + PV. It is analogue to first law of
thermodynamic Q = U + P V. It is a useful quantity for tracking chemical
reactions. In an exothermic reaction, energy is released to a system. It has
shown up in some measurable forms in terms of the state variables. An increase
in the enthalpy H = U + PV may be associated with an increase in internal
energy which can be measured by calorimeter, or with work done by the system,
or a combination of the two. If the process changes the volume, as in a chemical
reaction which produces gas, then work must be done to produce the change in
volume. For a constant pressure process the work done to produce a volume
change V is P V.
Gibbs free energy G states that G = U – TS + PV, where P is the absolute
pressure and V is the final volume. As discussed in enthalpy, an additional
amount of work PV must be done if the system is created from a very small
volume into large volume system. As discussed in Helmholtz free energy, an
environment at constant temperature T will contribute an amount TS to the
system, reducing the overall energy necessary for creating the system. This net
energy contribution for a system created in environment with temperature T
from an initial very small volume is Gibbs free energy.
The change in Gibbs free energy G, in a reaction is a very useful
parameter. It can be treated as the maximum amount of work obtainable from a
reaction. For example, in the oxidation of glucose, the change in Gibbs free
energy is G = 686 kcal = 2,870 kJ. This reaction is the main energy reaction in
living cells.
Example 8 Electrolysis of water in hydrogen and oxygen is a very good example for the
application of the above mentioned thermodynamic potentials. This process is
presumed is done at temperature 298K and one atmosphere pressure, and the
relevant values are taken from a table of thermodynamic properties shown
below.
Thermodynamics 120
The process must provide the energy for the dissociation and the energy to
expand the produced gases. Both of those are included in the change in enthalpy
which is shown in the table.
Quantity H2O H2 0.5 O2 Change
Enthalpy -285.83 kJ 0 0 H = 285.83 kJ
Entropy 69.91 J/K 130.68 J/K 0.5 x 205.14 J/K T S = 48.7 kJ
At temperature 298K and one atmosphere pressure, the system work is W =
P V = (101.3 x 103 Pa)(1.5 moles)(22.4 x 10
-3 m
3/mol)(298K/273K) = 3,715.0 J
Since the enthalpy H= U + PV, the change in internal energy U is then equal to
U = H - P V = 285.83 kJ - 3.72 kJ = 282.1 kJ.
This change in internal energy must be accompanied by the expansion of the
gases produced, so the change in enthalpy represents the necessary energy to
accomplish the electrolysis. However, it is not necessary to put in this whole
amount in the form of electrical energy. Since the entropy increases in the
process of dissociation, the amount T S can be provided from the environment
at temperature T. The amount must be supplied by the battery, which is actually
the change in the Gibbs free energy. i.e. G = H - T S = 285.83 kJ - 48.7 kJ =
237.1 kJ.
Since the electrolysis process results in an increase in entropy, the environment
contributes energy amounted to T S. Gibbs free energy tells the amount of
energy in other forms must be supplied to get the process to proceed.
Thermodynamics 121
7.10 Second Law of Thermodynamics
Ice cream melts and a cold can of soda get warm-up when they are left in hot
day. They never get colder when left in hot day. This spontaneous flow of heat
is the focus of one of the most profound law of science, which is the second law
of thermodynamics.
They are several ways that second law of thermodynamic can be defined.
Based on the above examples, the first statement of the second law of
thermodynamics - heat flows spontaneously from a hot to a cold body.
An explanation for this form of the second law can be obtained from
Newton's laws and the microscopic description of how heat flow through
conduction occurs when fast atoms collide with slow atoms, transferring some
of their kinetic energy in the process. One might wonder why the fast atoms
don't collide with the cool atoms and subsequently speed up, thereby gaining
kinetic energy as the cool atoms lose kinetic energy - this would involve the
spontaneous transfer of heat from a cool object to a hot object, in which it is the
violation of the second law. From the laws of conservation of momentum and
energy, in a collision between two objects, the faster object slows down and the
slower object speeds up.
It is possible to make a cool object in a warm place cooler like the
refrigerator but this involves the input of some external energy. As such, the
flow of heat is not spontaneous in this case.
The second form of the second law of thermodynamic states heat cannot be
completely converted into other forms of energy that places some practical
restrictions on the efficiency like internal combustion and steam powered
engines.
The third statement of second law of thermodynamic states that there exists
useful state variable called entropy, which is defined as the measure of number
of state variable for a system at a given time. The change in entropy *S is equal
to the heat transfer *Q divided by the temperature T.
*S = Sf – Si = *Q/T = 3f
iT
dQ 5 AvgT
Q (7.12)
Thermodynamics 122
The entropy of the system and the environment will remain a constant if the
process is reversible. If the initial and final states are denoted by Si and Sf then
Sf = Si for reversible system. An example of a reversible process would be
ideally forcing a flow through a constricted pipe. Ideal means no losses. As the
flow moves through the constriction, the pressure, temperature, and velocity
would change, but these variables would return to their original values
downstream of the constriction. The state of the gas would return to its original
conditions and the change of entropy of the system would be zero.
The second law also states that if the physical process is irreversible, the
entropy of the system and the environment MUST increase. The final entropy
must be greater than the initial entropy. An example of an irreversible process is
a hot object put in contact with a cold object. Eventually, they both attain the
same equilibrium temperature. If objects are separated after attaining thermal
equilibrium, they do not naturally return to their original different temperature
states. The process of bringing them to the same temperature is irreversible.
7.10.1 Heat Transfer from Hot Object to Cold Object
Let’s look at how the second law describes why heat is transferred from the hot
object with temperature T1 to the cold object with temperature T2. If heat is
transferred from the hot object to the cold object, the amount of heat transferred
is Q and the final equilibrium temperature for both objects is Tf. The
temperature of the hot object changes as the heat is transferred away from the
object. The average temperature Th of the hot object during the process is the
average of T1 and Tf. Similarly, for the cold object, the final temperature of the
cold object is Tf. The average temperature Tc during the process is the average
of Tf and T2. The entropy change for the hot object will be (-Q/Th), with the
minus sign applied because the heat is transferred away from the object. For the
cold object, the entropy change is (Q/Tc), positive value because the heat is
transferred into the object. So according to equation (7.12), the total entropy
change for the whole system would be given by the equation.
Sf - Si = -Q/Th + Q/Tc (7.13)
where Si and Sf being the final and initial values of the entropy.
Temperature Th is greater than temperature Tc, because T1 is greater than
T2. The term Q/Tc will always be greater than -Q/Th and therefore, Sf will be
greater than Si, as the second law predicts.
Thermodynamics 123
If the heat was being transferred from the cold object to the hot object, then
the final equation would be Sf = Si + Q/Th -Q/Tc. The signs on the terms would
be changed because of the direction of the heat transfer. Th would still be greater
than Tc, and this would result in Sf being less than Si. The entropy of the system
would decrease, which would violate the second law of thermodynamics.
7.10.2 Heat Engines
A heat engine is a device that uses heat to perform work. Essentially it has three
features. It has a hot reservoir that supplies heat. Part of the heat is used to
perform work by the working substance of the engine like gasoline- air mixture
in the automobile engine. The remaining part of the heat is rejected at
temperature lower than the input temperature called cold reservoir. The
schematic of a heat engine is shown in Fig. 7.12.
Figure 7.12: A schematic representation of a heat engine
For an engine to be efficient, it must produce a relatively large amount of work
from the input heat. The efficiency e of a heat engine is defined as the ratio of
the work done by the engine to the input heat QH. i.e. e = HQ
W. However, QH is
also equal to QH = QC + W, thus, the efficiency is
Thermodynamics 124
e = H
C
Q
Q1) (7.14)
Example 9 An automotive engine has an efficiency of 22.0% and produces 2,510 J of work.
How much heat is rejected?
Solution
From equation (7.12), The heat rejected is QC = QH(1- e). But QH is equal to QC
+ W. Thus, QC = (QC + W)(1 - e). This implies that QC = W/e - W =
2,510(1/0.22 - 1). = 8,899.0 J.
7.10.3 Carnot Engine
According to a French engineer Sadi Carnot, in order to get maximum
efficiency for an engine, the process within the engine must be reversible. This
shall mean that the process will return to its initial states with no wasteful
transfer of energy. Figure 7.13 shows the pressure-volume plot of the Carnot
engine.
Figure 7.13: Pressure-volume diagram of a Carnot engine
During the process step a to b, which is an isothermal expansion, heat QH is
absorbed by the working substance from the hot reservoir. Since the process is
an isothermal expansion process, the work done by the system is also equal to
heat absorbed. During the process step c to d, the working substance is releasing
heat QC to the cold reservoir. Since this is an isothermal compression, work is
Thermodynamics 125
done by the environment on the system. From concept of heat engine, there is
no heat transfer between hot and cold reservoir directly. Thus, the process step
bc and de must be the adiabatic process. The work done by the Carnot process
shall be the orange color area of the graph shown in Fig. 7.13.
According to first law of thermodynamic, *U = Q –W. Since the process is
reversible, therefore the change of internal energy *U is zero. Also for
differential change dQ = dW. The work done W shall be W = QH – QC.
From process step a to b there is a positive change of entropy *SH = QH/TH
and during the process step c to d, there is a negative change of entropy *SL =
QL/TL. There is no change of entropy for process step bc and da because there is
not heat absorbed or released. The change of entropy shall be *S = QH/TH -
QL/TL. Since Carnot engine is a reversible engine, the change of entropy shall
be zero. This implies that QH/TH = QL/TL. From the equation TH > TC, this shall
mean QH > QC. This analysis confirms the second statement of second law of
thermodynamic.
The efficiency of the Carnot engine shall be e = W/QH = (QH – QC)/QH = 1-
TL/TH.
7.10.4 Refrigeration
Refrigeration is a device that uses work to transfer energy from low temperature
reservoir to high temperature reservoir. Figure 7.14 shows the schematic of a
refrigerator.
Figure 7.14: The schematic of a refrigerator
Thermodynamics 126
A measure of the efficiency of a refrigerator is the coefficient of performance K,
which is defined as the ratio of heat extracted from the cold reservoir to the
work done. For an ideal Carnot refrigerator the coefficient of performance K
=LH
L
Q
). The change of entropy for ideal refrigerator shall be *S =
H
L
L
L
T
Q
T
Q-) .
Notice that TH >TL, implying the *S is a negative value, which is not allow for
second law of thermodynamic. Therefore, there is no perfect refrigerator.
Thermodynamics 127
Tutorials
1. A certain diet doctor encourages people to diet by drinking ice water. His
theory is that the body must burn off enough fat to raise the temperature of
water from 0.00C to the body temperature of 37.0
0C. How many liters of ice
water would have to be consumed to burn-off 454 g of fat, assuming that this
much fat required 3,500 Cal be transferred to ice water? Why this is not
advisable to follow this diet.
2. A tank of water has been outdoor in cold weather and a slab of ice 4.0 cm
thick has formed on its surface. The air above the ice is –100C. Calculate the
rate (centimeter per sec) of ice formation. Take the thermal conductivity of
the ice to be 0.0040 cal/s-cm-K and density 0.92 g/cm3. Assume the energy
is transferred through the wall or bottom of the tank is negligible.
3. Find the change of entropy when a 2.3 kg block of ice melts slowly
(reversible) at 273 K.
4. 1200 J of heat flowing spontaneously through a copper rod from a hot
reservoir at 650 K to a cold reservoir at 350 K. Determine the amount the
change in entropy for this irreversible process.
The change in entropy shall be -1200/650 + 1200/35 = 1.58 J/K.
5. A Carnot engine operates between temperature TH = 850 K and TL = 300 K.
The engine performs 1200 J of work for each cycle that takes 0.2 s.
Thermodynamics 128
(a). Find the efficiency of the engine.
(b). How much heat QH is extracted from high temperature reservoir for
each cycle? And QL delivered to cold reservoir?
(c). What is the entropy change of the working substance for the energy
transfer to it from the high temperature reservoir? From it to the low
temperature reservoir?
6. If an inventor claims that he has invented an engine that has an efficiency of
75% when operated between boiling point and freezing point. Is it possible?