A.J. WHITE
2014

(1)* MORAN, M.J. &
SHAPIRO, H.N.
FUNDAMENTALS OF ENGINEERING
THERMODYNAMICS
Wiley, 5 th
Edition (SI Units) 2006
VA 192
(2) SONNTAG, R.E.,
BORGNAKKE, C. & VAN
WYLEN, G.J.
(3) ROGERS, G.F.C &
MAYHEW, Y.R.
(4) CENGEL, Y.A. & BOLES,
M.A.
FUNDAMENTALS OF THERMODYNAMICS
Hoboken, NJ. 6 th Edition. 2003
ENGINEERING THERMODYNAMICS: WORK
AND HEAT TRANSFER
Harlow, 4 th Edition. 1992
THERMODYNAMICS: AN ENGINEERING
APPROACH
McGraw-Hill, 5 th Edition. 2006
VA 186
VA 178
VA 188
All of these texts give comprehensive coverage of Part 1A and Part 1B material. Copies are available in the CUED library and college libraries.
1A Thermodynamics 2014 2

Thermodynamics is the science of the relationships between heat , work and the properties of systems .
It brings together two subjects that hitherto you will probably have studied separately, namely
“Mechanics” and “Heat”. The fundamental ideas in thermodynamics are enshrined in four laws: the Zeroth, First, Second and Third Laws of Thermodynamics. We will cover the first three of these, focusing mainly on:
The First Law of Thermodynamics. This is really a statement of the conservation of energy. In fact, the First Law leads to the definition of energy as a property of systems.
The Second Law of Thermodynamics.
This is concerned with the fact that certain processes progress in only one direction. It is thus tied up with the concepts of reversibility and irreversibility . The Second Law leads to the definition of a property of systems known as the entropy. Unlike energy, entropy is not, in general, conserved.
The laws of thermodynamics were established in the 19 th century, their discovery stemming from efforts to understand and improve the performance of heat engines.
Heat engines are devices for converting heat into mechanical work; in the 19 th century they were mainly driven by steam. They will feature prominently in this course along with other similar devices involving heat and work interactions, such as refrigerators and heat pumps. However, the scope of thermodynamics goes far beyond this. Other applications include:
Chemical Processes.
Chemical reactions, and in particular equilibrium relations, are governed by thermodynamic considerations. A typical example is the conversion of chemical energy into electrical work in a fuel cell or battery.
Material Science.
In your Part 1B Materials course you will encounter “phase diagrams” for various materials (such as steel). These are determined by applying the first and second laws of thermodynamics to establish the limits of phase stability.
Biological Systems. Living organisms and biological processes are all subject to the laws of thermodynamics and can be studied by thermodynamic analysis. An example is the "osmotic pressure" that exists between two solutions of different concentrations separated by a semi-permeable membrane (such as a cell membrane).
In short, all systems (at least macroscopic ones) must obey the laws of thermodynamics.
1A Thermodynamics 2014 1

This course follows the classical approach to thermodynamics in which matter is treated as a continuum . We are only concerned here with the properties of macroscopic systems and not with the behaviour of individual atoms or molecules.
Even when applying the methods of calculus to an “infinitesimal” volume, dx × dy × dz , it may usually be assumed that vast numbers of molecules are involved, so that the material within the volume may be considered as having smoothed-out, average properties.
Example 1.1: Avogadro’s hypothesis states that 1 kmole of a gas (which contains 6.023
× 10 26 molecules) occupies a volume of 22.4 m 3 at standard atmospheric pressure and 0°C. How many air molecules are contained within a cubic volume of side 0.001 mm at these conditions?
The behaviour of matter at a molecular level is described by the techniques of kinetic theory and statistical mechanics. For example, in the simplest kinetic theory of gases, molecules are treated as hard spheres (as in the figure below) which collide elastically with each other and with the containing walls. The theory can be used to derive a relationship between the pressure, density and temperature of the gas (i.e., the equation of state ), and to determine the rate of diffusion of one gas into another. Classical thermodynamics cannot do this.
Occasional reference will be made to molecular models to support certain ideas, but for the most part you are encouraged (in this course) to forget atoms and molecules and to think instead in terms of continua.
1A Thermodynamics 2014 2

Many words such as system , work and heat have very precise definitions in thermodynamics.
From the outset it is important to recognise that thermodynamics does not tolerate any sloppiness: you should learn the definitions and adhere to them rigidly!
A thermodynamic system may be defined as an arbitrary geometrical portion of the universe with fixed or movable boundaries which may contain matter or energy or both.
This is a bit too general to be useful, so we will focus, to begin with, on closed systems.
A closed system is a fixed quantity of matter, around which a boundary can be drawn. Everything inside the boundary is the system and everything outside the boundary is the surroundings (or the environment ). Since no matter crosses the boundary, mass conservation is automatically satisfied for a closed system. Energy is allowed to cross the boundary in the form of heat and work (which are yet to be defined). If there is no heat or work exchange between a closed system and its surroundings then it is called an isolated system.
We will follow the tradition within CUED of using the word “system” to mean “closed system”.
Consider a system comprising a fixed mass of gas enclosed in a cylinder by a piston, as shown in the figure below (the dashed line is the system boundary).
This system possesses a number of thermodynamic properties such as pressure, volume and temperature, which together define its thermodynamic state . Thermodynamic properties are quantifiable characteristics of a system that depend only on the state of the system and not on how it arrived at that state.
You are already familiar with the properties pressure and volume, which have obvious physical meanings. Temperature is much more difficult to define precisely, and so this is left until section
6. For the present we simply note that temperature is a measure of the “degree of hotness” of a body. We will use the absolute Kelvin scale, which is related to the Celsius scale by:
T (K) = t (°C) + 273.15
1A Thermodynamics 2014 3

Thermodynamic properties fall into two categories:
Extensive properties.
These depend on the size (or the extent ) of the system. A typical example is volume: if two identical systems are brought together to form a single, larger system, the volume is doubled. Other extensive properties will be introduced in due course.
Intensive properties.
These do not depend on the size of the system. Typical examples are pressure and temperature: if two identical systems are brought together, the pressure and temperature are unchanged.
Specific properties are properties per unit mass and are thus a subset of intensive properties.
For example:
V (m 3 ) – volume (extensive) v (m 3 /kg) − specific volume (intensive).
By convention, small letters denote specific properties and capital letters extensive properties.
The specific volume is of course just the reciprocal of density, v = 1/ ρ .
The number of properties required to define the state of a system depends on its complexity. We will deal mainly with simple compressible systems which are typically comprised of pure substances (e.g. H
2
O) or non-reacting mixtures of gases (e.g. air). The important characteristics of a simple compressible system are that (i) its intensive properties are uniform throughout, and (ii) the effects of electricity, capillarity, gravity and magnetism can be ignored.
For simple compressible systems at rest, two independent intensive properties and the mass are sufficient to define the state, provided the system is in equilibrium .
For example, it is not possible to vary p, T and V independently for the system discussed above; once the temperature and volume are fixed, the pressure is also fixed via the equation of state.
Thus we may write:
p
= p ( T , v ) or
T
= T ( p , v ) or
v
= v ( p , T ) .
Note that the two properties chosen to define the state must be independent. Specifying v and ρ will not do!
The “two property rule” is a consequence of there being only two ways to change the state of a simple compressible system: by heat and work interactions with the surroundings.
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Thermodynamic equilibrium is defined by the following:
A thermodynamic system is in equilibrium when none of its thermodynamic properties are changing in time at a measurable rate.
The laws of thermodynamics can only furnish relationships connecting equilibrium states of a system.
Suppose we change the state of our gas-in-cylinder system by compressing it. This may be represented on a p-v diagram, as shown below. Initially the system is assumed to be in equilibrium at state 1. After moving the piston, the system is left to settle and will reach a new equilibrium at state 2. The laws of thermodynamics can be applied to relate these two end states, but cannot in general tell us much about the intervening process. In fact, it is not generally possible to plot the process on the p-v diagram because pressures and densities may vary from point to point in the system during the process.
If the compression is carried out slowly, however, departures from equilibrium may be kept very small and the process effectively passes through a series of equilibrium states. This is referred to as a quasi-equilibrium or quasi-static process, and it can be plotted on the p-v diagram.
How slow must a process be to qualify as quasi-equilibrium?
This cannot be answered by thermodynamics. The rate of the internal processes that restore equilibrium must be determined by either kinetic theory or by experiment. We will simply note that a number of real engineering processes can be treated as quasi-equilibrium without incurring serious error.
1A Thermodynamics 2014 5

A closed system, by definition, does not exchange matter with its surroundings, but it may exchange energy in the form of either heat or work . The First Law is a general statement of energy conservation. For a closed system it may be written as:
Q
−
W
= Δ
E
Q is the quantity of heat transferred to the system from the surroundings.
W is the work transfer from the system to the surroundings.
Δ E is the change in energy of the system.
IMPORTANT NOTES:
Heat and work are modes of energy transfer only. They do not reside in a system.
Heat and work are therefore not properties of a system.
The SI unit of heat, work and energy is the Joule (J). Rates of heat transfer and rates of work transfer (power) are measured in Joules per second or Watts (1W = 1J/s).
Take care with the sign convention!
Work transfer between a system and its surroundings may take a variety of forms, including mechanical , electrical and magnetic work. The formal definition is as follows:
Work is done by a system on its surroundings if the sole effect of everything external to the system could have been the raising of a weight.
This definition is a little hypothetical (“ could have been ”), so we will return to it later when we discuss electrical work. First we focus on mechanical work transfer which occurs when a system boundary moves by virtue of forces acting at the boundary:
W
=
F
⋅ ds .
Here F is the force (a vector) exerted by the system on an element of the boundary, and ds is the infinitesimal displacement (also a vector) of that element.
1A Thermodynamics 2014 6

“pdV” work
One type of mechanical work that we will encounter frequently is that due to fully restrained expansion of a gas. The force exerted by the pressure of gas acting over the surface of a piston
(see figure) is F = pA . If the piston is displaced to the right a small distance dx , whilst being resisted by an opposing force, the work done by the gas is:
Displacement work: δ W = pAdx = pdV
As the gas expands, the pressure may change, so the total work done between states 1 and 2 is: i.e., the area under the curve in the p-V diagram.
Although the above expression was derived for a piston in a cylinder, it is in fact valid for any geometry. This is because the pressure force is always normal to the system boundary, and when this boundary deforms the change in volume is given by the sum of each elemental area of boundary multiplied by its corresponding normal displacement. (This topic will be covered properly when you study Vector Calculus in next year’s maths course).
1A Thermodynamics 2014 7

Example 2.1: A quantity of dry air is slowly compressed in a cylinder by means of a piston, until its volume is halved. The initial pressure and volume are 1 bar (10 5 Pa) and 1 litre respectively, and during the compression the pressure varies inversely with the volume. Sketch the process on a p-V diagram and calculate the work transfer. Is this a positive or negative quantity?
Unrestrained Expansion
For the “pdV” expression to be valid, the expansion must be fully resisted. This is necessary in order that the pressure remains uniform throughout the system during the process.
If the piston motion were unrestrained, as in the adjacent figure, no useful work would be extracted during the expansion. This is easily shown by drawing the system boundary to enclose the final volume occupied by the gas and piston, as shown.
This system does not change its volume, so no displacement work is done when the piston lock is released. (A similar example is provided by replacing the piston by a fixed diaphragm which is then ruptured.)
1A Thermodynamics 2014 8

Shaft Work
Steam turbines, industrial gas turbines and other rotary “motors” do work by means of a rotating shaft. If the torque exerted by the shaft is T (not to be confused with temperature), then for a small angular displacement d θ ,
δ
W x
=
Td
θ
Electrical Work
When electrical energy crosses a system boundary it is always considered as a work transfer, irrespective of how it is then used. For example, in the left hand figure below, electricity is being supplied to a resistive heating element. However, with the system boundary as shown, the energy could have been used to drive an electric motor and winch (right hand figure). Provided the motor and winch are ideal (no losses) the sole effect is then the raising of a weight.
If the EMF (i.e. the driving voltage) of the electrical supply is ε , then the work done in passing an infinitesimal charge, dQ , around the circuit is:
δ
W e
= ε dQ .
Other Types of Work
We will be concerned chiefly with pdV work and shaft work, but there are a number of other kinds of work and you should be aware of them. Examples include the work required to stretch an elastic wire or to compress a spring (both negative quantities), magnetic work and capillary
(or surface tension) work. In most cases it is possible to write the infinitesimal work transfer in the form:
δ W = ϕ dX , where φ is some intensive property and X is the so-called “conjugate” extensive property. Be careful with the sign!
1A Thermodynamics 2014 9

Heat transfer is energy transfer that is not work. It occurs by virtue of a temperature difference, from hotter bodies to cooler bodies. Remember that heat transfer is defined as positive when it is into a system.
Mechanisms of Heat Transfer
Heat can cross a system boundary by two main mechanisms:
Conduction, which occurs in solids, liquids and gases, and may be thought of as energy transfer from more energetic to less energetic particles due to interactions between them, or (in gases and liquids) as diffusion of high energy particles to regions where they are less concentrated.
Radiation, which is emitted by matter due to changes in electronic configurations of the atoms or molecules. The energy is radiated as electromagnetic waves and can pass through a vacuum.
Convection is often considered as a third mode of heat transfer, but strictly it is just a redistribution of energy within either a system or its surroundings due to fluid motions.
Although Thermodynamics is not concerned with rates, it is worth noting that the rate of heat transfer between two bodies increases with the temperature difference between them. This gives a sufficient description of heat for our current purposes. Next year you will study heat transfer in much greater detail.
Energy, E , is a system property. Traditionally, changes in E are divided into three components:
Δ E = Δ KE + Δ PE + Δ U
is the change in kinetic energy of the system as a whole.
is the change in gravitational potential energy of the system as a whole. is the change in internal energy of the system.
In general, U may include contributions from magnetic, electrical and chemical energy. For a simple compressible system it is composed of random thermal energy of the atoms or molecules, plus any potential energy due to intermolecular force fields.
E, KE, PE and U are all extensive properties. Corresponding specific properties are obtained by dividing by the system mass; e.g. the specific internal energy is u = U / m.
1A Thermodynamics 2014 10

Adiabatic Processes
An adiabatic process is one that involves no heat exchange between system and surroundings. It is an idealised concept, requiring perfectly insulating walls or an infinitely short duration (so that there is no time for heat transfer). Nonetheless, many real processes (e.g., expansion of steam in a turbine) are approximately adiabatic.
The 1 st Law for an adiabatic process (with no changes in KE or PE) gives:
W = −Δ U = U
1
− U
2
Since U is a property, this tells us that the net work done on or by a system during an adiabatic process is independent of the path taken. An example is shown below.
The system can be taken from state 1 to 2 adiabatically either by:
(a) stirring at constant volume followed by expansion, or (b) stirring whilst also expanding at constant pressure. In each case the net work done (expansion work done by the system less shaft work done on the system) is the same.
Cyclic Processes
A cyclic process is one for which the system is returned to its original state. Since E is a property (a function of state), the change in E for a cyclic process is zero, and hence:

dQ
=
 dW , where the 
sign denotes the summation of all infinitesimal energy transfers around the cycle.
The net work output thus equals the net heat input for a cyclic process. Cyclic processes form the basis of many heat engines and heat pumps. Examples of systems undergoing a cyclic process include steam / water in a power station and the refrigerant in a domestic refrigerator.
1A Thermodynamics 2014 11

The laws of thermodynamics are valid for all systems, irrespective of the substances from which they are composed. However, to apply the laws we need to know how the various properties of the system are related to each other. For example, to apply the First Law to the expansion of a gas, a relationship between p, V and U is needed. This information is furnished either from tables (or charts) of properties compiled by experimentation, or from equations of state which may be empirical or may be based on molecular models.
We will deal only with pure substances which are defined to have uniform and unchanging chemical composition. They can, however, exist in different phases ; for example water can exist as ice, liquid water, and steam. A phase is a quantity of matter that is homogeneous throughout in both chemical composition and physical structure.
Examples:
Number of phases
Pure
(Y/N)
Salt water solution (liquid only)
Air at room temperature
A mixture of ice and water
A gin and tonic (ice, no slice)
A freezing saline solution
At this stage we will consider only single phase systems, mainly gases.
For simple systems composed of pure substances the two property rule applies. The empirical relationships between properties are thus conveniently expressed in tabular form. E.g., the specific volume and specific internal energy of ammonia (NH
3
) vapour are tabulated below as functions of pressure and temperature.
T (°C)
0
5
10
15
20
1.0 1.5
Pressure, bar (1 bar = 100 kPa)
2.0 2.5
1.3136
1.3392
1.3647
1.3900
0.8689
0.8864
0.9037
0.9210
0.64648
0.65989
0.67320
0.68640
0.51293
0.52393
0.53482
0.54560
1.4153 0.9382 0.69952
Specific Volume of Superheated Ammonia Vapour, v
0.55630
(m 3 /kg)
3.0
0.42382
0.43323
0.44251
0.45169
0.46078
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T (°C)
0
5
10
15
20
1.0
1341.00
1349.33
1.5
1337.68
1346.25
Pressure, bar (1 bar = 100 kPa)
2.0
1334.29
1343.11
2.5
1330.83
1339.91
1357.64
1365.95
1374.27
1354.78
1363.29
1371.79
1351.87
1360.59
1369.28
1348.91
1357.84
1366.72
Specific Internal Energy of Superheated Ammonia Vapour, u (kJ/kg)
3.0
1327.28
1336.64
1345.89
1355.05
1364.13
Example 3.1:
0.1kg of ammonia gas is maintained at a constant pressure of 2bar in a cylinder by means of a weighted piston. The ammonia is slowly heated with a Bunsen burner flame from an initial temperature of 10°C to a final temperature of 20°C. Calculate:
(a) The change in volume of the gas
(b) The work done during the expansion
(c) The heat addition during the process
Solution:
1A Thermodynamics 2014 13

In the above constant pressure process, for which the only work is “pdV” work, we have:
W
=
p dV
=
= p
Δ
V .
The First Law can thus be written in the form:
Q = Δ U + p Δ V
= Δ
( U
+ pV )
.
Since U, p and V are all properties of the system, the combination U + pV is also a property. It is called the enthalpy and is given the symbol H.
Thus for constant pressure heat addition:
Q
= Δ
H
= m
Δ h , where: H
=
U
+ pV , h
= u
+ pv .
The units of enthalpy are the same as for energy, but enthalpy is not energy. It is merely a shorthand for U + pV , and is introduced because this combination appears very frequently. Its usefulness will become more apparent when we study flow processes.
Heat transfer to a system will usually (though not always) increase its temperature. From this comes the notion of specific heat capacity , which is defined as the heat required to raise the temperature of 1kg of a substance by 1K. It was shown above that for a constant pressure process, Q = Δ H , whereas for a constant volume process (with no shaft work, or motion etc.) the
First Law gives Q = Δ U.
Accordingly, there are two specific heat capacities, defined as follows:
Constant volume (isochoric) specific heat capacity: c v
=
⎝⎜
⎛ ∂ u
∂
T
⎠⎟
⎞ v
Constant pressure (isobaric) specific heat capacity: c p
=
⎛
⎝⎜
∂ h
∂
T
⎞
⎠⎟ p
These definitions are valid for any substance. We will say more about c v relationship between them in the context of ideal gases.
and c p
and the
1A Thermodynamics 2014 14

For many gases at low density (so-called dilute gases ), the average separation of molecules is such that the intermolecular forces are insignificant, except during molecular collisions.
Experiments show that such gases obey the relation: pV = nRT where: n
N
N
A
= Number of kmols (kilogram moles) of the gas, N / N
A
= Number of molecules of the gas.
= Avogadro’s Number = 6.022
× 10 26 kmol − 1
= Universal gas constant = 8.3145 kJ kmol − 1 K − 1 .
This is the ideal gas equation of state . For engineering applications we usually work in terms of mass rather than moles, so for a particular gas of molar mass M (measured in kg kmol − 1 ): pV = mRT or pv = RT or p = ρ RT , where m =
R =
Worked Example 3.2:
Mass of gas =
The nM specific gas constant , R = R / M .
Dry air has a molar composition of 21% O
2
Calculate:
(M=32) and 79% “atmospheric” N
2
(M=28.15).
(a) The specific gas constant for air
(b) The mass of 1 litre of air at 1.013 bar and 25°C.
(c) The density of air at these conditions.
Solution:
(a) Molecular mass of air, M = 0.21 × 32 + 0.79 × 28.15 = 28.96 kg kmol − 1
∴ Specific gas constant, = 8.3145
× 10 3 / 28.96 = 287 J kg − 1 K − 1 . (Note units)
(b) In SI units, V = 0.001 m 3 , p = 1.013
× 10 5 Pa, T = 298.15 K.
Substituting these values into the ideal gas equation: m = pV / RT = 1.013
× 10 5 × 0.001 / (287 × 298.15) = 0.00118 kg
(c) Dividing the mass by the volume gives: ρ = m / V = 1.18 kg m − 3
(or we could use p = ρ RT directly).
(about 1.2g)
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Because intermolecular forces are negligible for ideal gases, there is no potential energy associated with molecular interactions. This means that the internal energy is independent of the separation between molecules, and thus depends only on the temperature. A similar argument applies to the enthalpy. Thus, for all ideal gases: u
= fn( T ) h
= fn( T ) or
Δ u = u
2
− u
1
Δ h
= h
2
− h
1
=
T
2
T
1 c v dT
,
=
T
2
T
1 c p dT where the right hand equations follow from the definitions of c v
and c p
respectively.
In these expressions, c v
and c p
will in general be functions of temperature. Ideal gases for which the specific heat capacities vary with temperature are called semi-perfect gases . Ideal gases for which the specific heat capacities are constant are called perfect gases .
Perfect gases only :
Δ u = c v
Δ T =
Δ h
= c p
Δ
T
= c v Δ ( pv ) c
R
R p Δ
( pv )
Over a limited range of temperature and pressure, the gases listed below behave as perfect gases
(i.e., they obey pv = RT , and have constant specific heat capacities).
Gas Molar Mass, M
(kg/kmol)
Gas constant, R ,
(kJ/kg K) c p
(kJ/kg K) c v
(kJ/kg K)
A
He
Air
N
2
* (Atmospheric N
2
)
N
2
O
2
H
2
CO
CO
2
SO
2
CH
4
C
2
H
6
40
4
29.0
28.15
28
32
2
28
44
64
16
30
0.208
2.080
0.287
0.295
0.297
0.260
4.120
0.297
0.189
0.130
0.520
0.277
NOTE: This data is also on p. 12 of the Thermofluids Data Book.
0.52
5.19
1.005
1.033
1.04
0.92
14.2
1.04
0.82
0.61
2.23
1.75
0.31
3.11
0.718
0.738
0.74
0.66
10.08
0.74
0.63
0.48
1.71
1.47
1.67
1.67
1.40
1.40
1.40
1.40
1.41
1.40
1.31
1.26
1.31
1.19
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Relationships between c p
and c v
:
Careful inspection of the above table reveals that:
(i) c p
− c v
= R exactly true for all ideal gases
(ii) c p c v
=
⎩⎪
1.67
for monatomic gases
1.40
for diatomic gases approximately true for most perfect gases
See Examples Paper 1/4 for the proof of (i).
The ratio c p
/ c v
crops up frequently and is given the symbol γ ( gamma ).
A Simple Molecular Model (aside):
The two values of γ given above can be derived from a simple statistical mechanical result which effectively states that: c v
=
R
×
DOF
2 where DOF is the number of “degrees of freedom” of the molecule. A monatomic molecule
(Helium, Argon, Neon etc.) has three degrees of freedom corresponding to translational kinetic energy in each of the x, y and z directions. Diatomic molecules have two additional degrees of freedom associated with rotational kinetic energy about two mutually perpendicular axes.
Combining the above expression for c v
with c p
− c v
= R gives:
γ =
DOF
+
2
DOF
This doesn’t really work for more complicated molecules because other energy modes (such as vibration and electronic excitation) come into play.
1A Thermodynamics 2014 17

Other important relations are:
H
=
U
+ pV h
= u
+ pv c v c p
=
⎛
⎝⎜
=
⎛
⎝⎜
∂
∂ u
T
∂ h
∂
T
⎞
⎠⎟
⎞
⎠⎟ p v
γ = c p c v c p
− c v
=
R definition of enthalpy for all gases definition of gamma ideal gases only u
= c v
T
+ u
0 h
= c p
T
+ u
0
=
= c
R v c p
R pv
+ u
0 pv
+ u
0
=
=
1
γ −
1 pv
+ u
0
γ
γ − 1 pv
+ u
0 perfect gases only
Note that the arbitrary constant u
0
is the same for perfect gas enthalpy and internal energy!
** WARNING:
Think carefully about whether it is appropriate before applying ideal / perfect gas relations!
1A Thermodynamics 2014 18

The aim of this section is to make use of the perfect gas relations to compute heat and work transfers and energy changes during simple processes. This serves not only to demonstrate application of the First Law but also forms the basis of the so-called air-standard cycles , which provide an approximate model of spark ignition and Diesel engines and of gas turbines. Once again we will neglect kinetic and potential energy changes (and magnetism, capillarity etc.) so that the First Law reduces to Q − W = Δ U.
Isobaric means constant pressure . If the only work is “pdV” work (i.e., no stirring) then:
W
=
p dV
= p ( V
2
−
V
1
)
= mR ( T
2
−
T
1
)
Δ U = mc v
( T
2
− T
1
)
Q = W + Δ U = mc p
( T
2
− T
1
)
Note that the expression for Δ U involves c v
even though this is not a constant volume process.
Make sure you understand why this is so – it is occasionally a source of confusion.
Isochoric means constant volume . In the absence of stirring:
W
=
p dV
=
0
Δ
U
= mc v
( T
2
−
T
1
)
Q
= Δ
U
= mc v
( T
2
−
T
1
)
Note the logical order in computing these terms: the heat transfer can often only be found by applying the First Law, once W and Δ U are known.
Isothermal means constant temperature . Using the ideal gas equation thus gives an inverse relation between p and V (as in example 2.1).
Assuming only fully resisted “pdV” work:
W
=
p dV
=
mRT dV
V
Δ
U
=
0
= mRT ln
⎛
⎝⎜
V
2
V
1
⎞
⎠⎟
Q
=
W
= mRT ln
⎛
⎝⎜
V
2
V
1
⎞
⎠⎟
Note: the unrestrained adiabatic expansion of an ideal gas is also isothermal (in the sense that the initial and final temperatures are equal), but there is no heat or work transfer.
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Example 4.1: 1 kg of air, initially at a temperature of 300K, undergoes the following cycle:
(i) Expansion at constant pressure until the volume is doubled.
(ii) Cooling at constant volume.
(iii) Isothermal compression to the initial state.
Plot the cycle on a p-v diagram and calculate (a) the heat and work transfers and internal energy changes during each stage of the cycle, (b) the net work produced and (c) the net work as a fraction of the positive heat addition .
Solution:
From Data Book, for air: R = 287 J / kg K p c v
= 718 J / kg K 1 2
(a) (i) w
=
p dv
=
Δ u
= c v
( T
2
−
T
1
) q
= w
+ Δ u p ( v
2
− v
1
)
=
=
=
=
R ( T
287
2
×
−
T
1
)
(600
−
300)
718
×
300
86.1
+
215.4
=
86.1 kJ v
= 215.4 kJ
=
301.5 kJ
1A Thermodynamics 2014 20

Notes:
(i) Because the process is cyclic, Δ U overall is zero and the net heat transfer balances the net work produced.
(ii) This is an example of a heat engine – i.e., a device that converts heat into work. It is, however, extremely impractical and very inefficient: most of the positive heat input is rejected to the environment.
Adiabatic processes (processes for which Q= 0) are very important in thermodynamics. This is partly because many processes happen so quickly that there is no time for heat transfer. Furthermore, it is often desirable to minimise heat exchanges with the environment and so many devices are thermally insulated. Consider a system comprising a perfect gas undergoing an infinitesimal adiabatic expansion or compression. The First Law gives:
δ Q = dU + δ W = 0 .
If the process is quasi-equilibrium then the temperature and pressure are uniform throughout the system and it is possible to write dU = mc v dT and δ W = pdV. Thus: mc v dT
+ pdV
=
0 .
Using p = mRT/V and dividing through by mT yields: c v dT
T
+
R dV
V
=
0 .
Noting that R/c v
= γ− 1 , integrating and taking inverse logs gives:
⎝⎜
⎛
T
2
T
1
⎠⎟
⎞
=
⎝⎜
⎛
V
V
2
1
⎠⎟
⎞
γ−
1
=
⎝⎜
⎛ v v
2
1
⎠⎟
⎞
γ−
1
=
⎛
⎝⎜
ρ
2
ρ
1
⎠⎟
⎞
γ−
1 or Tv
γ− 1 = constant.
Two similar relationships are obtained from this by straightforward substitution of pv = RT :
⎛
⎝⎜ p
2 p
1
⎞
⎠⎟
⎛
⎝⎜ p
2 p
1
⎞
⎠⎟
=
⎛
⎝⎜ v
1 v
2
⎠⎟
⎞
γ
=
=
⎛
T
2
⎝⎜ T
1
⎞
γ /( γ− 1)
⎠⎟
⎛
⎝⎜
ρ
2
ρ
1
⎞
γ
⎠⎟ or or pv
γ = constant. p / T
γ
/(
γ−
1) = constant.
Notes:
(i) These equations are in the Thermofluids Data Book where they are referred to as isentropic changes (we will later show that a quasi-equilibrium, adiabatic process has constant entropy).
(ii) Beware the assumptions involved: quasi-equilibrium , adiabatic and perfect gas !
1A Thermodynamics 2014 21

Worked Example 4.2: Filling of an Evacuated Vessel
A completely evacuated bottle of volume 0.1 m 3 is to be filled with helium from a main where the pressure is 5 bar and the temperature is 300K. The bottle is connected as shown, the valve is opened, helium enters the bottle and the valve is shut when the pressure in the bottle is 5 bar. If the bottle is well lagged so there is no heat loss to the surroundings, calculate the final temperature in the bottle and the mass entering. Assume helium behaves as a perfect gas with
R=2.08 kJ/kg K and γ = 1.67.
Solution:
With the system boundary as shown (note: no mass crossing the boundary), work is done on the system by the surrounding gas in the main. The pressure remains constant at p
1
on the moving boundary, so
Applying the 1 st Law:
∴
Rearranging: 1.67
× 300 = 501 K
We now simply apply the ideal gas law to the final state to find the mass of gas that has entered the bottle:
∴ = 0.048 kg
1A Thermodynamics 2014 22

Notes:
(i) Although adiabatic, this is not a quasi-equilibrium process, so only the end states can be related. This is because the helium expands into the empty bottle in an unrestrained manner.
(ii) The large temperature rise (201 K) comes from the work done by the gas in the main. This reduces the density and thereby restricts the amount of helium entering.
(iii) In practice there would probably be some heat transfer, so if the valve were left open long enough the helium would cool back down to 300 K and more of it would enter the bottle.
However, Thermodynamics cannot tell us how long this would take (this is the province of Heat
Transfer which you will study next year).
A compression or expansion for which the pressure and volume are related by: is known as a polytropic process. The parameter n is called the polytropic index and is assumed to remain constant during the process. Usually n will be in the range 1.0 to γ . For perfect gases:
(i) n = γ represents an isentropic process, as derived in section 4.4.
(ii) n = 1 represents an isothermal process (section 4.3).
(iii) 1 < n < γ usually (not always) represents a process with heat transfer. In this case the above relationship is purely empirical and cannot be derived from first principles.
(Note also that n = 0 is an isobaric process, and n = ∞ is isochoric.)
The work done on or by a system during a polytropic process is easily computed by integration:
W
=
=
2
1
p dV
=
2
1
p
1
V
1
− p
2
V
2 n
−
1 k
V n dV
=
⎡ k
V 1
− n
1 − n
⎤
⎦
V
2
V
1 where the last expression comes from noting that k = p
1
V
1 n = p
2
V
2 n .
(valid for )
To calculate the heat transfer we then apply the 1 st Law, noting that Δ U =
3.8 and examples paper 4, question 6)
P
2
V
2
− P
1
V
1
γ − 1
(see section
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The Second Law of Thermodynamics is concerned with reversibility and irreversibility and the distinction between heat and work. It is useful to first clarify these concepts with an example of an irreversible process.
Consider the unrestrained expansion of a gas, as discussed in section 2.2 and as illustrated by process (i) in the figure below. Assume that the container walls are perfectly insulating. The process is thus adiabatic and there is no work transfer so, according to the 1st Law, the internal energy in state 2 is the same as that in state 1.
Now consider how the system can be returned to its initial state. This might be achieved by the following processes:
(ii) Adiabatic compression back to the initial volume. This involves doing work on the system and therefore increases the internal energy. Thus, although the volume is restored, the gas in state 3 will be hotter than before.
(iii) Cooling at constant volume. The piston is now locked in place and (part of) the insulation removed to allow the gas to cool down to its original temperature. State 1 is now fully restored.
It is important to realise that although the system has been returned to its initial state, the surroundings have not: work has been extracted from them and an equal amount of heat has been received back in return. In fact, there is no conceivable way of restoring the system to its original state without affecting the surroundings, and so process (i) is said to be irreversible .
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There are two well-known statements of the Second Law of Thermodynamics (and many not so well-known ones).
The Kelvin-Planck Statement
It is impossible to construct a cyclic device whose sole effect is to produce positive work whilst receiving heat from a single thermal reservoir.
A “heat reservoir” is a body which acts as a source or sink of heat.
In the idealisations that are common in thermodynamics, it is often assumed to have infinite thermal capacity, so that its temperature remains constant as it gives out or receives heat. In practice, of course, heat sources are usually provided by burning a fuel or by a nuclear reaction.
The Kelvin-Planck statement is depicted in the adjacent figure. It tells us that we cannot convert all the heat from, say, burning a fuel into useful work: there is a limit to the efficiency of this process. Note that the reverse process – complete conversion of work into heat – is perfectly acceptable, as exemplified by the cyclic process described in 5.1, or by any frictional dissipation process.
The Clausius Statement
It is impossible to construct a cyclic device whose sole effect is the transfer of heat from a cooler to a hotter body.
NB In order to conform to the sign convention, one of the Qs in the adjacent figure should be negative and the other positive. However, in sections 5, 6 and 7 of these notes we will adopt the notation that energy transfers are positive in the direction of the arrows in the figures . Care must then be taken when writing the equations, but this is on the whole much less confusing than trying to stick rigidly to the convention.
The Clausius statement is perhaps more intuitive than the K-P statement, as it clearly describes the inherent irreversibility of nature. That heat cannot spontaneously pass from a cold to a hot body is a matter of common experience. Note, however, that if there is a work input to the cyclic device, it is perfectly possible to transfer heat from the cold to the hot reservoir: this is exactly what refrigerators and heat pumps do.
Notes:
(i) Neither of these impossible processes would violate the First Law.
(ii) The two statements are equivalent (violation of one implies violation of the other).
1A Thermodynamics Lent 2014 25

Thermal Efficiency:
A heat engine is a cyclic device for converting heat into work.
According to the Kelvin-Planck statement, complete conversion is not possible and some heat input must be rejected.
The figure shows the simplest configuration of heat engine, which operates between a high temperature source at T
1
and a low temperature sink at T
2
. Since the device is cyclic, there is no net change in its internal energy, so the First Law gives:
W
=
Q
1
−
Q
2
.
A concept of obvious practical importance for heat engines is the thermal (or cycle ) efficiency . This is defined as the fraction of the heat input that is converted to useful work:
η = what you get (useful work) what you pay for (heat input)
=
W
Q
1
.
Making use of the above expression for W , the thermal efficiency may be written in the alternative form:
η =
1
−
Q
2
Q
1
This form is often easier to compute in practice.
Example: A Power Station
A practical example of a cyclic heat engine is a power station. Most power stations are based on the steam cycle which you will study in detail next year. Briefly, however, it comprises the components shown below, with the working substance (water / steam) receiving heat at high
(but not constant) temperature in the boiler, and rejecting heat to the environment at low temperature from the condenser. The turbine and pump are more or less adiabatic.
The typical thermal efficiency of such a power station is about 40%. Is this good or bad?
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The Carnot Cycle
The French physicist Sadi Carnot was the first to establish the theoretical limits of efficiency for heat engines. He reasoned that, to achieve maximum efficiency, all processes within the cycle must be reversible, and that all heat transfer to the cycle should be at the temperature of the hot reservoir, whilst all heat transfer from the cycle should be at the temperature of the cold reservoir. The figures below provide an example of a Carnot cycle in which the working substance is a perfect gas.
The cycle consists of the following processes, all of which are assumed to be reversible:
A → B. The gas is brought into contact with a heat source at temperature T
1
and expands isothermally.
B → C. The heat source is removed and the expansion continues adiabatically until the temperature is T
2
.
C → D. The gas is brought into contact with a heat sink at temperature T
2
and is compressed isothermally.
D → A. The heat sink is removed and the gas is returned to its initial state by adiabatic compression.
Worked Example 5.1: Calculate the thermal efficiency of a Carnot cycle that uses a perfect gas as the working fluid and operates between thermal reservoirs at T
1
= 427°C and T
2
= 27°C.
Note that all four of the processes in the cycle involve work transfer, whereas only A → B and
C → D involve heat transfer. It is therefore easiest to compute the efficiency from η = 1 − Q
2
/Q
1
.
1A Thermodynamics Lent 2014 27

Process A → B (isothermal):
Process C → D, similarly:
W
AB
=
p dV
=
mRT
1
V dV
= mRT
1 ln( V
B
/ V
A
)
Δ
U
AB
=
0
Q
1
= Q
AB
= W
AB
= mRT
1 ln( V
B
/ V
A
)
Q
2
= −
Q
CD
= mRT
2 ln( V
C
/ V
D
)
The efficiency is thus: η =
1
−
Q
2
Q
1
=
1
− mRT
2 ln( V
C mRT
1 ln( V
B
/ V
D
)
/ V
A
)
However, processes B → C and D → A are adiabatic and reversible (which means they are also quasi-equilibrium) so, using Tv γ -1 = constant:
∴
( V
C
/ V
B
)
=
( T
1
/ T
2
)
1/( γ− 1) =
( V
D
/ V
A
)
( V
C
/ V
D
)
=
( V
B
/ V
A
)
Substitution into the expression for efficiency gives finally:
Carnot Efficiency: η = 1 −
T
2
T
1
Notes:
(i) For the given reservoir temperatures (700K and 300K) we find that the efficiency is 57%.
Thus 43% of the heat must be rejected to the heat sink and is effectively wasted. It would only be possible to extract further work from this waste heat if an even colder reservoir were available.
(ii) The expression for the Carnot efficiency depends only upon the reservoir temperatures and does not involve gas properties such as c p
and γ . In fact, it turns out that the same expression for efficiency would have been obtained even if the working substance had not been a perfect gas.
Two Theorems Concerning Cyclic Heat Engines
Theorem 1: The maximum efficiency of a cyclic heat engine operating between two thermal reservoirs is attained when the cycle is reversible.
Theorem 2: All reversible heat engines operating between the same thermal reservoirs are equally efficient.
1A Thermodynamics Lent 2014 28

Example 5.2: Prove Theorem 1 by showing that violation of this theorem is equivalent to violation of the Kelvin-Planck statement of the Second Law.
Note that Theorem 2 can be proved in a similar manner.
Reversible Cyclic Heat Engines
Since all reversible cyclic heat engines operating between the same two fixed temperatures have the same efficiency, and since a Carnot engine is such a reversible engine, it follows that:
The maximum thermal efficiency of a cyclic heat engine operating between T
1 and T
2
is
η = 1 −
T
2
T
1
, and this efficiency is attained when the cycle is reversible.
Furthermore, for all cyclic heat engines operating between T
1
and T
2
:
η =
1
−
Q
2
Q
1
.
Comparison of the two expressions for η shows that:
For all reversible cyclic heat engines operating between T
1 and T
2
:
Q
Q
2
1
=
T
2
T
1 or
Q
1
T
1
=
Q
2
T
2
This relationship is very important and we will return to it to define entropy later.
1A Thermodynamics Lent 2014 29

Refrigerators and heat pumps are devices for extracting heat from a cold thermal reservoir and delivering (a greater quantity of) heat to a hotter thermal reservoir. Such devices do not violate the Clausius statement of the Second Law because they have a work input (as shown in the figure).
The only distinction between the two types of device is that the purpose of a refrigerator is to remove as much heat as possible from a cold space, whereas the purpose of a heat pump is to deliver as much heat as possible to a hot space.
For a refrigerator, we pay for the work done and we want to maximise Q
2 defined as:
. Accordingly, the coefficient of performance is
Refrigerator: COP
R
= heat from cold space work input
=
Q
2
W
=
Q
1
Q
2
−
Q
2
For a heat pump, we pay for the work done and we want to maximise Q
1 coefficient of performance is defined as:
. Accordingly, the
Heat Pump: COP
P
= heat to hot space work input
=
Q
W
1
=
Q
1
Q
1
−
Q
2
=
1
+
COP
R
The COP of a fridge is usually greater than 1, the COP of a heat pump is always greater than 1.
It needs to be substantially greater than 1 to be worthwhile. Why?
Note: In past Tripos papers (prior to 2006), the COP of heat pumps is referred to as the performance energy ratio (PER) . This terminology is not used in text books or in industry.
As with heat engines, the best performance of a heat pump (or fridge) is achieved when the cycle is reversible. Similar arguments apply to those discussed in section 5.3 and, in particular, for reversible cycles :
Q
1 =
Q
2 ,
T
1
T
2 which is easily applied to find the maximum COP of a fridge or heat pump (see Examples 1/4).
A common pastime of small boys and misguided inventors is to attempt to outwit nature by devising perpetual motion machines. Perpetual Motion Machines of the First Kind are magical devices which violate the First Law, creating energy from nothing. Perpetual Motion Machines of the Second Kind (i.e. which break the Second Law) are more subtle, but nonetheless very useful. Such a device could draw heat from, say, the ocean and provide an endless supply of electricity. Despite the vast profits that stand to be made, it is recommended that you don't spend too much time on this.
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Much has been made of reversibility in the preceding sections, and it is obviously important to have a clear idea of what constitutes a reversible process.
A process is reversible if the system and its surroundings can be returned to their initial state.
An example of a reversible process is shown in the figure below which depicts a copper block exchanging heat with a thermal reservoir. An infinitesimal temperature difference is maintained between the block and the reservoir so that heat flows into the block. By increasing the temperature of the block very slightly, to T + δ T , the heat flow is reversed. Because δ T must be infinitesimally small, the process is very slow.
A common feature of all reversible processes is that they must not involve any substantial departures from equilibrium. Reversible processes are therefore quasi-equilibrium processes and must take place “slowly”. For example, we have frequently used the pdV expression for work done by a gas expanding in a cylinder. For this to be valid, the pressure must be uniform throughout the cylinder. Pressure non-uniformities are evened out by means of pressure waves that travel up and down the cylinder at the speed of sound. Thus, a necessary condition for reversibility is that the speed of the piston must be small compared to the speed of sound. In fact, this gives us considerable scope. (Try it for yourself: the speed of sound in air at 15°C is about 340 m/s. Compare this with the speed of a typical car engine piston with a stroke of 100 mm at an untypically high engine speed of 6000 RPM.)
Irreversible processes, by contrast, are characterised by non-uniformities
(gradients) in some property that subsequently decay via a dissipative mechanism. E.g. non-uniformities in temperature decay through conduction and non-uniformities in chemical concentration decay through diffusion.
In the adjacent figure, the two copper blocks have a finite temperature difference and are brought into thermal contact. The heat transfer clearly occurs irreversibly; once equilibrium has been restored, one would not expect a spontaneous flow of heat back to the top block.
Irreversibility invariably entails a lost opportunity for doing useful work.
For example, the two copper blocks could have been brought to equilibrium whilst driving a reversible heat engine, thereby by producing positive work.
1A Thermodynamics Lent 2014 31

In the preceding sections, we have used temperature as a property of systems, noting that it is a measure of the “hotness” of a body, and that heat tends to flow from high temperature to low temperature bodies. This can scarcely be considered a thorough definition. The purpose of this section is therefore to give temperature a firmer footing, and in particular to show that the
Second Law permits the definition of a thermodynamic temperature scale which is quite independent of the properties of any thermometric substance.
When two bodies are brought into thermal contact, it is observed that there is, in general, a change in the properties of those bodies. For example, if a hot copper block is brought alongside a cold one, the electrical resistance of the hot block is seen to fall, whereas that of the cold block rises. After a time, no further change occurs and the bodies are said to be in thermal equilibrium.
It is an observable fact that:
If systems B and C are each separately in thermal equilibrium with system A, then they would be in thermal equilibrium with each other if brought into thermal contact.
This statement might seem both trivial and blindingly obvious, but it is impossible to prove, and without it temperature cannot be defined and the whole subject of
Thermodynamics would collapse. It has therefore been granted the status of a Thermodynamic Law. Since the
First Law had already been established (following on from the Second Law) it clearly had to be called the
Zeroth Law of Thermodynamics.
Bodies that are in thermal equilibrium must have some common property, which we recognise as being the temperature.
1A Thermodynamics Lent 2014 32

An empirical temperature scale is established by:
(i) Selecting a thermometric substance for which some property (the thermometric property ) varies with temperature in a “well-behaved” fashion. Typical examples would be the length of a mercury column in a glass capillary tube or the electrical resistance of a fine platinum wire.
(ii) Specifying the values of temperature at two fixed, reproducible points. For example, centigrade scales are set by fixing the ice and steam point temperatures (the freezing and boiling temperatures of pure water at standard atmospheric pressure) to 0° and 100° respectively.
Intermediate readings of temperature will, however, depend on the selected thermometric property. For example, if the mercury column and platinum resistance are both assumed to vary linearly with temperature, then they will indicate 50° for bodies which are actually at slightly different temperatures.
The perfect gas temperature scale is an empirical scale based on the (absolute) pressure of a sample of gas maintained at constant volume, as shown in the figure.
This scale was considered superior to other empirical scales because many different gases would give the same readings for temperature, provided the density of the gas was sufficiently low.
Furthermore, temperature defined in this manner has a clear physical interpretation on the basis of kinetic theory of gases: it is proportional to the mean translational kinetic energy of molecules.
1A Thermodynamics Lent 2014 33

Absolute Zero
The perfect gas scale permits the definition of a non-arbitrary absolute zero of temperature. To see this, we suppose that “perfect gas” temperature, θ , is proportional to the pressure of a sample of gas maintained at constant volume:
θ ∝ p
(Note that we use “ θ ” because “T” is reserved for thermodynamic temperature.) It follows that:
θ
θ s i
= p s p i
, where “s” and “i” denote the steam and ice points respectively. If we now set the size of the unit of temperature by choosing and extrapolate back between the steam and ice points, the “absolute zero” is found, as shown in the graph.
When this process is carried out with different gases, slight differences emerge, but the results collapse onto a single line when progressively smaller amounts of gas (and hence lower absolute pressures) are used. In the low pressure limit, absolute zero is found to lie 273.15 (degrees) below the ice point.
1A Thermodynamics Lent 2014 34

Despite its advantages, the perfect gas scale still relies on the properties of gases. The Second
Law, however, allows us to define a temperature scale that is completely independent of any thermometric substance. The starting point for this is the fact that all reversible heat engines operating between the same fixed temperatures have the same efficiency. Consequently,
Q
1
Q
2
= fn ( T
1
, T
2
) (reversible heat engine)
Example 6.1: Show that the above function must take the form :
Q
Q
2
1
=
Φ ( T
1
Φ
( T
2
)
)
The choice of the function Φ ( T ) is arbitrary, and so we choose the simplest form consistent with our notion of “hotness” – i.e. Φ ( T ) = T. This gives a definition of “thermodynamic temperature” which is numerically identical to perfect gas temperature:
Q
1
Q
2
=
T
1
T
2
=
θ
1
θ
2
A single fixed point is required to define the size of the temperature unit (the Kelvin), and this is taken to be the triple point (not the ice point) of pure water. The triple point is the point at which ice, water and steam coexist in equilibrium. It is given the exact value of 0.01°C, or
273.16 K on the thermodynamic scale.
1A Thermodynamics Lent 2014 35

Measurement of thermodynamic temperature is hampered by the fact that no heat engine is perfectly reversible. The constant volume gas thermometer provides accurate measurements over a wide range of temperatures, but it is a bulky and cumbersome device, and cannot be applied in cases where the gas either liquefies or dissociates.
The following five devices provide practical means of temperature measurement, and you should be able to explain how each of these works.
1.
Mercury-in-glass thermometers.
2.
Thermocouples. These consist of a junction between two dissimilar metals, such as platinum-rhodium or copper-constantan. The junction generates an e.m.f. (a voltage) which is temperature dependent.
3.
Platinum resistance thermometers.
The resistance of platinum increases with temperature in a very precise manner. Thermometers of this type may also be made small and responsive.
4.
Thermistors. These are small pieces of semi-conductor whose resistance varies exponentially with temperature. Over certain ranges they are thus very sensitive.
5.
Optical pyrometers.
These devices rely on the relationship between temperature and the wavelength of emitted radiation. They are used at very high temperatures.
To assist with the calibration of such devices, an International Practical Scale of
Temperature has been established, based on a number of fixed points. In all, there are 11 fixed points, ranging from the triple point of hydrogen ( − 259.34°C) up to the normal freezing point of gold (1064.43°C).
1A Thermodynamics Lent 2014 36