# THERMAL PHYSICS 2012

```BUSE
Virtual Open and
Distance Learning
STUDY GUIDE
PH002
THERMAL PHYSICS
Bindura University of Sc ience Education
Physics and Mathematics Department
BUSE
BUSE-NDLOVU.S. 2012
Bindura University of Sc ience Education
Physics and Mathematics Department
Bindura University Of Science Education
Private Bag 1020
Zimbabwe
Fax: +263 079 715035
E-mail: info@buse.ac.zw
Website: www.www.buse.ac.zw
BUSE
Acknowledgements
The Bindura University of Sc ience Education Physics and Mathematics Department wishes to thank
those below for their contribution to this STUDY GUIDE:
PH002
Contents
1
How this STUDY GUIDE is structured ........................................................................... 1
Course overview
3
Welcome to THERMAL PHYSICS PH002 ..................................................................... 3
THERMAL PHYSICS PH002 is this course for you? ..................................................... 3
Course outcomes ............................................................................................................... 4
Timeframe ......................................................................................................................... 4
Study skills ........................................................................................................................ 4
Need help? ........................................................................................................................ 6
Assignments ...................................................................................................................... 6
Assessments ...................................................................................................................... 7
Getting around this STUDY GUIDE
8
Margin icons ..................................................................................................................... 8
Unit 1
9
Temperature:
.......................................................................................................................................... 9
Introduction ............................................................................................................. 9
1.1 Temperature scales
1.11.2 Practical thermometers……………………………………………………………...1
1.2.3 Thermistor thermometer……………………………………………………………..1
1.2.4 Thermocouple thermometer Centigrade / Celsius scale
1.1.2 Thermodynamic / Absolute scale
.................................................................................................................................................... 9
1.2 Practical thermometers
1.2.1 Liquid – in – glass thermometer
1.2.2 Resistance thermometer
.................................................................................................................................................. 10
1.2 Practical thermometers
ii
Contents
1.2.5 Constant volume gas thermometer
Unit summary ................................................................................................................. 13
Assignment ..................................................................................................................... 14
Assessment...................................................................................................................... 15
Unit 2
17
[Thermal properties of materials ................................Error! Bookmark not defined.
Introduction ........................................................................................................... 17
2.1 Internal energy……………………………………………………………………………..1
2.2 Zeroth law of thermodynamics
………………………………………………………….Error! Bookmark not defined.
2.3 First law of thermodynamics …………………………………………………………….1
2.4 Heat capacity……………………………………………………………………………….1
2.5 Specific heat capacity .........................................................Error! Bookmark not defined.
2.5.1 Determination of the specific heat capacity of a solid by electrical methods…………...1
2.5.2 Determination of the specific heat capacity of a liquid by electrical methods…………..1
2.6 Latent heat………………………………………………………………………………….1
2.7 Specific latent heat…………………………………………………...…………………….1
2.7.1 Determination of the specific latent heat by electrical methods…………………………1
Unit summary ................................................................................................................. 31
Assignment ..................................................................................................................... 31
Assessment...................................................................................................................... 35
Unit 3
39
Ideal gases ......................................................................Error! Bookmark not defined.
Introduction ........................................................................................................... 39
3.1 Gas laws ………………………….………………………………………………………..1
3.1.1 Boyle’s law……………………..……………………………………………………1
3.1.2 Charles’ 1st law………………………………………………………………………1
3.1.3 Charles’ 2nd law / pressure law……………………………………………………...1
3.2 Ideal gas equation / equation of state .................................Error! Bookmark not defined.
PH002
3.2.1 Applications of the ideal gas equation
3.3 Kinetic theory of gases
3.2.1 Assumptions
3.2.2 Derivation of the kinetic theory equation
3.2 Kinetic theory of gases
3.2.3 Applications of the kinetic theory equation
3.3 Kinetic theory of gases
3.2.3 Root mean square
3.2.4 Kinetic energy of (a) molecule(s) ...............................Error! Bookmark not defined.
3.3 Kinetic theory of gases
3.2.5 Mean kinetic energy of (a) molecule(s)………………………………………………..1
3.4 Work done by an ideal gas,
3.5 Calculations of work done from p – V graphs
3.5.1 Isothermal changes
Unit summary ................................................................................................................. 60
Assignment ..................................................................................................................... 61
Assessment...................................................................................................................... 65
Unit 4
69
Thermal energy transfer ..............................................Error! Bookmark not defined.
Introduction ........................................................................................................... 69
4.1 Thermal equilibrium
...............................................................................Error! Bookmark not defined.
4.2 Thermal conduction
4.3 Thermal Convection
............................................................................................................................... 80
iv
Contents
Unit summary ................................................................................................................. 83
Assignment ..................................................................................................................... 84
Assessment...................................................................................................................... 85
PH002
PH002 THERMAL PHYSICS has been produced by Bindura University
of Sc ience Education. All STUDY GUIDEs produced by Bindura
University of Sc ience Education are structured in the same way, as
outlined below. The information was obtained from notes given by my
general physics and chemistry textbooks.
How this STUDY GUIDE is
structured
The course overview
The course overview gives you a general introduction to the course.
 If the course is suitable for you.
 What you will already need to know.
 What you can expect from the course.
 How much time you will need to invest to complete the course.
The overview also provides guidance on:
 Study skills.
 Where to get help.
 Course assignments and assessments.
 Activity icons.
 Units.
We strongly recommend that you read the overview carefully before
The course content
The course is broken down into units. Each unit comprises:
 An introduction to the unit content.
 Unit outcomes.
 New terminology.
 Core content of the unit with a variety of learning activities.
 A unit summary.
 Assignments and/or assessments, as applicable.
1
Temperature:
Resources
For those interested in learning more on this subject, we provide you with
a list of additional resources at the end of this STUDY GUIDE; these
may be books, articles or web sites.
After completing PH002 we would appreciate it if you would take a few
moments to give us your feedback on any aspect of this course. Your
 Course content and structure.
 Course reading materials and resources.
 Course assignments.
 Course assessments.
 Course duration.
 Course support (assigned tutors, technical help, etc.)
Your constructive feedback will help us to improve and enhance this
course.
2
PH002
Course overview
Welcome to PH002 THERMAL
PHYSICS
This is the observational science dealing with heat and work. It
was developed based on empirical observations without
assumptions about the make up of matter. It describes
macroscopic quantities, such as heat, work, internal energy, etc. A
good example of a thermodynamic system is a gas confined by a
piston in a cylinder. If the gas is heated, it will expand, doing
mechanical work on the piston; this is one example of how a
thermodynamic system can do work.
Thermal properties are the response of matter to applied heat or
sources of different temperature. These properties are controlled
largely by interatomic motions or kinetic energy.
Thermal equilibrium is an important concept in thermodynamics.
When two systems are in thermal equilibrium, there is no net heat
transfer between them. This occurs when the systems are at the
same temperature. In other words, systems at the same temperature
will be in thermal equilibrium with each other.
PHOO2 THERMAL PHYSICS
The aims of this course are to:
1. Provide learners with a base to the study of thermal physics;
2. Develop abilities and skills that are relevant to the study and
practice of thermal physics;
3. Develop attitudes relevant to science such as integrity, the skills
of enquiry and inventiveness.
4. Promote awareness that the study and practice of thermal physics
are cooperative and cumulative activities and are subject to social,
economic, technological, ethical and cultural influences and
limitations.
5. Stimulate students and create a sustained interest in thermal
physics.
3
Course overview
Temperature:
The student should have done physical science or physics at ordinary
level with ZIMSEC, Cambridge or any recognised examination board. He
or She should have passed mathematics at ordinary level.
Course outcomes
Upon completion of PH002 THERMAL PHYSICS you will be able to:
1. Describe the first law in terms of heat and work interactions.
2. Identify the difference between internal and total energy.
Outcomes
including
3. Identify specialized thermodynamic equations and tell when they
can be applied.
4. Describe ideal gas thermodynamic properties.
5. Students should be able apply thermodynamic principles in
there daily life.
Timeframe
48 hours
[How much formal study time is required?]
[How much self-study time is expected/recommended?]
How long?
Study skills
As an adult learner your approach to learning will be different to that
from your school days: you will choose what you want to study, you will
have professional and/or personal motivation for doing so and you will
most likely be fitting your study activities around other professional or
domestic responsibilities.
Essentially you will be taking control of your learning environment. As a
consequence, you will need to consider performance issues related to
time management, goal setting, stress management, etc. Perhaps you will
also need to reacquaint yourself in areas such as essay planning, coping
with exams and using the web as a learning resource.
Your most significant considerations will be time and space i.e. the time
you dedicate to your learning and the environment in which you engage
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PH002
in that learning.
We recommend that you take time now—before starting your selfstudy—to familiarize yourself with these issues. There are a number of
excellent resources on the web. A few suggested links are:
 http://www.how-to-study.com/
The “How to study” web site is dedicated to study skills resources.
You will find links to study preparation (a list of nine essentials for a
good study place), taking notes, strategies for reading text books,
using reference sources, test anxiety.
 http://www.ucc.vt.edu/stdysk/stdyhlp.html
This is the web site of the Virginia Tech, Division of Student Affairs.
You will find links to time scheduling (including a “where does time
go?” link), a study skill checklist, basic concentration techniques,
control of the study environment, note taking, how to read essays for
analysis, memory skills (“remembering”).
 http://www.howtostudy.org/resources.php
Another “How to study” web site with useful links to time
getting the most out of doing (“hands-on” learning), memory building,
tips for staying motivated, developing a learning plan.
The above links are our suggestions to start you on your way. At the time
of writing these web links were active. If you want to look for more go to
www.google.com and type “self-study basics”, “self-study tips”, “selfstudy skills” or similar.
5
Course overview
Temperature:
Need help?
http://physics.bu.edu/~duffy/py105/Firstlaw.html
Help
eeF1y
http://www.chm.davidson.edu/vce/gaslaws/boyleslawcalc.html
http://www.chemguide.co.uk/physical/kt/idealgases.html#top
Is there a course web site address?
Mr.S. NDLOVU, 0772938446, sndlovu@buse.ac.zw
Is there a teaching assistant for routine enquiries? Where can s/he be
located (office location and hours, telephone/fax number, e-mail
Is there a librarian/research assistant available? Where can s/he be located
(office location and hours, telephone/fax number, e-mail address)?
Is there a learners' resource centre? Where is it located? What are the
opening hours, telephone number, who is the resource centre manager,
what is the manager's e-mail address)?
Who do learners contact for technical issues (computer problems, website
access, etc.)
Assignments
Three
Two
Mr S Ndlovu
Assignments
Special dates
[What is the order of the assignments? Must they be completed in the
order in which they are set?]
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PH002
Assessments
ThreeHow many assessments will there be in this course?
In class testAre they self-assessments or teacher-marked assessments?
Assessments
Selected datesWhen will the assessments take place?
Two hoursHow long will the assessments be?
Two weeksHow long will learners be allowed to complete the
assessment(s)?
Two weeksHow long will it take a teacher to mark the assessment(s)?
7
Getting around this STUDY GUIDE
Temperature:
Getting around this STUDY GUIDE
Margin icons
While working through this STUDY GUIDE you will notice the frequent
use of margin icons. These icons serve to “signpost” a particular piece of
to find your way around this STUDY GUIDE.
A complete icon set is shown below. We suggest that you familiarize
yourself with the icons and their meaning before starting your study.
8
Activity
Assessment
Assignment
Case study
Discussion
Group activity
Help
Note it!
Outcomes
Reflection
Study skills
Summary
Terminology
Time
Tip
PH002
Unit 1
Temperature:
Introduction
We all have a feel for what temperature is. We even have a shared
language that we use to qualitatively describe temperature. The cup of tea
feels hot or cold or warm. The weather outside is chilly or steamy. We
certainly have a good feel for how one temperature is qualitatively different
than another temperature. We may not always agree on whether the room
temperature is too hot or too cold or just right. But we will likely all agree
that we possess built-in thermometers for making qualitative judgments
Upon completion of this unit you will be able to:
 Define temperature
 Distinguish between temperature and heat .
Outcomes
 Calculate temperature using the absolute scale
Temperature:
Terminology
Table graphic Removing
rows from the table graphic

A measure of the average
kinetic energy of the particles in a
sample of matter, expressed in terms
of units or degrees designated on a
standard scale.
Heat
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
What is Temperature?
Despite our built-in feel for temperature, it remains one of those concepts
in science that is difficult to define. It seems that a tutorial page exploring
9
Unit 1 Temperature:
the topic of temperature and thermometers should begin with a simple
definition of temperature. But it is at this point that a familiar resource, the
internet was used where a variety of definitions that vary from the toosimple to the too-complex were found. These are listed below:

The degree of hotness or coldness of a body or
environment.

A measure of the warmth or coldness of an object or
substance with reference to some standard value.

A measure of the average kinetic energy of the
particles in a sample of matter, expressed in terms of units
or degrees designated on a standard scale.

A measure of the ability of a substance, or more
generally of any physical system, to transfer heat energy to
another physical system.

Any of various standardized numerical measures of
this ability, such as the Kelvin, Fahrenheit, and Celsius scale.
For certain, we are comfortable with the first two definitions - the degree or
measure of how hot or cold and object is. But our understanding of
temperature is not furthered by such definitions. The third and the fourth
definitions that reference the kinetic energy of particles and the ability of a
substance to transfer heat are scientifically accurate.
Practical thermometers
There are a variety of types of thermometers. The type that most of us are
familiar with from science class is the type that consists of a liquid encased
in a narrow glass column. Older thermometers of this type used liquid
mercury. In response to our understanding of the health concerns
associated with mercury exposure, these types of thermometers usually
use some type of liquid alcohol. These liquid thermometers are based on
the principal of thermal expansion. When a substance gets hotter, it
expands to a greater volume. Nearly all substances exhibit this behavior of
thermal expansion. It is the basis of the design and operation of
thermometers. As the temperature of the liquid in a thermometer
increases, its volume increases. The liquid is enclosed in a tall, narrow
glass (or plastic) column with a constant cross-sectional area. The increase
in volume is thus due to a change in height of the liquid within the column.
The increase in volume, and thus in the height of the liquid column, is
proportional to the increase in temperature.
The relationship between the temperature and the column's height is linear
over the small temperature range for which the thermometer is used. This
linear relationship makes the calibration of a thermometer a relatively easy
task.eg Suppose that a 20-degree increase in temperature causes a 2-cm
increase in the column's height. Then a 40-degree increase in temperature
will cause a 4-cm increase in the column's height. And a 60-degree
increase in temperature will cause s 6-cm increase in the column's height.
Calibrating a thermometer
The calibration of any measuring tool involves the placement of divisions or
marks upon the tool to measure a quantity accurately in comparison to
known standards. The tool needs divisions or markings; for instance, a
meter stick typically has markings every 1-cm apart or every 1-mm apart.
These markings must be accurately placed and the accuracy of their
placement can only be judged when comparing it to another object known
to have an accurate length.
10
PH002
A thermometer is calibrated by using two objects of known temperatures.
The typical process involves using the freezing point and the boiling point
of water. Water is known to freeze at 0&deg;C and to boil at 100&deg;C at an
atmospheric pressure of 1 atm. By placing a thermometer in mixture of ice
water and allowing the thermometer liquid to reach a stable height, the 0degree mark can be placed upon the thermometer. Similarly, by placing the
thermometer in boiling water (at 1 atm of pressure) and allowing the liquid
level to reach a stable height, the 100-degree mark can be placed upon the
thermometer. With these two markings placed upon the thermometer, 100
equally spaced divisions can be placed between them to represent the 1degree marks. Since there is a linear relationship between the temperature
and the height of the liquid, the divisions between 0 degree and 100
degree can be equally spaced.
Temperature Scales
The thermometer calibration process described above results in what is
divisions or intervals between the normal freezing point and the normal
boiling point of water. Today, the centigrade scale is known as the Celsius
scale, named after the Swedish astronomer Anders Celsius who is credited
with its development. The Celsius scale is the most widely accepted
temperature scale used throughout the world
The Kelvin temperature scale, which is the standard metric system of
temperature measurement and perhaps the most widely used temperature
scale among scientists. The Kelvin temperature scale is similar to the
Celsius temperature scale in the sense that there are 100 equal degree
increments between the normal freezing point and the normal boiling point
of water. However, the zero-degree mark on the Kelvin temperature scale
is 273.15 units cooler than it is on the Celsius scale. So a temperature of 0
Kelvin is equivalent to a temperature of -273.15 &deg;C. Observe that the
degree symbol is not used with this system. So a temperature of 300 units
above 0 Kelvin is referred to as 300 Kelvin and not 300 degree Kelvin; such
a temperature is abbreviated as 300 K. Conversions between Celsius
temperatures and Kelvin temperatures (and vice versa) can be performed
using one of the two equations below.
&deg;C = K - 273.15&deg;
K = &deg;C + 273.15
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Unit 1 Temperature:
Absolute zero
The zero point on the Kelvin scale is known as absolute zero. It is the
lowest temperature that can be achieved. The concept of an absolute
temperature minimum was promoted by Scottish physicist William
Thomson (a.k.a. Lord Kelvin) in 1848. Thomson theorized based on
thermodynamic principles that the lowest temperature which could be
achieved was -273&deg;C.
Measurements of the variations of pressure and volume with changes in
the temperature could be made and plotted. Plots of volume vs.
temperature (at constant pressure) and pressure vs. temperature (at
constant volume) reflected the same conclusion that is,” the volume and
the pressure of a gas reduce to zero at a temperature of -273&deg;C”. Since
these are the lowest values of volume and pressure that are possible, it is
reasonable to conclude that -273&deg;C was the lowest temperature that was
possible as shown on the graphs below.
Thomson referred to this minimum lowest temperature as absolute zero.
Scientists have been able to cool matter down to temperatures close to
-273.15&deg;C, but never below it. In the process of cooling matter to
temperatures close to absolute zero, a variety of unusual properties have
been observed. These properties include superconductivity, superfluidity,
etc
12
PH002
Unit summary
In this unit you learned that Temperature is what the thermometer reads.
Summary
But what exactly is temperature a reflection of? The concept of an absolute
zero temperature is quite interesting and the observation of remarkable
physical properties for samples of matter approaching absolute zero makes
one ponder the topic more deeply. Is there something happening at the
particle level which is related to the observations made at the macroscopic
level? Is there something deeper to temperature than simply the reading
on a thermometer? As the temperature of a sample of matter increases or
decreases, what is happening at the level of atoms and molecules?
13
Unit 1 Temperature:
Assignment
1. The readings of a resistance thermometer are 20.0 Ω at ice point, 28.2
Assignment
Ω at steam point and 23.1 Ω at an unknown temperature. Calculate the
unknown temperature on the Celsius scale of the thermometer.
2. Room temperature is 25 0C. What is it on the thermodynamic
temperature scale?
3. S A mercury-in-glass thermometer is to be used to measure the
temperature of some oil. The oil has a mass 32.0 g and specific heat
capacity 1.40 Jg-1K-1. The actual temperature of the oil is 54.0 0C. The
bulb of the thermometer has a mass 12.0 g and an average specific heat
capacity of 0.180 Jg-1K-1. Before immersing the bulb in the oil, the
thermometer reads 19.0 0C. The thermometer bulb is placed in the oil and
. (i) Give any three reasons why is mercury used as the liquid.
(ii) Determine the steady temperature recorded on the thermometer,
the ratio =
change in temperature of oil
initial temperature of oil
(iii) Suggest, with an explanation, a type of thermometer that would be
likely to give a smaller value for the ratio calculated in (b)(ii)2.
(iv) The mercury-in-glass thermometer is used to measure the boiling
point of a liquid. Suggest why the measured value of the boiling point
will not be affected by the thermal energy absorbed by the thermometer
bulb.
(c) The constant volume gas thermometer was formerly used as a
standard to calibrate other practical thermometers.
(i) What is the thermometric property and temperature range of a
constant volume gas thermometer?
(ii) State any three advantages of using the constant volume gas
thermometer to calibrate other practical thermometers.
14
R  R0
x100 0c
R100  R0
1. 


23.1  20.0
x100 0c
28.2  20.0
  37.8 0c
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2.
T=θ + 273.15
T= 25 +273.15 K
T = 298.15 K
3.(i) it is easily seen; it expands almost uniformly with increase in
temperature; it does not stick to glass; its temperature range -20
350
to
is sufficient for normal use.
(ii). heat lost by oil = heat gained by thermometer;
32 x 1.4 x (54-T) = 12 x 0.18 x (T-19); T = 52.4
either ratio
 1.6 
 54   0.030
 
or
 1.6 
 327   0.005


(iii). Thermistor thermometer because it has a small specific
heat capacity, thereby responding quickly to rapid changing
temperatures.( allow resistance thermometer – sensitive to
small temperature changes.)
(iv). Boiling point temperature is constant.eg heating the
bulb would affect only the rate of boiling
c (i) thermometric property: pressure of a fixed mass of gas
at cte volume.
(ii). it is extremely accurate; it is very sensitive; the cte
volume gas thermometer has a very wide temperature
range.
Assessment
1 (a) Define
(i) Temperature,
Assessment
(ii) Triple point of water.
2. The e.m.f. generated in a thermocouple thermometer may be used for
the measurement of temperature. Figure 2.1 shows the variation with
temperature T of the e.m.f. E
.
15
Unit 1 Temperature:
E / mV
1.5
1.0
0.5
0
300
400
500
600
700
T /K
Figure 2.1
(i)
(ii)
By reference to Figure 2.1, state two advantages of using
this thermocouple when the e.m.f. is about 1.0 mV.
An alternative to the thermocouple thermometer is the
resistance
thermometer. State two advantages that a
thermocouple has over a resistance thermometer.
Possible solutions
1.(i) The degree of hotness or coldness of a body or
environment.

A measure of the warmth or coldness of an object or
substance with reference to some standard value.

A measure of the average kinetic energy of the
particles in a sample of matter, expressed in terms of units
or degrees designated on a standard scale.

A measure of the ability of a substance, or more
generally of any physical system, to transfer heat energy to
another physical system.
(ii). The triple point of water is that unique temperature at
which pure ice, pure water and pure water vapour can exist
together in equilibrium.
2.(i).variation is non linear, two possible temperatures
(ii) Can measure the temperature of small object and its
small specific heat capacity enables it to measure rapidly
fluctuating temperatures; readings are taken at a point /
physically small; no power supply is needed.
16
PH002
Unit 2
Thermal properties of materials
Introduction
In the late 1700s, Sir Benjamin Thompson, (otherwise known as
Count Rumford) who was an army officer, noticed that a great deal
of heat was generated when boring out the hole in a cylinder of iron
in order to make cannon. In fact, enough heat was generated to boil
water. From his private research on heat generated by friction he
came to the conclusion that heat was a form of motional energy, as
opposed to the then current view that it was a material substance.
He was right; we now know that it is kinetic energy transferred to
the atoms and molecules of a solid, liquid or gas.
As any boy or girl scout will tell you that it is at least in principle
possible to start a fire by twisting the end of a dry stick on dry
leaves placed on a dry log. The first law of thermodynamics relates
changes in internal energy to heat added to a system and the work
done by a system. The first law is simply a conservation of energy
law.
Upon completion of this unit you will be able to:
1. Describe the first law in terms of heat and work interactions.
2. Describe the second law in terms of adiabatic and reversible
processes.
Outcomes
3. Identify the difference between internal and total energy.
4. Identify specialized thermodynamic equations and tell when they
can be applied.
5. Describe ideal gas thermodynamic properties.
.
Terminology
system:
The system is a general term which refers to
an object or collection of objects either solid,
liquid, or gas, or a combination of these.
heat:
Is simply the transfer of energy from a hot object to
a colder object.
Thermal
equilibrium:
Thermal equilibrium is simply another way of
saying that two or more objects are at the same
17
Unit 2 Thermal properties of materials
temperature.
2.1 Internal energy
Internal energy is defined as the energy associated
with the random, disordered motion of molecules. It
is separated in scale from the macroscopic ordered
energy associated with moving objects; it refers to
the invisible microscopic energy on the atomic and
molecular scale. For example, at room temperature,
a glass of water sitting on a table has no apparent
energy, either potential or kinetic. But on the
microscopic scale it is a seething mass of high
speed molecules travelling at hundreds of meters
per second. If the water were tossed across the
room, this microscopic energy would not
necessarily be changed when we superimpose an
ordered large scale motion on the water as a whole.
U
18
is the most common symbol used for internal energy.
PH002
For an ideal monoatomic gas, this is just the translational kinetic energy
of the linear motion of the &quot;hard sphere&quot; type atoms, and the behavior of
the system is well described by kinetic theory. However, for polyatomic
gases there is rotational and vibrational kinetic energy as well.
2.2 Zeroth law of thermodynamics
The &quot;zeroth law&quot; 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 A and C are in thermal equilibrium with B, then A is in thermal
equilibrium with B. Practically this means that all three are at the same
temperature, and it forms the basis for comparison of temperatures.
It is observed that a higher temperature object which is in contact with a
lower temperature object will transfer heat to the lower temperature
object. The objects will approach the same temperature, and in the
absence of loss to other objects, they will then maintain a constant
temperature. They are then said to be in thermal equilibrium
The reason for this is law to be called the zeroth law of
thermodynamics is that physicists first found the first and second
laws, then realised that there is a more fundamental law so they
decided to give it the number zero
2.3 First law of thermodynamics
When you pump up the tyre of a bicycle very quickly, you will
feel that the valve of the tyre is very hot. That energy came from
19
Unit 2 Thermal properties of materials
the work that you did in pushing air through the valve. In this case,
the system is the valve/tyre, and the work was done by you. If
however you press on the valve to let the air out of the tyre, you
will notice that the valve becomes very cold. In this case the
valve/tyre system was itself doing work on the air inside it.
The first law states that the energy in these processes is
conserved, and that heat can be converted into work and work into
heat. You can’t get more work out than heat you put in, and you
can’t get more heat out than work you put in.
Here are other statements of the first law:
• The energy of a closed system can only be changed through
heat and work interactions.
• Heat and work are equal.
where U is the internal energy, Q is the heat added to a system,
and W is the work done by the system. In other words, the net
energy transferred to the system, Q - W, equals the change in the
internal energy, U.
Internal energy is simply the sum of the kinetic and potential
energies of all the molecules of the system. For the applications
that we will be looking at, an increase in the internal energy of a
system means that its temperature will rise. A decrease in the
internal energy corresponds to a decrease in temperature. Note that
work, W, can be done on or by the system. Heat, Work, can be
added to or given out by the system. To take these into account we
adopt a sign convention to be used with the above equation. That
is:
Q &gt; 0 if heat is added to the system
Q &lt; 0 if heat is given out by the system
W &gt; 0 if work is done by the system
W &lt; 0 if work is done on the system
The first law of thermodynamics is the application of the conservation of
energy principle to heat and thermodynamic processes.
20
PH002
The First Law identifies both heat and work as methods of energy
transfer which can bring about a change in the internal energy of a
system. After that, neither the words work or heat has any
usefulness in describing the final state of the system - we can speak
only of the internal energy of the system.
Example:
When you pop open a champagne bottle, a little fog develops in the
opening. This is because the gas inside expands so quickly that
heat from the outside can't get to it fast enough to supply the
energy needed for expansion, so it uses up it's own internal energy.
This means that the kinetic energy of the gas particles has been
reduced, which implies that the temperature has also been reduced.
The reduction in temperature can be low enough to condense the
moisture in the air in the opening of the bottle, which forms the
little fog.
In terms of the first law of thermodynamics, we say that the gas in
the bottle has done work on its environment by expanding. This
means that W has a positive value in the above equation. Since no
heat enters in this short time we can set Q = 0. So the first law of
thermodynamics equation now becomes
U  0 W  W
This means that the change in internal energy is negative i.e a
reduction in internal energy, therefore a reduction in temperature.
2.4 Heat Capacity
If you light a match and put the flame to a small object such as a
needle or a pin then you know that after a few seconds you will not
be able to hold the needle because it has become too hot. On the
other hand, if you put the flame of the match to a massive object,
then instinctively you know it will not make one bit of difference
to its temperature. Although in both cases the temperature of the
flame is the same. So somehow the mass of an object is related to
the temperature it will reach in a certain time.
In the olden days people defined the concept of heat capacity
because they thought that somehow an object can have a capacity
for being &quot;filled&quot; with heat just as a bucket has a capacity for
holding water. This is not a correct point of view since it would be
theoretically possible to keep transferring heat to the object without
limit. Although in practice the object might eventually vaporise.
21
Unit 2 Thermal properties of materials
However, we will define it here anyway:
The ratio of the energy transferred and the change in temperature is
called the heat capacity
c 
Q
T
where T=Tfinal - Tinitial is the temperature change and Q is the
amount of heat transferred.
2.5 Specific heat capacity
It's a little cumbersome to use heat capacity since this &quot;constant&quot;
keeps changing as the mass of the object changes. So people went
further and defined something slightly better, The Specific heat
capacity.
The specific heat is the amount of heat per unit mass required to raise the
temperature by one degree Celsius. The relationship between heat and
temperature change is usually expressed in the form shown below where
c is the specific heat and is expressed in units of J.kg-1 K-1
Q  cmT
The relationship does not apply if a phase change is encountered,
because the heat added or removed during a phase change does not
change the temperature.
The specific heat of water is 4.186 joule/gram &deg;C which is higher than
any other common substance. As a result, water plays a very important
role in temperature regulation. The specific heat per gram for water is
much higher than that for a metal,
Example: 3 equal masses of Al (c=900 J/kg.K), Cu (c=386
J/kg.K), Pb (c=128 J/kg.K). All three masses are heated to the same
temperature. If each mass was plunged in a separate beaker
containing the same amount of water as the other beakers, then
wouldn't we expect the temperature of the water in the three beaker
to be raised to the same final temperature. Well many people
would, however a glance at the equation
22
PH002
Q  cmT
will show that the amount of heat transferred, Q, is dependent on
the size of the specific heat c. The smaller the c, the smaller the
amount of heat transferred and since all the beakers contain the
same amount of water, m, the smaller the temperature change
2.5.1 Determination of the specific heat capacity of a solid and
liquid by electrical methods.
Measurement of specific heat capacities
There are several simple methods for measuring the specific heat
capacities of both solids and liquids, such as the method of mixtures, but
we will consider here only electrical methods. Since the specific heat
capacity varies with temperature, we have seen it is important to record
the mean temperature at which the measurement is made.
Electrical calorimeters
Figure 1(a) and 1(b) show possible arrangements for electrical
calorimeters for a solid and a liquid specimen.
Determination of the specific heat capacity of a solid by
electrical methods
23
Unit 2 Thermal properties of materials
Determination of the specific heat capacity of a liquid by
electrical methods
The material under investigation is heated by an electrical immersion
heater and the input energy (Q) and the rise in temperature that this
produces are measured. If the mass of the specimen (solid or liquid) is m
and its specific heat capacity C, then: The electrical energy supplied to
the heater coil (Q = V I t) may be found readily with a joulemeter or with
an ammeter, voltmeter and a stop watch.
Electrical energy supplied (Q) = V1 I1 t1 = m1 C (T) + q
Where T is the temperature change of the specimen and q is the heat
loss. Using the cooling correction, the value of q may be found. This
simple method can be used for liquids or solids, although in the case of a
liquid, allowance has to be made for the thermal capacity of the
container, and the liquid should also be stirred to allow an even
distribution of the heat energy throughout its volume. This is necessary
since liquids are such poor thermal conductors
24
PH002
Metal
Specific Thermal
Electrical
Density
Heat
Conductivity
Conductivity
k
cp
watt/cm K
cal/g&deg; C
g/cm3
1E6/Ωm
Brass
0.09
1.09
8.5
Iron
0.11
0.803
7.87
11.2
Nickel
0.106
0.905
8.9
14.6
Copper
0.093
3.98
8.95
60.7
Aluminum 0.217
2.37
2.7
37.7
0.352
11.2
0.0305
2.6 Latent heat
Heats of Fusion and Vaporisation
From the equation containing the specific heat, it seems that we
can keep on transferring heat to an object and raising the
temperature indefinitely. In fact, the equation seems to suggest that
if we put in an infinite amount of heat we well get an infinite
increase in temperature. However, instinctively we know that this
is not totally true since at 100 &ordm;C water starts to boil and stays at
this temperature until it changes to a gas. So the equation doesn't
fully account for changes of state, which are also known as phase
changes.
In general, the temperature stays constant during any phase change.
That is, from solid to liquid and liquid to gas, although energy (i.e
heat) is still transferred.
25
Unit 2 Thermal properties of materials
If you take a cup of water and heat it under atmospheric
conditions, the heating curve will look like one shown above. In
the figure, the horizontal axis shows time (or more energy
added), and the vertical axis is temperature. Notice that the
temperature of water stays constant with added energy during
phase transformation. This is because all of the energy is used
to transform one state to another that no energy is available for
heating the water.
There are some definitions we must add at this stage. These are:
Melting point - temperature at which a solid turns to a liquid or
vice versa.
Boiling point - temperature at which a liquid turns to a gas (or
vapour) or vice versa.
These points can be changed by adding impurities to the water /
ice.
Example: Put a piece of string on an ice cube, which is straight out
of the fridge. Sprinkle some salt on the string/ice cube. Wait a few
seconds, you will then be able to lift the ice cube by the string. The
string seems to have become glued to the cube.
The explanation is this: The salt lowers the freezing temperature of
ice. That is, the ice has a good chance of melting if it isn't too cold.
So the melted ice soaks into the string. The salt now becomes less
concentrated as it diffuses out of the region of the string so the
freezing temperature is raised again. Since the rest of the ice cube
is still at a temperature below freezing, the water will re-freeze
including that which has soaked into the string. So the string will
essentially be frozen to the cube.
Phase diagram description of water
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PH002
Consider the phase diagram of water given below. This type of
diagram is very common in material science and helps us
determine the phase of the material for different temperatures and
pressures. There are three regions: solid, liquid, and vapour,
separated by boundaries. How do you read this diagram? As an
example, say you were trying to find out whether water is a solid,
liquid or gas at a temperature of 50 &ordm;C and pressure of 50 kPa.
Draw a line up from 50 &ordm;C and a line across from 50 kPa; the
intersection is in the liquid region of the diagram.
Using the same diagram, draw a vertical line up from 0 &ordm;C. You'll
notice that you will eventually cross from the solid to the liquid
region. This means that as you increase the pressure on ice, it will
eventually turn to liquid. This is useful if you're an ice skater
because it means that the increase in pressure, due to your body
weight, melts the ice underneath the blade which helps you glide
more smoothly. Three States of Water
Under normal conditions, water exists in one of three states: solid,
liquid, and gas. Water in its solid state is what we call ice. Pure water
at sea level freezes at 0&deg;C. Liquid water is most often used in
cooking, and the temperature varies between 0&deg;C and 100&deg;C. Water
boils at 100 &deg;C (at sea level) and becomes vapor. The vapor state is
more commonly called steam. The transition between two states is
called a phase transformation.
27
Unit 2 Thermal properties of materials
Boiling
Boiling water is one of the most commonly used heat source in
cooking. Boiling water is undergoing liquid-to-gas transition, and
because of this it stays at a constant temperature of approximately
100&deg;C. This provides a convenient standard for us to control the
cooking process.
As liquid water is heated, the molecules become increasingly mobile.
Some of the molecules acquire enough energy to escape as vapor.
These molecules exert a force on the atmosphere, called the vapor
pressure. The vapor pressure is opposed by another force, created by
a column of air pushing down on the pot.
Figure: Pressure and Boiling
This pressure is the atmospheric pressure. The two pressures are
illustrated in Figure . Water begins to boil when the vapor pressure
overcomes the atmospheric pressure. This means that the majority
of water molecules have become energetic enough to escape the
surface.
The temperature at which water boils is related to the vapor
pressure required for boiling, which is equal to the atmospheric
pressure. The implication of this is that as the atmospheric
pressure changes, the boiling point of water changes as well.
When you go up a mountain, the air pressure is lower (the
column of air pushing down is smaller). Therefore, water boils
at a lower temperature, and food takes longer to cook. For
every 1000 ft. in altitude, the boiling point of water decreases
A clever appliance designed to take advantage of the pressureboiling point relation is the pressure cooker. A pressure cooker
is a tightly sealed pot which uses the steam from water to
28
PH002
increase the internal pressure of the pot. As the pressure inside
the pot increases, the boiling point rises, and a higher
temperature can be achieved for cooking. The internal
temperature can be determined by controlling the vapor
pressure inside the pot. Since the internal temperature can be
raised substantially above 100&deg;C, food cooks faster.
The boiling point of water can also be changed by adding
impurities in the water. Impurities include salt, sugar, and other
dissolving molecules. Generally, impurities increase the boiling
point of water. A simple explanation of this is that the impurities
dilute the concentration of water (the number of water
molecules per unit volume decreases), and the number of
molecules that can vaporize at any give temperature decreases.
The result is that a higher temperature is required to achieve
the same vapor pressure. Concentrated sugar-water solutions
that are used for making candies and caramel boil at
temperatures exceeding 150 &deg;C.
Freezing
Water freezes when the molecules have slowed down enough to
develop bonds upon collision. The rate at which freezing occurs
is governed by nucleation and growth.
Nucleation is the formation of small solids in a liquid. The
clusters of solids are called the nuclei. The rate at which new
nuclei form (number of nuclei per second) is the nucleation rate.
Once the nuclei have formed, they become the landing sites for
other molecules to attach onto. The growth rate is the rate at
which the radius of a nucleus grows after formation. The
solidification rate is determined by the combination of
nucleation and growth rates.
The size of crystals formed during solidification is determined by
the nucleation/growth processes. A solidification process with
fast nucleation rate and/or slow growth rate will result in many
small crystals forming. Larger crystals form from slow
nucleation rate.
Most liquids decrease in volume upon solidification. Water,
however, has a rather unique property of expanding during
liquid-to-solid transformation. This property comes from the
hexagonal structure of ice crystals; water molecules form a
hexagonal crystal structure, which actually takes up more
volume than if the molecules were freely slipping past one
another. Consequently, ice cubes float in water.
The freezing point of water at sea level is 0&deg;C. This temperature
can be changed, however, by adding impurities in water.
Sprinkling salt on road surfaces on an icy day melts the ice by
lowering the melting temperature. Salt is also used in simple ice
cream machines during cooling of the cream. In an ice cream
machine, the vessel containing the ice cream mixture is cooled
by concentrated brine (salt-water solution) which has a
temperature that is lower than the freezing point of ice cream
mixture. Another consequence of the decrease in freezing point
due to impurities is the soft texture of ice cream. As ice cream
29
Unit 2 Thermal properties of materials
freezes, the remaining liquid becomes more and more
concentrated with sugar and other impurities. The concentrated
liquid has a much lower freezing temperature than water. As a
result, ice cream never completely freezes, and retains the
characteristic soft texture.
Energy is acquired or released when a material changes phase. For
example, energy is required to melt ice and vaporise water.
However, energy is given out if water vapour condenses or water
freezes. The heat acquired or released is called the latent heat.
During a phase change there is no change in temperature so we
cannot use the equation containing the specific heat to determine
the amount of heat transferred. The formal definition of latent heat
is the energy given out or absorbed without a change in
temperature and is given by
Q  mL
Q is the heat, m is the mass, L is a constant for a certain material
and is called the Latent heat of fusion (for melting) or vaporisation
(for boiling). Latent heat of fusion is the energy required to melt 1
kg of a solid. The units are J.kg-1.
Latent heat of vaporisation is the energy required to evaporate 1
kg of a liquid. The units are J.kg-1.
30
PH002
Unit summary
Summary
In this unit you learned that Internal energy is defined as the energy
associated with the random, disordered motion of molecules. The Zeroth
law of thermodynamics: If system A is in thermal equilibrium with
system B, and system B is in thermal equilibrium with system C,
then system A is in thermal equilibrium with system C. Importance:
This gives rise to the idea of temperature. A, B, C are at the same
temperature.
A system has internal energy U which can be changed by
transferring heat Q or by doing work W. The first law of
thermodynamics : ΔU = Q+W
There are a number of possible demonstrations of the First Law. Simple and
dramatic ones include commercial devices that let you compress a cylinder of air
rapidly and ignite a small wad of cotton. A more conventional alternative is to
compress the air in a bicycle pump and to observe the rise in temperature.
[Continue
here]
31
Unit 2 Thermal properties of materials
Assignment
1. (i) When are two bodies said to be in thermal equilibrium?
(iii)
Assignment
State the Zeroth law of thermodynamics
Possible solutions.
Two bodies are said to be in thermal equilibrium if their
temperatures are the same.
Thermal equilibrium is simply another way of saying
that two or more objects are at the same temperature.
The &quot;zeroth law&quot; 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
2. Explain how the refrigerator works?
Possible Solution: Your refrigerator has a system of pipes in the
back that contain a gas (which used to be freon before the ozone
depletion problems) that is compressed in narrow pipes in the back
of the refrigerator. Just like the heat produced when you compress
air through a bicycle tyre valve, the narrow pipes heat up when the
gas is compressed in them by the compressor of the refrigerator
(just feel the pipes in the back and you'll find that they are warm).
The compressed gas is then aloud to expand quickly through a
larger diameter pipe that is embedded in the back of the freezer.
This adiabatic expansion reduces the internal energy of the gas and
therefore its temperature. How cold does the pipe become? That
pipe is what cools the top chamber of the refrigerator i.e the
freezer. The result is the sub-zero temperatures in the freezer.
3.(a)Explain what is meant by a temperature gradient.
(b) One end of a uniform metal rod is maintained at 100 &ordm;C and the other
at room temperature. Sketch a labelled graph to show how the
temperature gradient varies with distance along the rod when its sides are:
(i)
efficiently lagged,
(ii)
unlagged,
(iii) explain the shape of each graph.
(c) Write down an equation relating the rate of flow of heat through a
thin slice of a solid to the temperature gradient across it. State the
meaning of any other symbol which appears in the equation.
(d) Outline two processes by which heat may be conducted through a
solid
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PH002
Possible solutions
. Temperature gradient is the rate of change of temperature with
distance  dT 

d 
dx 
b. (i). Temperature gradient of efficiently lagged metal rod
θ
100 &ordm;C
direction of heat flow in the rod
x
Heat
(ii) Temperature gradient of unlagged metal rod
θ
100 &ordm;C
x
Heat
(ii). Explanation of graphs
33
Unit 2 Thermal properties of materials
Graph in (i)
 heat loss from metal surface is negligible
slope of graph is constant
Graph in (ii)
 heat escapes from the sides
 the temperature gradient varies along the length of the bar
dQ
d
 kA
c. dT
dx ; where
dQ
- Rate of heat flow
dT
k - Thermal conductivity
A - Cross-section area
d
dx
d.
Free electrons diffusion in which electrons with high kinetic
energy diffuse to the cooler end.
Transfer of kinetic energy of thermal vibrations from
atom/molecule to the other; as atoms vibrate about the mean fixed
positions only;
34
PH002
Assessment
Assessment
Demonstration: Manufacturers claim that microwaves give out a
certain amount of power, usually around 700 or 800 watts. By
placing a litre of water in the microwave and heating it for a certain
time, by measuring the temperature of the water before and after
heating we can work out whether the manufacturer's claim is true.
Power = (Energy Transferred)/time
P
mcT
time
for 1 litre of water the mass is 1 kilogram, specific heat is c = 4190
J/kg.K.
How many 20 g ice cubes, whose initial temperature is -10 &ordm;C, must
be added to 1.0 L of hot tea, whose initial temperature is 90 &ordm;C, in
order that the final temperature of the mixture be 10 &ordm;C? Assume all
the ice melts in the final mixture and the specific heat of tea is the
same as that of water.
Latent heat of fusion of ice = Lv = 333 kJ. kg-1 = 333000 J. kg-1
Specific heat of water = cwater = 4190 J. kg-1 K-1
Specific heat of ice = cice = 2100 J. kg-1 K-1
Assume that the tea has the same properties as water. Note that 1
litre of water has a mass of 1 kg.
35
Unit 2 Thermal properties of materials
Solution:
Let mice be the mass of ice required, mtea be the mass of tea = mass
of water with the same volume = 1kg . Use conservation of energy.
The energy required to melt the ice, then heat it to 10 &ordm;C, must
come from the tea.
The following are the three stages the ice must go through to reach
the final temperature of 10 &ordm;C:
(i) ice heats up from -10 &ordm;C to 0 &ordm;C
(ii) ice melts at 0 &ordm;C
(iii) melted ice heats up from 0 &ordm;C to 10 &ordm;C
(iv) energy for stages (i), (ii), and (iii) must come from tea.
Now conserve energy: i.e
Energy required for (i) + (ii) + (iii) = Energy lost by tea
The expressions for the different stages are given by
(i) mice cice T(i) = mice (2100 J. kg-1 K-1) (0 &ordm;C - (-10 &ordm;C)) = mice 2100 x
10 = 21000 mice
(ii) mice Lv = mice 333000
(iii) mice cwater T(iii) = mice (4190 J. kg-1 K-1) (10 &ordm;C - 0 &ordm;C) = mice41900
(iv) mtea cwater Ttea= (1 kg) (4190 J. kg-1 K-1) (90 &ordm;C - 10 &ordm;C) = 335200
Using the conservation of energy equation above we have
21000 mice + mice 333000 + mice41900 = 335200
Now solving for mice we get mice = 0.847 kg = 847 g of ice are
needed. Divide this by the mass of one ice cube (20 g) to find out
how many cubes are needed.
847
20  43 ice cubes.
(i) Explain what is meant by the internal energy of a substance.
(ii) State and explain, in molecular terms, whether the internal energy
of the following increases, decreases or does not change.
1. a lump of iron as it is cooled,
2. Some water as it evaporates at constant temperature.
(i)
36
Define specific latent heat of fusion.
PH002
(ii) A mass of 24 g of ice at -15 0C is taken from a freezer and placed in
a beaker containing 200 g of water at 28 0C. Data for ice and water are
given in Table 1.1.
Table 1.1
specific heat capacity
/ Jkg-1K-1
specific latent heat of
fusion / Jkg-1
ice
2.1 x 103
3.3 x 105
water
4.2 x 103
_
1. Calculate the quantity of thermal energy required to convert
the ice at -15 0C to water at 0 0C.
2. Assuming that the beaker has negligible mass, calculate the
final temperature of the water in the beaker.
4.{a) State the first law of thermodynamics in terms of the increase in
internal energy ΔU, the heating q of the system and the work done on the
system.
(b) The volume occupied by 1.00 mol of liquid water at 100 0C is 1.87 x
10-5 m3. When the water is vaporized at an atmospheric pressure of 1.03
x 105 Pa, the water vapour has a volume of 2.96 x 10-2 m3. The latent heat
required to vaporize 1.00 mol of water at 100 0C and 1.03 x 105 Pa is 4.05
x 104 J.
Determine, for this change of state,
(i)
(ii)
(iii)
The work w done on the system,
The heating q of the system,
The increase in internal energy ΔU of the systm.
(c) A kettle is rated 2.3 kW. A mass of 750 g of water at 20 0C is
poured into the kettle. When the kettle is switched on, it takes
2.0 minutes for the water to start boiling. In a further 7.0 minutes,
one half of the mass of water is boiled away.
(i)
Define the term specific heat capacity.
(ii)
Estimate, for this water:
1. the specific heat capacity,
2. the specific latent heat of vaporization.
37
Unit 2 Thermal properties of materials
(d) The volume of some air, assumed to be an ideal gas, in the cylinder
of a car engine is 540 cm3 at a pressure of 1.1 x 105 Pa and a temperature
of 270C. The air is suddenly compressed, so that no thermal energy enters
or leaves the gas, to a volume of 30 cm3. The pressure rises to 6.5 x 106
Pa.
Determine the temperature of the gas after the compression.
(i)
Use the first law of thermodynamics to explain why the
temperature of the air changed during the compression.
Possible solution:
3.(a) the increase in internal energy of a system ( U) is
equal to the heat flowing into the system (Q) and the
workdone (W) on the system.
b (i) W =
W = 1.03 x105 x( 2.96 x 10-2 – 1.87 x 10-5) = - 3050 J
Q = 4.05 x 104 J;
(ii)
(iii)
104 – 3050)J = 37500J
The specific heat capacity, c, of a substance is the amount of
heat required to raise the temperature of a unit mass of that
substance by 1 K.
(iii)
Q = mc
but also (Q = Pt)
c. (i)
2300 
0.75xcx(100  20)
; c = 4600Jkg-1K-1;
120
(ii) Q = mL; 2300 = (0.375/420) x L; L = 2.6 x 106 Jkg-1
d (i)
PV
 cte ; T = (6.5 x 105 x 30 x300)/ (1.1 x 105 x
T
540) = 985 K.
(ii) Q is zero
= W and U increases.
U is rise in kinetic energy of atoms and mean kinetic
energy T.
38
PH002
Unit 3
Unit 3: Ideal gases
Introduction
The kinetic theory of gases is the study of the microscopic behavior of
molecules and the interactions which lead to macroscopic relationships
like the ideal gas law.
The study of the molecules of a gas is a good example of a physical
situation where statistical methods give precise and dependable
results for macroscopic manifestations of microscopic phenomena.
For example, the pressure, volume and temperature calculations
from the ideal gas law are very precise. The average energy
associated with the molecular motion has its foundation in the
Boltzmann distribution, a statistical distribution function. Yet the
temperature and energy of a gas can be measured precisely.
What is the work done in compressing a gas? The parameters
associated with a gas are pressure and volume. So how do we get
force times distance out of these? Answer: work done on a gas is
simply pressure times volume and is related to the temperature of
the gas by the ideal gas equation
A physical or chemical process can result in heat energy being
generated or consumed. If heat is generated the process is
exothermic. If heat is consumed, the process is endothermic. If the
system undergoing the process is thermally connected to the
outside world, then the outside world acts as a thermal bath, and
maintains the system at a constant temperature, by absorbing any
heat energy that is generated, or by supplying heat to make up for
39
Unit 3
the heat energy consumed. So when the system is connected to a
thermal bath the process the system undergoes is isothermal. For
example, most reactions of an acid and base mixed together to form
a salt are exothermic. If a strongly acidic solution is rapidly poured
into a strongly basic solution in a beaker (glass cup) that is wrapped
with thermal insulation, the heat generated cannot escape, and the
resulting system of solution plus salt (either dissovled or
precipiated) heats up. This is not an isothermal process. But if the
thermal insulation is removed from the beaker of basic solution,
and the beaker is set into a large tub of water, and the temperature
of the basic solution is allowed to equilibrate with the water bath
(come to the same temperature), and the acid solution is added
slowly, then the heat of reaction will have time to move through the
beaker glass wall into the large bath of water, and the temperature
of the solution in the beaker will remain at the temperature of the
bath as the acid is slowly poured into the beaker and the acid and
base react. This is an isothermal process.
Upon completion of this unit you will be able to:

Use Boyle's Law to predict the final pressure for a gas
that has undergone an expansion.[verb] [complete the
sentence].
Outcomes
 Calculate volume using Charles’ law
An isothermal process is a thermodynamic process in which the
temperature of the system remains constant. The heat transfer into or
out of the system typically must happen at such a slow rate that the
thermal equilibrium is maintained.
 An adiabatic process is a thermodynamic process in which there is
no heat transfer (Q) into or out of the system. In other words Q =
0.
.
Terminology
40
PH002
[Term]:
[Term description]
Gas Laws
Calculations using Boyle's Law
Concepts
Boyle's Law states that the product of the pressure
and volume for a gas is a constant for a fixed amount
of gas at a fixed temperature. Written in
mathematical terms, this law is
P V = constant
A common use of this law is to predict how a change
in pressure will alter the volume of the gas or vice
versa. Such a problem can be regarded as a two state
problem: the initial state (represented by subscript i)
and the final state (represented by subscript f). If a
sample of gas initially at pressure Pi and volume Vi is
subjected to a change that does not change the
amount of gas or the temperature, the final pressure
Pf and volume Vf are related to the initial values by
the equation
Pi Vi = Pf Vf
Statement
For a fixed mass of gas at constant temperature,
the volume is inversely proportional to the
pressure.
That means that, for example, if you double the pressure,
you will halve the volume. If you increase the pressure 10
times, the volume will decrease 10 times.
You can express this mathematically as
41
Unit 3
PV = constant
Is this consistent with PV = nRT ?



You have a fixed mass of gas, so n (the number of
moles) is constant.
R is always constant - it is called the gas constant.
Boyle's Law demands that temperature is constant as
well.
That means that everything on the right-hand side of pV =
nRT is constant, and so PV is constant - which is what we
have just said is a result of Boyle's Law.
Simple Kinetic Theory explanation
I'm not going to try to prove the relationship between
pressure and volume mathematically - I'm just showing that it
is reasonable.
This is easiest to see if you think about the effect of
decreasing the volume of a fixed mass of gas at constant
temperature. Pressure is caused by gas molecules hitting the
walls of the container. With a smaller volume, the gas
molecules will hit the walls more frequently, and so the
pressure increases.
You might argue that this isn't actually what Boyle's Law says
- it wants you to increase the pressure first and see what
effect that has on the volume. But, in fact, it amounts to the
same thing. If you want to increase the pressure of a fixed
mass of gas without changing the temperature, the only way
you can do it is to squeeze it into a smaller volume. That
causes the molecules to hit the walls more often, and so the
pressure increases.
Charles' Law
Statement
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PH002
For a fixed mass of gas at constant pressure, the
volume is directly proportional to the kelvin
temperature.
That means, for example, that if you double the kelvin
temperature from, say to 300 K to 600 K, at constant
pressure, the volume of a fixed mass of the gas will double
as well.
You can express this mathematically as
V = constant x T
Is this consistent with PV = nRT ?



You have a fixed mass of gas, so n (the number of
moles) is constant.
R is the gas constant.
Charles' Law demands that pressure is constant as
well.
If you rearrange the PV = nRT equation by dividing both
sides by p, you will get
V = nR/p x T
But everything in the nR/p part of this is constant.
That means that V = constant x T, which is Charles' Law.
Gay-Lussac's Law
. Gay-Lussac's most important contributions to the study of gases,
were experiments he performed on the ratio of the volumes of
gases involved in a chemical reaction. Gay-Lussac studied the
volume of gases consumed or produced in a chemical reaction
because he was interested in the reaction between hydrogen and
oxygen to form water. He argued that measurements of the weights
of hydrogen and oxygen consumed in this reaction could be
influenced by the moisture present in the reaction flask. The third
gas law from the Combined Gas Law has been named the
Gay-Lussac law. The law is the relationship of pressure and
temperature with constant volume (V1 = V2.) the pressure
and absolute temperature of a gas are directly proportional.
43
Unit 3
P1 V1 = T1
P2 V2 T2
And so we get the third law, the relationship between the
pressure and temperature of a gas.
P1 = T1
P2 T2
KNOW THIS
It can be arranged so that it appears in the same form you
see in most books.
P1 = P2
T1 T2
To get a feel for the third Law, consider an automobile tyre.
With a tyre gauge measure the pressure of the tyre before
and immediately after a long trip. When cool, the tyre has a
lower pressure. As the tyre turns on the pavement, it alters
its shape and becomes hot. There is some expansion of the
air in the tyre, as seen by the tire riding slightly higher, but
we can ignore that small effect. If you were to plot the
temperature versus pressure of a car tire, would zero
pressure extrapolate out to absolute zero? Remember what
you are measuring. The pressure of a car tyre is actually the
air pressure above atmospheric pressure. If you add
atmospheric pressure to your tire gauge, you would
certainly come closer to extrapolating to absolute zero.
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PH002
3.2.1 Applications of the ideal gas equation
Ideal Gas Law
An ideal gas is defined as one in which all collisions between
atoms or molecules are perfectly elastic and in which there are no
intermolecular attractive forces. One can visualize it as a collection
of perfectly hard spheres which collide but which otherwise do not
interact with each other. In such a gas, all the internal energy is in
the form of kinetic energy and any change in internal energy is
accompanied by a change in temperature.
An ideal gas can be characterized by three state variables: absolute
pressure (P), volume (V), and absolute temperature (T). The
relationship between them may be deduced from kinetic theory and
is called the






n = number of moles
R = universal gas constant = 8.3145 J/mol K
N = number of molecules
k = Boltzmann constant = 1.38066 x 10-23 J/K = 8.617385 x 10-5
eV/K
k = R/NA
NA = Avogadro's number = 6.0221 x 1023 /mol
The ideal gas law can be viewed as arising from the kinetic
pressure of gas molecules colliding with the walls of a container in
accordance with Newton's laws. But there is also a statistical
element in the determination of the average kinetic energy of those
molecules. The temperature is taken to be proportional to this
average kinetic energy; this invokes the idea of kinetic temperature.
One mole of an ideal gas at STP occupies 22.4 liters.
45
Ideal Gas Law
Kinetic Theory assumptions about ideal gases
There is no such thing as an ideal gas, of course, but many
gases behave approximately as if they were ideal at ordinary
working temperatures and pressures. Real gases are dealt
with in more detail on another page.
The assumptions are:





Gases are made up of molecules which are in
constant random motion in straight lines.
The molecules behave as rigid spheres.
Pressure is due to collisions between the molecules
and the walls of the container.
All collisions, both between the molecules themselves,
and between the molecules and the walls of the
container, are perfectly elastic. (That means that there
is no loss of kinetic energy during the collision.)
The temperature of the gas is proportional to the
average kinetic energy of the molecules.
And then two absolutely key assumptions, because these are
the two most important ways in which real gases differ from
ideal gases:

There are no (or entirely negligible) intermolecular
forces between the gas molecules.
The volume occupied by the molecules themselves is entirely
negligible relative to the volume of the container.
Simple Kinetic Theory explanation
Again, I'm not trying to prove the relationship between
pressure and volume mathematically - just that it is
reasonable.
Suppose you have a fixed mass of gas in a container with a
moveable barrier - something like a gas syringe, for example.
The barrier can move without any sort of resistance.
46
PH002
The barrier will settle so that the pressure inside and outside
is identical.
Now suppose you heat the gas, but not the air outside.
The gas molecules will now be moving faster, and so will hit
the barrier more frequently, and harder. Meanwhile, the air
molecules on the outside are hitting it exactly as before.
Obviously, the barrier will be forced to the right, and the
volume of the gas will increase. That will go on until the
pressure inside and outside is the same. In other words, the
pressure of the gas will be back to the same as the air again.
So we have fulfilled what Charles' Law says. We have a fixed
mass of gas (nothing has been added, and nothing has
escaped). The pressure is the same before and after (in each
case, the same as the external air pressure). And the volume
increases when you increase the temperature of the gas.
http://misterguch.brinkster.net/gaslawworksheets.html
Each of these relationships is a special case of a more general
relationship known as the ideal gas equation.
PV = nRT
In this equation, R is a proportionality constant known as the ideal
gas constant and T is the absolute temperature. The value of R
depends on the units used to express the four variables P, V, n, and
T. By convention, most chemists use the following set of units.
P: atmospheres
47
Ideal Gas Law
T: kelvin
V: liters
n: moles
3.4 Work done by an ideal gas,
Work is simply a force multiplied by the distance moved in the
direction of the force. A good example of a thermodynamic system
that can do work is the gas confined by a piston in a cylinder, as
shown in the diagram.
If the gas is heated, it will expand and push the piston up, thereby
doing work on the piston. If the piston is pushed down, on the other
hand, the piston does work on the gas and the gas does negative
work on the piston. This is an example of how work is done by a
thermodynamic system. The gas is heated, expanding it and
moving the piston up. If the volume occupied by the gas doubles,
how much work has the gas done?
An assumption to make here is that the pressure is constant. Once
the gas has expanded, the pressure will certainly be the same as
before because the same free-body diagram applies. As long as the
expansion takes place slowly, it is reasonable to assume that the
pressure is constant.
If the volume has doubled, then, and the pressure has remained the
same, the ideal gas law tells us that the temperature must have
doubled too.
The work done by the gas can be determined by working out the
force applied by the gas and calculating the distance. However, the
force applied by the gas is the pressure times the area, so:
48
PH002
W=Fs=PAs
and the area multiplied by the distance is a volume, specifically the
change in volume of the gas. So, at constant pressure, work is just
the pressure multiplied by the change in volume:
This is positive because the force and the distance moved are in the
same direction, so this is work done by the gas.
Types of thermodynamic processes
There are a number of different thermodynamic processes that can
change the pressure and/or the volume and/or the temperature of a
system. To simplify matters, consider what happens when
something is kept constant. The different processes are then
categorized as follows :
1. Isobaric - the pressure is kept constant. An example of an
isobaric system is a gas, being slowly heated or cooled,
confined by a piston in a cylinder. The work done by the
system in an isobaric process is simply the pressure
multiplied by the change in volume, and the P-V graph
looks like:
2. Isochoric - the volume is kept constant. An example of this
system is a gas in a box with fixed walls. The work done is
zero in an isochoric process, and the P-V graph looks like:
49
Ideal Gas Law
3. Isothermal - the temperature is kept constant. A gas
confined by a piston in a cylinder is again an example of
this, only this time the gas is not heated or cooled, but the
piston is slowly moved so that the gas expands or is
compressed. The temperature is maintained at a constant
value by putting the system in contact with a constanttemperature reservoir (the thermodynamic definition of a
reservoir is something large enough that it can transfer heat
into or out of a system without changing temperature).
If the volume increases while the temperature is constant,
the pressure must decrease, and if the volume decreases the
pressure must increase.
removed from the system.
The isothermal and adiabatic processes should be examined in a
little more detail.
Isothermal processes
In an isothermal process, the temperature stays constant, so the
pressure and volume are inversely proportional to one another. The
P-V graph for an isothermal process looks like this:
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PH002
The work done by the system is still the area under the P-V curve,
but because this is not a straight line the calculation is a little tricky,
and really can only properly be done using calculus.
The internal energy of an ideal gas is proportional to the
temperature, so if the temperature is kept fixed the internal energy
does not change. The first law, which deals with changes in the
internal energy, thus becomes 0 = Q - W, so Q = W. If the system
does work, the energy comes from heat flowing into the system
from the reservoir; if work is done on the system, heat flows out of
the system to the reservoir. In general, during an isothermal process
there is a change in internal energy, heat energy, and work.
The internal energy of an ideal gas, however, depends solely on the
temperature, so the change in internal energy during an isothermal
process for an ideal gas is also 0.
In an adiabatic process, no heat is added or removed from a system.
The first law of thermodynamics is thus reduced to saying that the
change in the internal energy of a system undergoing an adiabatic
change is equal to -W. Since the internal energy is directly
proportional to temperature, the work becomes:
An example of an adiabatic process is a gas expanding so quickly
that no heat can be transferred. The expansion does work, and the
temperature drops. This is exactly what happens with a carbon
dioxide fire extinguisher, with the gas coming out at high pressure
and cooling as it expands at atmospheric pressure.
An adiabatic process is generally obtained by surrounding the entire
system with a strongly insulating material or by carrying out the
51
Newton's Laws and Collisions
process so quickly that there is no time for a significant heat transfer to
take place.


A system that expands under adiabatic conditions does positive
work, so the internal energy decreases.
A system that contracts under adiabatic conditions does negative
work, so the internal energy increases.
There is often, though not always, a change in temperature associated
with the change in internal energy.
The compression and expansion strokes in an internal-combustion
engine are both approximately adiabatic processes. What little heat
transfers outside of the system is negligible and virtually all of the
energy change goes into moving the piston
In an adiabatic process, no heat is added or removed from a system.
The first law of thermodynamics is thus reduced to saying that the
change in the internal energy of a system undergoing an adiabatic
change is equal to -W. Since the internal energy is directly
proportional to temperature, the work becomes:
With liquids and solids that are changing temperature, the heat
associated with a temperature change is given by the equation:
Newton's Laws and Collisions
Applying Newton's Laws to an ideal gas under the assumptions of
kinetic theory allows the determination of the average force on
container walls. This treatment assumes that the collisions with the
walls are perfectly elastic.
52
PH002
In this development, an over bar indicates an average quantity. In
the expression for the average force from N molecules, it is
important to note that it is the average of the square of the velocity
which is used, and that this is distinctly different from the square of
the average velocity.
Gas Pressure from Kinetic Theory
Under the assumptions of kinetic theory, the average force on container
walls has been determined to be
and assuming random speeds in all
directions
The average force and
pressure on a given wall
depends only upon the
components of velocity Then the pressure in a container can be
toward that wall. But it
expressed as
can be expressed in terms
of the average of the
entire translational kinetic
energy using the
assumption that the
53
Kinetic Temperature
molecular motion is
random.
Expressed in terms of average molecular kinetic energy:
This leads to a concept of kinetic temperature and
to the ideal gas law.
Kinetic Temperature
The expression for gas pressure developed from kinetic theory
relates pressure and volume to the average molecular kinetic
energy. Comparison with the ideal gas law leads to an expression
for temperature sometimes referred to as the kinetic temperature.
The more familiar form expresses the average molecular kinetic energy:
It is important to note that the average kinetic energy used here is
limited to the translational kinetic energy of the molecules. That is,
they are treated as point masses and no account is made of internal
degrees of freedom such as molecular rotation and vibration. This
distinction becomes quite important when you deal with subjects
like the specific heats of gases. When you try to assess specific
heat, you must account for all the energy possessed by the
molecules, and the temperature as ordinarily measured does not
account for molecular rotation and vibration. The kinetic
temperature is the variable needed for subjects like heat transfer,
because it is the translational kinetic energy which leads to energy
transfer from a hot area (larger kinetic temperature, higher
molecular speeds) to a cold area (lower molecular speeds) in direct
collisional transfer.
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PH002
Molecular Speeds
From the expression for kinetic temperature
Calculation
substitution gives the root mean square (rms) molecular velocity:
Maxwell speed distribution this speed as well as the average and
From the
most probable speeds can be calculated
An ideal gas - a microscopic approach
In 1827 the English botanist Robert Brown discovered that pollen
suspended in water shows a continuous random motion when
viewed under a microscope (known as Brownian motion). At first
these motions were considered a form of life, but it was soon found
that small inorganic particles behave similarly. Until that time,
there was a great deal of debate as to whether atoms actually
existed. The random motion of the pollen gave indirect evidence
that matter is made from discrete particles (i.e atoms) and that they
are colliding in continuous random motion. An exact quantitative
description of Brownian motion was finally given by Albert
Einstein.
Interesting experiment:
Suspend a mirror from a string in a chamber that has all the air
taken out (i.e a vacuum). Now shine a light beam on the mirror so
that it reflects onto a screen. Look at the light beam spot on the
screen and you'll find that nothing out of the ordinary happens.
Now let a very small amount of air into the chamber. You'll find
that the spot on the screen is now moving in a random motion.
What we are looking at is simply the random collisions of the air
molecules with the mirror. We don't see this phenomenon when we
do the experiment at normal air pressure because there are so many
55
From the Maxwell speed distribution this speed as well as the
average and most probable speeds can be calculated
collisions from all sides of the mirror that it remains stationary. At
low pressures there are not many molecules around so that there is
an imbalance in the number of particles striking both sides of the
mirror, so that it develops a net force in a particular direction. This
experiment was carried out in 1827 by Kappler.
So on the atomic level, if all that is happening is the random
motion of atoms, then how can we describe concepts such as
pressure and temperature?
Did you ever do this? Close your mouth and blow out your
cheeks. What holds out your cheeks? If you said pressure then
basically you have said nothing since pressure is just a word we
use to describe a more fundamental process. Your cheeks moved
outwards because something pushed them. Keep in mind that the
air in your mouth is made of atoms which are hitting the insides of
your cheeks many times a second. So some how we must define
pressure in terms of the rate at which atoms are hitting the inside of
Let's examine what we mean by pressure. An atom heads
towards a wall with momentum mv. It strikes the wall and returns
in the opposite direction with momentum -mv. This change in
momentum, mv - (-mv) = 2mv, divided by the collision time gives
the force applied to the wall. Now imagine billions and billions of
these collisions per second over a certain area of the wall. This is
what we define as pressure i.e the force exerted by the change in
momentum of the atoms divided by the area. It can be shown
that the ideal gas equation can be rewritten in terms of more
fundamental parameters
where n is the number of moles, M is the mass of one mole, and
&lt;v2&gt; is the average of the speed squared of the gas particles. So for
the first time we see that the ideal gas equation can be written in a
form that depends on the microscopic parameters of atoms i.e their
mass and velocity. But really, nothing has changed, the above
equation must still give the same results as the more familiar ideal
gas equation, which deals with macroscopic parameters such as
temperature. So if we equate the two we have
PV(macroscopic) = PV(microscopic)
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PH002
rearranging this we get
where vrms is known as the root mean square velocity, which is
another way of defining the average velocity.
How fast do gas atoms go?
What atomic speeds are we talking about: Find the root mean
square speed of 1 mole of nitrogen molecules at a pressures of 101
kPa and temperature of 300 K. Note that 1 mole of nitrogen has a
mass of 28 grams = 0.028 kg.
Solution:
This is faster than the speed of sound!!
Now we are in a position to get the average translational kinetic
energy of gas particles.
where m is the mass of one particle (atom or molecule). The molar
mass M is defined as the mass of one particle multiplied by
Avogadro's number = NA = 6.023 x 1023. That is
M = mNA
substituting this in the expression for kinetic energy we get
57
From the Maxwell speed distribution this speed as well as the
average and most probable speeds can be calculated
for convenience we write
as one constant i.e
k is known as Boltzman's constant.
Boltzman's constant is named in honour of Ludwig Boltzman
who began the branch of physics know as statistical mechanics. He
combined classical mechanics with the statistical description of the
motion of atoms as in Brownian motion. He also had theories about
the irreversibility of many processes. A large section of the
scientific community did not accept them mostly because they
could not understand them. Finally, ill and depressed, took his own
life on September 5 1906. Since then, all of his theories have been
fully justified.
From the above equation for kinetic energy, we can see that what
we call temperature is simply the kinetic energy divided by a
constant. So temperature is a direct measure of the kinetic energy
of atoms and molecules.
Although we have talked about the average kinetic energy, let's
look at the actual speeds of the atoms / molecules in a gas, which
are given by the following
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PH002
This is known as a maxwellian distribution of velocities. Note
that there is a continuous distribution of velocities from zero to
very high values. So some particles are not moving while others
are well and truly supersonic. However, there are not too many
particles at the extremes of the distribution. As we have seen, the
average in the square of the speeds defines the temperature. Note
that the peak moves to higher velocities as the temperature is
increased.
59
From the Maxwell speed distribution this speed as well as the
average and most probable speeds can be calculated
Unit summary
In this unit you learned As has been discussed, a gas enclosed by a
Summary
piston in a cylinder can do work on the piston, the work being the
pressure multiplied by the change in volume. If the volume doesn't
change, no work is done. If the pressure stays constant while the
volume changes, the work done is easy to calculate. On the other
hand, if pressure and volume are both changing it's somewhat
harder to calculate the work done.
As an aid in calculating the work done, it's a good idea to draw a
pressure-volume graph (with pressure on the y axis and volume on
the x-axis). If a system moves from one point on the graph to
another and a line is drawn to connect the points, the work done is
the area underneath this line. We'll go through some different
thermodynamic processes and see how this works.
60
PH002
Assignment
1. Consider a gas in a cylinder at room temperature (T = 293 K),
with a volume of 0.065 m3. The gas is confined by a piston with a
weight of 100 N and an area of 0.65 m2. The pressure above the
piston is atmospheric pressure.
Assignment
(a) What is the pressure of the gas?
This can be determined from a free-body diagram of the piston.
The weight of the piston acts down, and the atmosphere exerts a
downward force as well, coming from force = pressure x area.
These two forces are balanced by the upward force coming from
the gas pressure. The piston is in equilibrium, so the forces balance.
Therefore:
Solving for the pressure of the gas gives:
2.(a) The pressure p of an ideal gas is given by the expression
p
1 Nm 2
c .
3 V
The ideal gas has a density of 2.4 kgm-3 at a pressure of 2.0 x 105 Pa and a
temperature of 300 K.
(i)Determine the root-mean-square (r.m.s.) speed of the gas
atoms at 300K.
(ii)Calculate the temperature of the gas for the atoms to have
an r.m.s. speed that is twice that calculated in.
(b) Calculate the quantity of heat conducted through 2 m2 of a brick wall
12 cm thick in 1 hour if the temperature on one side is 8 &ordm;C and on the
other side is 28 &ordm;C.
(c) The tungsten filament of an electric lamp has a length of 0.5 m and a
diameter6 x 10-5 m. The power rating of the lamp is 60 W. Assuming the
radiation from the filament is equivalent to 80 % that of a perfect black
the filament.
3.(a) State what is meant by an ideal gas.
(b) The product of the pressure p and volume V of an ideal gas of density
ρ at temperature T is given by the expressions
61
From the Maxwell speed distribution this speed as well as the
average and most probable speeds can be calculated
1
p   c2
3
and
pV  NkT ,
where N is the number of molecules and k is the Boltzmann
constant.
(i) State the meaning of the symbol c 2 .
(ii) Deduce that the mean kinetic energy EK of the molecules of an ideal
gas is given by the expression
EK 
3
kT.
2
(c) In order for an atom to escape completely from the Earth’s
gravitational field, it must have a speed of approximately 1.1 x 104 ms-1 at
the top of the Earth’s atmosphere.
(i)
Estimate the temperature at the top of the atmosphere such
that helium, assumed to be an ideal gas, could escape
from the Earth. The mass of a helium atom is 6.6 x 10-27
kg.
(ii)
Suggest why some helium atoms will escape at
temperatures below that calculated in (i).
(d) The mean kinetic energy EK of an atom of an ideal gas is given by
EK 
3
kT ,
2
where k is the Boltzmann constant and T is the thermodynamic
temperature.
Using the equation v 2  2Rg , where v is the escape speed, g is the
acceleration due to gravity, R is the radius of a planet, estimate the
temperature at the Earth’s surface such that helium atoms of mass 6.6 x
10-27 kg could escape to infinity.
You may assume that helium gas behaves as an ideal gas and that
the radius of the Earth is 6.4 x 106 m.
62
PH002
(e) The air in a car tyre has a constant volume of 3.1 x 10-2 m3. The
pressure of this air is 2.9 x 105 Pa at a temperature of 17 0C. The air may
be considered to be an ideal gas.
(i)
Calculate the amount of air, in mol, in the tyre.
(ii) The pressure in the tyre is to be increased using a pump. On each
stroke of the pump, 0.012 mol of air is forced into the tyre. Calculate the
number of strokes of the pump required to increase the pressure to 3.4 x
105 Pa at a temperature of 27 0C.
(d) Distinguish between boiling and evaporation.
Possible solution
a.(i)
1
P   c2
3
c2 
r.m.s speed=
(ii)
New
c2
c
2
3P


3x2,0 x105
 2.5x105 m2 s 2
2.4
c2  500ms 1
= 1.0 x 106 ie
c2
increases by a factor of 4.
is proportional temperature or
2
3kT m c

2
2
 1.0 x106 
(b) T  
5 
 2.5x10 
28  8
Km1 and t= 3600s
=
2
12 x10
28  8
Q  0.13x2 x3600 x(
)J
12 x102
Q = 156 kJ
0.8x 5.7x10-8x2∏ x 3x10-5 x 0.5 x T4 where surface area of the
cylindrical wire is = 2 rh.
1
60

4
T 
8
5 
 0.4 x5.7 x10 x2 x3x10 
here]
63
From the Maxwell speed distribution this speed as well as the
average and most probable speeds can be calculated
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PH002
Assessment
Some water in a saucepan is boiling.
Assessment
(a) Explain why
(i) external work is done by the boiling water;
(ii) there is a change in the internal energy as water changes to
steam.
By reference to the first law of thermodynamics and your answers in (a),
show that thermal energy must be supplied to the water during the boiling
process.
(c) (i) The kinetic theory of gases leads to the equation
1
3
m c 2  kT.
2
2
Explain the significance of the quantity
1
m c2 .
2
(ii) Use the equation to suggest what is meant by the absolute zero
of temperature.
(d) Two insulated gas cylinders A and B are connected by a tube of
negligible volume, as shown in Figure 4.1.
cylinder A
cylinder B
65
From the Maxwell speed distribution this speed as well as the
average and most probable speeds can be calculated
Figure 4.1: Two insulated gas cylinders A and B connected by a
tube of negligible volume.
Each cylinder has an internal volume of 2.0 x 10-2 m3. Initially, the
tap is closed And cylinder A contains 1.2 mol of an ideal gas at a
temperature of 370 C. Cylinder B contains the same ideal gas at pressure
1.2 x 105 Pa and temperature 370 C.
(i) Calculate the amount, in mol, of the gas in cylinder B.
(ii) The tap is opened and some gas flows from cylinder A to
cylinder B. Using the fact that the total amount of gas is constant,
determine the final pressure of the gas in the cylinders.
(e) A gas cylinder contains 4.00 x 104 cm3 of hydrogen at a pressure of
2.50 x 107 Pa and a temperature of 290 K.
The cylinder is to be used to fill balloons. Each balloon, when
filled, contains 7.24 x 103cm3 of hydrogen at a pressure of 1.85 x 105 Pa
and a temperature of 290 K.
Given that the equation
PV = constant x T
relates pressure p and volume V of a gas to its thermodynamic
temperature T. Assuming that the hydrogen obeys this equation,
calculate,(i) the total amount of hydrogen in the cylinder,
(ii) the number of balloons that can be filled from the cylinder.
4 (a)i Volume increases on evaporation as a result work is
done pushing back the atmosphere.
(ii) Ek of atoms is cte, since temperature is cte.
Ep changes because separation of a toms changes, so
internal energy changes because
U = E k + Ep
(b)
U  W  Q
is positive since Ep increases (separation of atom
increases)
W is negative because water has lost energy in pushing
back the atmosphere, so
is positive, ie the water has
gained energy by heating.
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C(i)
1
m c2
2
is the mean kinetic energy of the atoms/
molecules/ particles.
(ii)At absolute zero, atoms have no kinetic energy.
D(i) PV  nRT ; n = ( 1.2 x 105 x 2,0 x 10-2)/8.31 x 310)
= 0.93 mol.
(ii) total amount = (1.20 + 0.93) = (4.0 x 10-2 x )/ (8.31 x
310)
= 1.37 x 105 Pa
E(i) n 
PV
RT
n = (2.5 x 107 x 4.00 x 104 x 1.0 x10-6) /
(8.31 x 290) = 415 mol.
(ii) Volume of gas at 1.85 x 105 Pa =
 2.5x107 x4.00 x104 
6
3

  5.41x10 cm
5
1.85
x
10


So 5.41 x 106 = 4.00 x 104 + 7.24 x 103 N
Therefore N = 741
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Unit 4
Unit 4: Thermal energy transfer
Introduction
It is observed that a higher temperature object which is in contact with a
lower temperature object will transfer heat to the lower temperature
object. The objects will approach the same temperature, and in the
absence of loss to other objects, they will then maintain a constant
temperature. They are then said to be in thermal equilibrium.
Upon completion of this unit you will be able to:
 [verb] [complete the sentence].
 [verb] [complete the sentence].
Outcomes
 [verb] [complete the sentence].
 [verb] [complete the sentence].
 [verb] [complete the sentence].
 [verb] [complete the sentence].
Terminology
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
[Term]:
[Term description]
4.1 Thermal equilibrium
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Unit 4 Unit 4: Thermal energy transfer
Types of equilibrium
Thermal equilibrium
Thermal equilibrium is achieved when two systems in thermal
contact with each other cease to exchange energy by heat. It
follows that if two systems are in thermal equilibrium, then their
temperatures are the same.
Thermal equilibrium occurs when a system's macroscopic thermal
observables have ceased to change with time. For example, an ideal
gas whose distribution function has stabilised to a specific
Maxwell-Boltzmann distribution would be in thermal equilibrium.
This outcome allows a single temperature and pressure to be
attributed to the whole system. Thermal equilibrium of a system
does not imply absolute uniformity within a system; for example, a
river system can be in thermal equilibrium when the macroscopic
temperature distribution is stable and not changing in time, even
though the spatial temperature distribution reflects thermal
pollution inputs.
Quasistatic equilibrium
Quasistatic equilibrium is the quasi-balanced state of a
thermodynamic system near to thermodynamic equilibrium, in
some sense. In a quasistatic or equilibrium process, a sufficiently
slow transition of a thermodynamic system from one equilibrium
state to another occurs such that at every moment in time the state
of the system is close to an equilibrium state. During a quasistatic
process, the system reaches equilibrium much faster, almost
instantaneously, than its physical parameters vary.
Non-equilibrium
Non-equilibrium thermodynamics is a branch of thermodynamics
that deals with systems that are not in thermodynamic equilibrium.
Most systems found in nature are not in thermodynamic
equilibrium because they are not in stationary states, and are
continuously and discontinuously subject to flux of matter and
energy to and from other systems. The thermodynamic study of
non-equilibrium systems requires more general concepts than are
dealt with by equilibrium thermodynamics. Many natural systems
still today remain beyond the scope of currently known
macroscopic thermodynamic methods.
Now let's imagine a third situation. Suppose that a small metal cup
of hot water is placed inside of a larger Styrofoam cup of cold
water. Let's suppose that the temperature of the hot water is
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initially 70&deg;C and that the temperature of the cold water in the outer
cup is initially 5&deg;C. And let's suppose that both cups are equipped
with thermometers (or temperature probes) that measure the
temperature of the water in each cup over the course of time. What
do you suppose will happen? Before you read on, think about the
question and commit to some form of answer. When the cold water
is done warming and the hot water is done cooling, will their
temperatures be the same or different? Will the cold water warm up
to a lower temperature than the temperature that the hot water cools
down to? Or as the warming and cooling occurs, will their
temperatures cross each other?
Fortunately, this is an experiment that can be done and in fact has
been done on many occasions. The graph below is a typical
representation of the results.
As you can see from the graph, the hot water cooled down to
approximately 30&deg;C and the cold water warmed up to
approximately the same temperature. Heat is transferred from the
high temperature object (inner can of hot water) to the low
temperature object (outer can of cold water). If we designate the
inner cup of hot water as the system, then we can say that there is a
flow of water from the system to the surroundings. As long as there
is a temperature difference between the system and the
surroundings, there is a heat flow between them. The heat flow is
more rapid at first as depicted by the steeper slopes of the lines.
Over time, the temperature difference between system and
surroundings decreases and the rate of heat transfer decreases. This
is denoted by the gentler slope of the two lines. Eventually, the
system and the surroundings reach the same temperature and the
heat transfer ceases. It is at this point, that the two objects are said
to have reached thermal equilibrium.
Now in this chapter we learn a similar principle related to the flow
of heat. A temperature difference between two locations will cause
a flow of heat along a (thermally) conducting path between those
two locations. As long as the temperature difference is maintained,
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Unit 4 Unit 4: Thermal energy transfer
a flow of heat will occur. This flow of heat continues until the two
objects reach the same temperature. Once their temperatures
become equal, they are said to be at thermal equilibrium and the
flow of heat no longer takes place.
Example: You and I have never met. Not even shaken hands. Yet
if we are in good health you can bet that our body temperatures are
at 37 &ordm;C. We are both in thermal equilibrium. Ignoring the fact that
our extremities (e.g hands, feet and nose!) may be colder than the
rest of our body.
Methods of Heat Transfer
Temperature is a measure of the average amount of kinetic energy
possessed by the particles in a sample of matter. The more the
particles vibrate, translate and rotate, the greater the temperature of
the object. You have hopefully adopted an understanding of heat as
a flow of energy from a higher temperature object to a lower
temperature object. It is the temperature difference between the two
neighbouring objects that causes this heat transfer. The heat
transfer continues until the two objects have reached thermal
equilibrium and are at the same temperature.
Conduction - A Particulate View
Lets begin our discussion by returning to our thought experiment in
which a metal can containing hot water was placed within a
Styrofoam cup containing cold water. Heat is transferred from the
hot water to the cold water until both samples have the same
temperature. In this instance, the transfer of heat from the hot water
through the metal can to the cold water is sometimes referred to as
conduction. Conductive heat flow involves the transfer of heat
from one location to another in the absence of any material flow.
There is nothing physical or material moving from the hot water to
the cold water. Only energy is transferred from the hot water to the
cold water. Other than the loss of energy, there is nothing else
escaping from the hot water. And other than the gain of energy,
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there is nothing else entering the cold water.
In solids, atoms are bound to each other by a series of bonds,
analogous to springs as shown in Figure 1.1. When there is a
temperature difference in the solid, the hot side of the solid
experiences more vigorous atomic movements. The vibrations are
transmitted through the springs to the cooler side of the solid.
Eventually, they reach equilibrium, where all the atoms are
vibrating with the same energy.
Solids, especially metals, have free electrons, which are not bound
to any particular atom and can freely move about the solid. The
electrons in the hot side of the solid move faster than those on the
cooler side. This scenario is shown in Figure 1.2. As the electrons
undergo a series of collisions, the faster electrons give off some of
their energy to the slower electrons. Eventually, through a series of
random collisions, equilibrium is reached, where the electrons are
moving at the same average velocity. Conduction through electron
collision is more effective than through lattice vibration; this is why
metals generally are better heat conductors than ceramic materials,
which do not have many free electrons.
Figure 1.1 Conduction by lattice vibration
73
Unit 4 Unit 4: Thermal energy transfer
Figure 1.2 Conduction by particle collision
The effectiveness by which heat is transferred through a material is measured
by the thermal conductivity, k. A good conductor, such as copper, has a high
conductivity; a poor conductor, or an insulator, has a low conductivity.
Conductivity is measured in watts per meter per Kelvin (W/mK). The rate of
heat transfer by conduction is given by:
(Eq. 1.1)
where A is the cross-sectional area through which the heat is conducting, T is
the temperature difference between the two surfaces separated by a distance
Δx (see Figure 1.3). In heat transfer, a positive q means that heat is flowing
into the body, and a negative q represents heat leaving the body. The negative
This is sometimes not obvious: Like when you shake hands with a
person with cold hands. The conclusion that many people make is
that cold has travelled from that person to you. It is only heat that
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travels. The coldness that you feel is simply the heat leaving your
hand.
Simple Experiment: Put a block of wood and a bowl of water in
the fridge. Allow the water to freeze. Then take both of them out
and feel them. Which feels &quot;colder&quot;? Most will say the ice. So
which has the lowest temperature? If you say the ice, then you are
wrong! They both have the same temperature. It feels colder
because the ice conducts heat faster than wood. What you feel as
&quot;colder&quot; simply means there is more heat leaving your hand every
second than when touching the wood. So our concept of hot or cold
does not just depend on temperature but also on how fast heat
travels in different materials.
So how fast does heat travel?
Heat travels at different rates in different materials. The quantity of
heat transferred per unit time (in other words the rate of heat
transfer) is given by:
(7)
where k is the thermal conductivity, A is the cross-sectional area, L
is the length of the object, TH is the higher temperature at one end
of the solid, Tc is the lower temperature at the other end.
Demonstration: Three metal strips of the same length are heated
by the same flame at the same time. Matches placed at the end of
these strips do not light up at the same time. The reason is that the
three metal strips are made from 3 different materials: stainless
steel (k=14 W/mK), copper (k=401 W/mK) and Brass 220
(W/mK). Since copper is the most conducting, the match on it will
light up first and so on.
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Unit 4 Unit 4: Thermal energy transfer
Heat Transfer by Convection
Is conduction the only means of heat
transfer? Can heat be transferred
across through the bulk of an object
in methods other than conduction?
The answer is yes. The model of heat
transfer through the ceramic coffee
mug and the metal skillet involved
conduction. The ceramic of the
coffee mug and the metal of the skillet are both solids. Heat
transfer through solids occurs by conduction. This is primarily due
to the fact that solids have orderly arrangements of particles that
are fixed in place. Liquids and gases are not very good conductors
of heat. In fact, they are considered good thermal insulators. Heat
typically does not flow through liquids and gases by means of
conduction. Liquids and gases are fluids; their particles are not
fixed in place; they move about the bulk of the sample of matter.
The model used for
explaining heat transfer
through the bulk of liquids
and gases involves
convection. Convection is
the process of heat transfer
from one location to the next
by the movement of fluids.
The moving fluid carries
energy with it. The fluid
flows from a high
temperature location to a low
temperature location.
To understand convection in fluids, let's consider the heat transfer
through the water that is being heated in a pot on a stove. Of course
the source of the heat is the stove. The metal pot that holds the
water is heated by the stove. As the metal becomes hot, it begins to
conduct heat to the water. The water at the boundary with the metal
pan becomes hot. Fluids expand when heated and become less
dense. So as the water at the bottom of the pot becomes hot, its
density decreases. Differences in water density between the bottom
of the pot and the top of the pot results in the gradual formation of
circulation currents. Hot water begins to rise to the top of the pot
displacing the colder water that was originally there. And the
colder water that was present at the top of the pot moves towards
the bottom of the pot where it is heated and begins to rise. These
circulation currents slowly develop over time, providing the
pathway for heated water to transfer energy from the bottom of the
pot to the surface.
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Convection is the main method of heat transfer in fluids such as
water and air. It is often said that heat rises in these situations. The
more appropriate explanation is to say that heated fluid rises. The
convection method of heat transfer always involves the transfer of
heat by the movement of matter. Our model of convection
considers heat to be energy transfer that is simply the result of the
movement of more energetic particles.
The driving force of the circulation of fluid is natural - differences
in density between two locations as the result of fluid being heated
at some source. (Some sources introduce the concept of buoyant
forces to explain why the heated fluids rise. We will not pursue
such explanations here.) Natural convection is common in nature.
The earth's oceans and atmosphere are heated by natural
convection. Natural convection (or free convection) refers to a case
where the fluid movement is created by the warm fluid itself. The
density of fluid decrease as it is heated; thus, hot fluids are lighter
than cool fluids. Warm fluids surrounding a hot object rises, and
are replaced by cooler fluid. The result is a circulation of air above
the warm surface, as shown in Figure 1.4.
Radiation is the transfer of heat by means of electromagnetic
waves. To radiate means to send out or spread from a central
location. Whether it is light, sound, waves, rays, if something
radiates then it spreads outward from an origin. The transfer of heat
by radiation involves the carrying of energy from an origin to the
space surrounding it. The energy is carried by electromagnetic
waves and does not involve the movement or the interaction of
matter. Thermal radiation can occur through matter or through a
region of space that is void of matter (i.e., a vacuum). In fact, the
heat received on Earth from the sun is the result of electromagnetic
waves travelling through the void of space between the Earth and
the sun.
All objects radiate energy in the form of electromagnetic waves.
The rate at which this energy is released is proportional to the
Kelvin temperature (T).
The hotter the object, the more it radiates. A huge fire obviously
radiates off more energy than a hot cup of tea. The temperature also
affects the wavelength and frequency of the radiated waves. As the
77
Unit 4 Unit 4: Thermal energy transfer
temperature of an object increases, the wavelengths within the
spectra of the emitted radiation also decrease. Hotter objects tend to
emit shorter wavelength, higher frequency radiation. Objects at
typical room temperatures radiate energy as infrared waves. They
are invisible to the human eye. An infrared camera is capable of
detecting such radiation. Perhaps you have used a remote control of
a TV or have seen thermal photographs or videos of the radiation
surrounding a person or animal. The energy radiated from an object
is usually a range of wavelengths. This is usually referred to as an
emission spectrum. The tungsten filament of an incandescent light
bulb emits electromagnetic radiation in the visible (and beyond)
range. This radiation not only allows us to see, it also warms the
glass bulb that contains the filament. Put your hand near the bulb
(without touching it) and you will feel the radiation from the bulb
as well.
Thermal radiation is a form of heat
transfer because the electromagnetic
radiation emitted from the source carries
energy away from the source to
surrounding (or distant) objects. This
energy is absorbed by those objects,
causing the average kinetic energy of
their particles to increase and causing
the temperatures to rise. In this sense,
energy is transferred from one location to another by means of
electromagnetic radiation. The image at the right was taken by a
thermal imaging camera. The camera detects the radiation emitted
by objects and represents it by means of a color photograph.
(Images courtesy Peter Lewis and Chris West of Standford's
SLAC.)
Electromagnetic radiation is categorized into types by their
wavelengths. Radiations with shorter wavelengths are more
energetic, evident by the harmful gamma and x-rays on the shorter
end of the spectrum. Radio waves, which are used to carry radio
and TV signals, are much less energetic; however, they can pass
through walls with no difficulty due to their long wavelengths.
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PH002
System Work
When work is done by a thermodynamic system, it is ususlly a gas
that is doing the work. The work done by a gas at constant pressure
is:
Example
For non-constant pressure, the work can be visualized as the area
under the pressure-volume curve which represents the process
taking place. The more general expression for work done is:
Work done by a system decreases the internal energy of the system,
as indicated in the First Law of Thermodynamics. System work is a
major focus in the discussion of heat engines.
79
Heat and Work Example
Unit 4: Thermal energy transfer
Heat and Work Example
This example of the interchangeability of heat and work as agents
for adding energy to a system can help to dispel some
misconceptions about heat. I found the idea in a little article by
Mark Zemansky entitled &quot;The Use and Misuse of the Word 'Heat'
in Physics Teaching&quot;. One key idea from this example is that if you
are presented with a high temperature gas, you cannot tell whether
it reached that high temperature by being heated, or by having work
done on it, or a combination of the two.
To describe the energy that a high temperature object has, it is not a
correct use of the word heat to say that the object &quot;possesses heat&quot; it is better to say that it possesses internal energy as a result of its
molecular motion. The word heat is better reserved to describe the
process of transfer of energy from a high temperature object to a
lower temperature one. Surely you can take an object at low
internal energy and raise it to higher internal energy by heating it.
But you can also increase its internal energy by doing work on it,
and since the internal energy of a high temperature object resides in
random motion of the molecules, you can't tell which mechanism
was used to give it that energy.
In warning teachers and students alike about the pitfalls of
misusing the word &quot;heat&quot;, Mark Zemansky advises reflecting on
the jingle:
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Zemanzky's plea
Don't refer to the &quot;heat in a body&quot;,
or say &quot;this object has twice as
&quot;Teaching thermal physics much heat as that body&quot;. He also
Is as easy as a song:
objects to the use of the vague term
You think you make it simpler &quot;thermal energy&quot; and to the use of
When you make it slightly the word &quot;heat&quot; as a verb, because
wrong.&quot;
they feed the misconceptions, but it
is hard to avoid those terms. He
would counsel the introduction and
use of the concept of internal
energy as quickly as possible.
Zemansky points to the First Law of Thermodynamics as a
clarifying relationship. The First Law identifies both heat and work
as methods of energy transfer which can bring about a change in
the internal energy of a system. After that, neither the words work
or heat have any usefulness in describing the final state of the
sytem - we can speak only of the internal energy of the system.
Mechanical Equivalent of Heat
Constant Pressure Work
From the definition
of work W=Fd for a
constant force F
acting along a
distance d:
For a constant
pressure process
only, the work is:
PV = nRT
where P is the pressure, V the volume, R is a constant = 8.314
Jmol-1K-1, T is the temperature in Kelvins, and n is the number of
81
Constant Pressure Work
Unit 4: Thermal energy transfer
moles defined as number of gas particles (atoms or molecules)
divided by Avogadro's number (NA = 6.023 X 1023).
The above equation is counterintuitive since it says that all gases
contain the same number of atoms or molecules for the same
pressure, volume, and temperature regardless of the type of gas.
This is definitely not the case for solids and liquids. Recall such
thermal properties as linear expansion for solids and liquids. The
expansion coefficient depends strongly on the type of material.
However, we find that for gases this is not the case since the
volume occupied by the gas does not rely on the type of gas. So for
the first time we see that there are material properties which are
material
independent.
Although the product of pressure and volume gives the work
done on or by a gas, in practice this may not be a constant
quantity throughout a particular experiment. For example a plot of
pressure versus volume may look something like this
So how can we measure the work done on a gas from this graph?
Answer: the work is the area underneath the graph.
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Just be careful to give the correct sign to the work. As we saw in
the section on the first law of thermodynamics, the work done on a
system is negative, and the work done by the system is positive. In
this case, the gas is expanding, as indicated by the arrow on the
curve, so the gas itself is doing work, which means the work done
by it is given a positive sign.
Although the ideal gas equation is quite convenient for solving
many problems, it is still a macroscopic approach and does not
give us any deeper understanding of what pressure and temperature
actually are. A little bit like driving a car but not knowing how it
works. Let's take a quick look under the bonnet (or hood).
Unit summary
In this unit you learned Now in this chapter we learn a principle
Summary
related to the flow of heat. A temperature difference between two
locations will cause a flow of heat along a (thermally) conducting
path between those two locations. As long as the temperature
difference is maintained, a flow of heat will occur. This flow of
heat continues until the two objects reach the same temperature.
Once their temperatures become equal, they are said to be at
thermal equilibrium and the flow of heat no longer takes place.
Heat flow and work are both ways of transferring energy. As
illustrated in the heat and work example, the temperature of a gas
can be raised either by heating it, by doing work on it, or a
combination of the two.
83
Constant Pressure Work
Unit 4: Thermal energy transfer
In a classic experiment in 1843, James Joule showed the energy
equivalence of heating and doing work by using the change in
potential energy of falling masses to stir an insulated container of
water with paddles. Careful measurements showed the increase in
the temperature of the water to be proportional to the mechanical
energy used to stir the water. At that time calories were the
accepted unit of heat and joules became the accepted unit of
mechanical energy.
Assignment
Explain why Zimbabwean women set huge fire adjacent to place of
winnowing?
Assignment
Possible solution: A fire placed on the floor of the place warms up
the air in the surrounding of the place. Air present near the fire is
warm up. As the air warms up, it expands, becomes less dense and
begins to rise. The hot air rises. The cold air moves to the bottom to
replace the hot air that has risen. As the colder air approaches the
fire, it becomes warmed up by the fire and begins to rise. Once
more, convection currents are slowly formed. Air travels along
these pathways, carrying energy with it from the fire throughout the
winnowing place. Thus providing wind for winnowing.
When is something neither Hot nor Cold?
Answer: When there is no heat transfer between you and the
object. That is when H = 0 i.e when the object is at the same
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Assessment
Assessment
An aluminium pot contains water that is kept steadily boiling (100
&ordm;C). The bottom surface of the pot, which is 12 mm thick and
in area, is maintained at a temperature of
by
an electric heating unit. Find the rate at which heat is transferred
through the bottom surface. Compare this with a copper based pot.
The thermal conductivities for aluminium and copper are kAl = 235
Wm-1K-1 and kCu = 401 Wm-1K-1 respectively.
Solution:
The following is a schematic diagram of the pot.
The rate of heat conduction across the base is given by equation .
For the aluminium base:
TH = 102 &ordm;C, TC = 100 &ordm;C, L=12 mm = 0.012 m, k = kAl = 235 Wm1 -1
K
Base area = A =
= 0.015 m2.
Substituting these into the above equation:
Js-1 (or Watts)
For the copper base k = kCu = 401 Wm-1K-1. So the rate of heat
85
Constant Pressure Work
Unit 4: Thermal energy transfer
conduction across the base is
Js-1 (or Watts)
So the copper based pot transfers 1.7 times more energy every
second compared with the aluminium pot. Generally copper bottom
pots are more expensive.
86
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