BUSE Virtual Open and Distance Learning STUDY GUIDE PH002 THERMAL PHYSICS Bindura University of Sc ience Education Physics and Mathematics Department BUSE Copyright BUSE-NDLOVU.S. 2012 Bindura University of Sc ience Education Physics and Mathematics Department Bindura University Of Science Education Private Bag 1020 Chimurenga Road 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 About this STUDY GUIDE 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 3.5.2 Adiabatic 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 4.4 Thermal Radiation ............................................................................................................................... 80 iv Contents Unit summary ................................................................................................................. 83 Assignment ..................................................................................................................... 84 Assessment...................................................................................................................... 85 PH002 About this STUDY GUIDE 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 lecturers during college days, downloaded from internet and from the 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. Information contained in the course overview will help you determine: 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 starting your study. 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 About this STUDY GUIDE 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. Your comments 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 feedback might include comments on: 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 4 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 management, efficient reading, questioning/listening/observing skills, 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 http://answers.yahoo.com/question/index?qid=20080324094219AA eeF1y http://physics.about.com/od/glossary/g/phasediagram.htm 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 address)? 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?] 6 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 text, a new task or change in activity; they have been included to help you 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 Reading 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 about relative temperatures. 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 Adding extra rows to the 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°C and to boil at 100°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 known as a centigrade thermometer. A centigrade thermometer has 100 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 °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. °C = K - 273.15° K = °C + 273.15 11 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°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°C”. Since these are the lowest values of volume and pressure that are possible, it is reasonable to conclude that -273°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°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 the steady reading on the thermometer is taken. . (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 PH002 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. Temperature range: from about -250 to about 1500 (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 "hard sphere" 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 "zeroth law" states that if two systems are at the same time in thermal equilibrium with a third system, they are in thermal equilibrium with each other. If 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 > 0 if heat is added to the system Q < 0 if heat is given out by the system W > 0 if work is done by the system W < 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 "filled" 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 "constant" 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 cmT 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 °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 cmT 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° 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 Lead 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 ºC 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 26 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 ºC and pressure of 50 kPa. Draw a line up from 50 ºC 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 ºC. 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°C. Liquid water is most often used in cooking, and the temperature varies between 0°C and 100°C. Water boils at 100 °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°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 by about 1 °C. 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°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 °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°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] your body text 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 "zeroth law" states that if two systems are at the same time in thermal equilibrium with a third system, they are in thermal equilibrium with each other 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 ºC 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 32 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 ºC direction of heat flow in the rod x Heat (ii) Temperature gradient of unlagged metal rod θ 100 ºC 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 - Temperature gradient 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 mcT 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 ºC, must be added to 1.0 L of hot tea, whose initial temperature is 90 ºC, in order that the final temperature of the mixture be 10 ºC? 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 ºC, must come from the tea. The following are the three stages the ice must go through to reach the final temperature of 10 ºC: (i) ice heats up from -10 ºC to 0 ºC (ii) ice melts at 0 ºC (iii) melted ice heats up from 0 ºC to 10 ºC (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 ºC - (-10 ºC)) = mice 2100 x 10 = 21000 mice (ii) mice Lv = mice 333000 (iii) mice cwater T(iii) = mice (4190 J. kg-1 K-1) (10 ºC - 0 ºC) = mice41900 (iv) mtea cwater Ttea= (1 kg) (4190 J. kg-1 K-1) (90 ºC - 10 ºC) = 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. [Add introductory text here] 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 42 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. 44 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. 4. Adiabatic - in an adiabatic process, no heat is added or 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: 50 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. Adiabatic processes 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. This leads to the expression 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. 54 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 your cheeks. 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 <v2> 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) 56 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 58 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 ºC and on the other side is 28 ºC. (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 body radiator at the same temperature, estimate the steady temperature of 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 x300 1200K Temperature gradient (b) T 5 2.5x10 28 8 Km1 and t= 3600s = 2 12 x10 Q =kAt x temperature gradient 28 8 Q 0.13x2 x3600 x( )J 12 x102 Q = 156 kJ (c )When the temperature is steady, power radiated from filament = power received = 60 W 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 1933K [Add assignment text 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 64 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. 66 PH002 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 67 PH002 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 69 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 70 PH002 initially 70°C and that the temperature of the cold water in the outer cup is initially 5°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°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, 71 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 ºC. 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, 72 PH002 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 sign in Eqn. 1.1 ensures that this convention is obeyed. 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 74 PH002 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 "colder"? 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 "colder" 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. 75 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. 76 PH002 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. Heat Transfer by Radiation 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. 78 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 [Second topic heading] 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 "The Use and Misuse of the Word 'Heat' in Physics Teaching". 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 "possesses heat" 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 "heat", Mark Zemansky advises reflecting on the jingle: 80 PH002 Zemanzky's plea Don't refer to the "heat in a body", or say "this object has twice as "Teaching thermal physics much heat as that body". He also Is as easy as a song: objects to the use of the vague term You think you make it simpler "thermal energy" and to the use of When you make it slightly the word "heat" as a verb, because wrong." 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. 82 PH002 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 temperature as your hand. 84 PH002 Assessment Assessment An aluminium pot contains water that is kept steadily boiling (100 ºC). 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 ºC, TC = 100 ºC, 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