Thermal Physics I PDF

PHYSICS –PHY101 VU Summary of Lecture 36 – THERMAL PHYSICS I 1. The ancient view was that heat is a colourless, weightless, fluid which occupies no volume has no smell, etc. This imagined...

PHYSICS –PHY101 VU Summary of Lecture 36 – THERMAL PHYSICS I 1. The ancient view was that heat is a colourless, weightless, fluid which occupies no volume has no smell, etc. This imagined substance - called phlogiston - was supposedly stored in objects and transferred between objects. It took a long time to reject this notion. Why is it wrong? Because (as we will see) heat can be created and destroyed, whereas liquids keep their volume and cannot be created or destroyed. 2. We are all familiar with an intuitive notion of temperature. We know that hotter things have higher temperature. But let us try to define temperature more rigorously. In the diagram below, the three bodies A,B,C are in contact with each other. After sufficient time passes, one thing will be common to all three - a quantity that we call temperature. A B C 3. Now let us understand heat. Heat is energy, but it is a very special kind of energy: it is that energy which flows from a system at high temprature to a system at low temperature. hi gh T heat l ow T Stated in a slightly different way: heat is the flow of internal energy due to a temperature difference. (Note that we do not have to know about atoms, molecules, and the internal composition of a body to be able to define heat - all that will come later). 4. The term "thermal equilibrium" is extremely important to our understanding of heat. If some objects (say, a glass of water placed in the open atmosphere) are put in thermal contact but there is no heat exchange, then we say that the objects are in thermal equilibrium. So, if the glass of water is hot or cold initially, after sufficient time passes it will be in thermal equilibrium and will neither receive nor lose heat to the atmosphere. 5. We have an intuitive understanding of temperature, but how do we measure it? Answer: by looking at some physical property that changes when the temperature changes. So, for example, when the temperature rises most things expand, the electrical resistance changes, some things change colour, etc. These are called "thermometric properties". © Copyright Virtual University of Pakistan 117 PHYSICS –PHY101 VU 6. Let's take a practical example: the constant volume gas thermometer. Here the reference level is kept fixed by raising or lowering the tube on the right side (the tube below is made of rubber or some flexible material). So the gas volume is fixed. The gas pressure is   g  h, and so the pressure is known from measuring h. To find the temperature of a substance, the gas flask is placed in thermal contact with the substance. When the temperature is high, the pressure is large. From the graph of pressure versus temperature, you can easily read off the temperature. Note that we are using two fixed points, which we call 00C and 1000C. This is called the Centigrade scale, and the two fixed points correspond to the freezing and boiling of water. 7. There is a temperature below which it is not possible to go. In other words, if you cool and cool there comes a point after which you cannot cool any more! How do we know this? Take different gases and plot how their pressure changes as you cool them down. You can see from the graph that all the lines, when drawn backwards, meet exactly at one point. This point is the absolute zero of temperature and lies at about -273.150C. This is also called 00 K, or zero degrees Kelvin. 8. Temperature Scales. If you had been born 300 years ago, and if you had discovered a reliable way to tell different levels of "hotness", then maybe today there would be a temperature scale named after you! The relation between Centigrade, Fahrenheit, and Kelvin scales is illustrated to your right. You can see that absolute zero corresponds to -4600 K , - 273.150 C , and 00 K. The Kelvin scale is the most suited for scientific purposes, and you should be careful to use this in all heat related calculations. © Copyright Virtual University of Pakistan 118 PHYSICS –PHY101 VU Conversion between degrees Celsius and degrees Fahrenheit: TF  ( 95 F  / C  )TC  32  F Conversion between degrees Fahrenheit and degrees Celsius: TC  ( 95 C  / F  )(TF  32  F ) Conversion between Celsius and Kelvin temperatures: T  TC  273.15 9. How hot is hot, and how cold is cold? Whenever a scientist says something is large or small, it is always relative to something. The hottest thing ever was the universe when it just came into existence (more about this in the last lecture). After this comes the hydrogen bomb (108 K). The surface of the sun is not so hot, only 5500K. Copper melts around 1000K, water turns to steam at 373K. If you cool further, then all the gases start to solidify. The lowest temperature that has ever been achieved is 10-7 K , which is one tenth of one millionth of one degree! Why not still lower? We shall later why this is not possible. 10. When you rub your hands, they get hot. Mechanical work has been converted into heat. The first person to investigate this scientifically was Joule. In the experiment below, he allowed a weight to drop. This turned a paddle that stirred up the water and caused the temperature to rise. The water got hotter if the weight was released from a greater height. Joule established the units for the mechanical equivalent of heat. The units we use today are: 1 calorie (1 cal) raises the temperature of 1 g of water by 1 °C 1 cal = 4.186 Joule, 1 kilocalorie (1 kcal) = 1000 cal Remember also that joule is the unit of work: when a force of one newton acts through a distance of one metre, the work done is one joule. © Copyright Virtual University of Pakistan 119 PHYSICS –PHY101 VU 11. Let's now consider one important effect of heat - most things expand when heated. Of course, our world is 3-dimensional but if there is a thin long rod then the most visible effect of heating it is that the rod increases in length. By how much? L0 at T0 L at T L Look at the above diagram. Call the length of the rod at T0 as L0. When the rod is heated to T , then the length increases to L  L0 1   T  T0  . The difference of the lengths is, L  L0   L0 T  T0  , or L   L0 T. Here  is just a dimensionless number that tells you how a particular material expands. It is called the coefficient of linear expansion. If  was L / L0 zero, then the material would not expand at all. You can also write it as  . T V / V0 Similarly, define a coefficient of volume expansion  call it   as  . T 12. There is a relation between  and  , the linear and volume coefficients. Let's look at the change in volume due to expansion: V   ( L  L)3  ( L   LT )3  L3  3 L3T  3 2 L3T 2   3 L3T 3  L3  3 L3T  V  3V T We are only looking for small changes, so the higher terms in T can be safely dropped. Hence V  3V T. From the definition, this we immediately see that   3. Just to get an idea, here is what  looks like for various different materials: Steel Quartz Glass Al Hg Air 106 10 5 10 4 10 3 Coefficient of volume expansion  (1/°C ) 13. We can use the fact that different metals expand at different rates to make thermostats. For example, you need a thermostat to prevent an electric iron from getting too hot or a © Copyright Virtual University of Pakistan 120 PHYSICS –PHY101 VU refrigerator from getting too cold. In the diagram below, one metal is bonded to another. When they expand together, one expands less than the other and the shape is distorted. This breaks off a circuit, as shown here. 14. Water behaves strangely - for certain temperatures, it contracts when heated (instead of expanding)! Look at the graph: between 00C and 40C, it exhibits strange or anomalous behaviour. The reason is complicated and has to do with the molecular structure of water. If water behaved normally, it would be very bad for fish in the winter because they rely upon the bottom of the lake or sea to remain liquid even though there the surface is frozen. Q 15. We define the "heat capacity" of a body as C  , where T is the increase in T temperature when an amount of heat Q is added to the body. Heat capacity is always positive; Q and T have the same sign. The larger the heat capacity, the smaller is the change in the body's temperature when a fixed amount of heat is added. In general, Q  mcT , where Q  heat added , m  mass , c  specific heat , and T = change in temperature. Water has a very large specific heat c, c  1.0 cal /(º C g ); this means it takes one calorie to raise the temperature of 1 gm of water by 1 degree Celsius. In joules per kilogram this is the same as 4186. In the table are the specific heats of various common materials. You can see that metals have small c, which means that it is relatively easy to raise or lower their temperatures. The opposite is true of water. Note also that steam and ice have smaller c ' s than water. This shows that knowing the chemical composition is not enough. © Copyright Virtual University of Pakistan 121 PHYSICS –PHY101 VU 16. A 0.5-kg block of metal with an initial temperature of 30.0 C is dropped into a container holding 1.12 kg of water at 20.0 C. If the final temperature of the block-water system is 20.4 C , what is the specific heat of the metal? SOLUTION: Write an expression for the heat flow out of the block Qblock  mb cb (Tb  T ). Do the same for water, Qwater  mwcw (T  Tw ). Now use the fact that all the energy that is lost by the block is gained by the water: Qblock  Qwater  Tmbcb (Tb  T )  mwcw (T  Tw )  0 mwcw (T  Tw ) (1.12kg )[4186 J /(kg  K )](20.4 C  20.0 C ) From this, cb   mb (Tb  T ) (0.500kg )(30.0 C  20.4 C )  391J /(kg  K ) 17. When lifting a "daigchee" from a stove, you would be wise to use a cloth. Why? Because metals transfer, or conduct, heat easily whereas cloth does not. Scientifically we define conductivity using experiments and apparatus similar to the following: Heat flows from the hotter to the colder plate. Let us use the following symbols: k = thermal conductivity Q = heat transferred A = cross sectional area t = duration of heat transfer L = length T = temperature difference  T  Then the heat transferred in time t is, Q  kA   t. This formula allows us to measure  L  k if all the other quantities in it are measured. 18. Conduction is one possible way by which heat is transferred from one portion of a system to another. It does not involve physical transport of particles. However, there is another way by which heat can be transferred - convection. In convection, heat is carried by a moving fluid. So when you heat a pot of water, molecules at the bottom move up, and the ones at the top come down - the water has currents inside it that transfer heat. Another mechanism for transferring heat is through radiation. We have already talked about this while discussing blackbody radiation and the Stefan-Boltzman Law. © Copyright Virtual University of Pakistan 122 PHYSICS –PHY101 VU Summary of Lecture 37 – THERMAL PHYSICS II 1. Let us agree to call whatever we are studying the "system" (a mixture of ice and water, a hot gas, etc). The state of this system is specified by giving its pressure, volume, temperature, etc. These are called "thermodynamic variables". The relation between these variables is called the "equation of state". This equation relates P,V,T. So, for example: P  f (V , T ) (knowing V and T gives you P) V  g ( P, T ) (knowing P and T gives you V ) T  h( P,V ) (knowing P and V gives you T ) For an ideal gas, the equation of state is PV  Nk BT. Here N is the number of molecules in the gas, T is the temperature in degrees Kelvin, and k B is called the Boltzman cosntant. This EOS is easy to derive, although I shall not do it here. In principle it is possible to mathematically derive the EOS from the underlying properties of the atoms and molecules in the system. In practice, however, this is a very difficult task (except for an ideal gas). However the EOS can be discovered experimentally. 2. Thermodynamics is the study of heat and how it flows. The First Law of Thermodynamics is actually just an acknowledgment that heat is a form of energy (and, of course, energy is always conserved!). Mathematically, the First Law states that: E  q  w where: E  internal energy of the system q  heat transferred to the system from the surroundings w  work done on the system by the surroundings. In words, what the above formula says is this: if you do an amount of work w and also transfer an amount of heat q, then the sum of these two quantities will be the additional amount of energy E that is stored in the system. There could be nothing simpler! Note: q and w are positive if heat is added to or work is done on a system, and negative if heat is removed from the system, or if the system does work on the environment. 3. Work and heat are called path variables - their values depend on the steps leading from one state to another. To give an obvious example: suppose you drag a heavy box on the floor in a straight line from point A to point B, and then in a very roundabout way from A to B again. Obviously the work done by you will be very different. On the other hand, the internal energy does not depend upon the path. Example: a gas has internal energy proportional to its temperature. It makes no difference whether the gas had been slowly heated or rapidly; it will have the same internal energy. Since the internal energy does not  State 2 depend on the path we can write dE  E2  E1  E. State1 © Copyright Virtual University of Pakistan 123 PHYSICS –PHY101 VU 4. Let us calculate the work done by an expanding gas. In the diagram below, suppose the piston moves out by distance dx, then the work done is dw  Fdx where F is the force exerted on the gas. But F  P  A   Pext  A. (The minus sign is important to understand; remember that the force exerted by the gas on the piston is the negative of the force that the piston exerts on the gas). So dw  Fdx   Pext  A  dx   Pext dV. The work done by the V2 expanding gas is w    Pext dV where V1 and V2 are the initial and final volumes. V1 Closed system Compressed ideal gas Pint Pext Note that w depends on the path. Let's consider 3 different paths. a)Suppose that the gas expands into the vacuum, Pext  0. Then w  0. So, if in the figure below, the valve is opened, then gas will flow into the vacuum without doing work. Gas Vacuum b)Suppose the pressure outside has some constant value. Then, w  Pext  (V2  V1 ) or, w  Pext V. You can see that this is just the area of the rectangle below. P  P0 ,V0  W  P V P V V c)We can also let the gas expand so that the internal and external pressures are almost the same (this is called reversible expansion). Then the work done on the P  P1 , V1  system is: dw   Pext dV. Using PV  Nk BT gives, V2 Nk BT V w   dV   Nk BT (ln V2  ln V1 )   Nk BT ln 2 V1 V V1  P2 ,V2  Of course, the work done by the gas as it expands is positive since V2 is larger than V1. Remember that the log function is positive if its argument is bigger than 1. © Copyright Virtual University of Pakistan 124 PHYSICS –PHY101 VU 5. The internal energy of a gas depends only on the number of molecules it contains and 3 on the temperature, E  Nk BT. In a free expansion of gas (it doesn't do any work in this 2 case), the initial and final internal energies are equal, E f = E i. 6. Imagine that a substance is heated while keeping its volume constant. Obviously, you have to supply more heat as you raise the temperature higher, QV  CV T , where CV is given the name "specific heat at constant volume". No work is done since there is no expansion, W  0. Hence, using the First Law, QV  E  W  E , and so E  CV T. 7. Now take the same system as above, but allow it to expand at constant pressure as it is heated. Then you will have to supply an amount of heat, QP  CP T. Here, CP is called the specific heat at constant pressure, QP  E  w  E  PV. Hence, CP T  U  PV. If we supply only a small amount of heat then, CP dT  dU  PdV. This gives, CP dT  CV dT  PdV. If we consider the special case of an ideal gas, then PV  Nk BT. Hence, PdV  Nk B dT and CP dT  CV dT  Nk B dT. Hence, CP  CV  Nk B. 3 dU 3 Since the internal energy is E  Nk BT , it follows that CV   Nk B. From this, 2 dT 2 3 5 the specific heat at constant pressure is: CP  Nk B  Nk B  Nk B. From the equation 2 2 Nk B CP  CV C CP  CV  Nk B , it follows that . Let us define a new symbol, P  . CV CV CV 8. We can find the EOS of an ideal gas when it expands without losing any heat (this is called adiabiatic expansion). In this case, dQ  dE  dW  0. Hence, CV dT  PdV  0, dV dT Nk B dV dT dV or CV dT  Nk BT  0. Dividing by dT gives,   0 or     1  0. V T CV V T V Integrating this gives, ln T     1 ln V  Constant. Equivalently, ln TV  1   Constant. A more convenient form is: TV  1  Constant. This holds for the entire adiabatic expansion. 9. Almost nothing beats the importance of the Second Law Of Thermodynamics. There are many equivalent ways of stating it. The one I like is: "There can be no process whose only final result is to transfer thermal energy from a cooler object to a hotter object". This seems extremely simple (and almost useless), but in fact it tells you, among other things, that no perpetual motion machine (such as that which generates electricity without any fuel input) can ever be built. If someone could build such a machine, then it could also be used to transfer heat from a cold object to a hot object - a contradiction with the Second Law. © Copyright Virtual University of Pakistan 125 PHYSICS –PHY101 VU 10. Before we consider some implications of the Second Law, I want to introduce the concept of a thermal reservoir. Suppose you put a thermometer in your mouth. You will transfer a small amount of heat to the thermometer, but because your body is so big this will make no measureable difference to your body temperature. Your body therefore is a thermal reservoir as far as the thermometer is concerned. More generally, any large mass of material will act as a thermal reservoir if it is at constant temperature. A heat engine (e.g. a car engine, refrigerator, etc) operates between reservoirs at two different temperatures. 11. A heat engine works between a high temperature TH Hot reservoir at temperature T h and a low temperature TC. It aborbs heat Qin , h from the hot reservoir and rejects heat Qout , c into the cold Qh reservoir. Since the internal energy of the engine does not change, U  0 and the First Law gives simply, Heat engine W Q  U  W  W. The work done by the engine is W  Qin , h  Qout , c. Now define the efficiency of the work done by engine W Qc machine as   . Then heat put in Qin , h Cold reservoir at temperature Tc Q  Qout , c Q   in , h  1  out , c. Now, the heat that is Qin , h Qin , h transferred out of or into a reservoir is proportional to its absolute temperature. Hence we TC arrive at the important result that   1 . This number is always less than one, showing TH that no machine can convert all the input heat into useful work. As an example, a nuclear reactor has temperature 300 0 C at the core and rejects heat into a river at 30 0 C. The 30  273 maximum effic iency it can have is   1   0.471 300  273 12. A refrigerator is a heat engine working in reverse. Hot reservoir at temperature Th Work is done (by an electric motor) to pump heat from a cold reservoir (inside of refrigerator) to the Qh hot exterior. The Second Law can be shown to imply the following: it is impossible for a refrigerator to Refrigerator W produce no other effect than the transfer of thermal energy from a cold object to a hot object. Again, the Qc efficiency of a refrigerator is always less than one. Cold reservoir at temperatureTc © Copyright Virtual University of Pakistan 126 PHYSICS –PHY101 VU Summary of Lecture 38 – THERMAL PHYSICS III 1. In the two previous lectures, I concentrated exclusively on heat as a form of energy that flows from a hot to a cold body. I made no mention of the fact that all matter is made of atoms and molecules; the notion of heat would exist even if we did not know this. But, the modern understanding of heat is that it is the random kinetic energy of atoms. So, for example, the difference between a cold and hot gas is illustrated below. cold gas hot gas Longer arrows denote atoms that are moving faster. The hotter gas has, on the average, faster moving atoms. 2. The study of heat, considered as arising from the random motion of the basic constituents of matter, is an area of physics called statistical mechanics. Its goal is the understanding, and prediction of macroscopic phenomena, and the calculation of macroscopic properties from the properties of individual molecules. Temperature is the average energy related to the speed of atoms in an object, and heat is the amount of energy transferred from one object to another. 3. Imagine a gas so dilute that atoms rarely collide with each other (this is also called an ideal gas). We can readily understand why the pressure is directly proportional to the temperature for a gas confined to a box: increasing the temperature T makes gas molecules move faster, striking the walls of the container harder and more often, thus giving an increase in pressure P, i.e. P  T. On the other hand, for an ideal gas at constant pressure, the volume is directly proportional to the temperature : V  T. So, as the air in a balloon is heated up, its volume will increase in direct proportion because of the impact of the atoms on the balloon walls. Note that this is where the "absolute" or Kelvin scale comes from: at 0° K an ideal gas would have zero volume because the atoms would not be moving. 4. Pressure is related to the outward force per unit area exerted by the gas on the container wall. Newton's v p 2nd Law is: F . Here p is to be understood as t the momentum destroyed in time t. Using this, we can p do a little calculation to find the pressure in an ideal gas. v © Copyright Virtual University of Pakistan 127 PHYSICS –PHY101 VU Suppose there are N atoms in a box of volume, and the average speed of an atom in the x direction is v x. By symmetry, half are moving in the  x and half in the  x directions. N 1 So the number of atoms that hit the wall in time t is v x tA. Hence the change in 2V 1 N  N momentum is p   2mv x    v x tA   mv 2x At. From this, we can calculate the 2V  V F 1 p N 1  pressure P    mv 2x. Hence PV  Nmv 2x  2 N  mv 2x . A A t V  2  av 5. Let us return to our intuitive understanding of heat as the random energy of small particles. 1  Now,  mv 2x  is the average kinetic energy of a particle on account of its motion in the 2  av 1  x direction. This will be greater for higher temperatures, so  mv 2x   T , where T is the 2 av 1  absolute temperature. So we write  mv 2x   k BT , which you can think of as the 2  av definition of the Boltzmann constant, k B. So we immediately get PV  Nk BT , the equation obeyed by an ideal gas (dilute, no collisions). Now, there is nothing special about any     v    v  and therefore the average total velocity is particular direction, so v 2x av 2 y av 2 z av  v    v    v    v   3 v . So the total average kinetic energy of one atom is, 2 av 2 x av 2 y av 2 z av 2 x av 1  3 1  3 K av   mv 2   k BT , and the energy of the entire gas is K  N  mv 2   Nk BT. 2 av 2 2 av 2   3k T From this, the mean squared speed of atoms in a gas at temperature is v 2  B. av m 6. Heat, which is random kinetic energy, causes changes of phase in matter: a) Water molecules attract each other, but if water is heated then they can escape and water becomes steam. b) If steam is heated, the water molecules break up and water becomes separated hydrogen and oxygen atoms. c) If a gas of atoms gets hot enough, then the atoms collide so violently that they lose electrons (i.e. get ionized). d) More heat (such as inside the core of a star) will break up atomic nuclei into protons and neutrons. © Copyright Virtual University of Pakistan 128 PHYSICS –PHY101 VU 7. Evaporation: if you leave water in a glass, it is no longer there after some time. Why? Water molecules with high enough KE can break free from the water and escape. But how does a water molecule get enough energy to escape? This comes from random inter- molecular collisions, such as illustrated below. Two molecules collide, and one slows down while the other speeds up. The faster molecule might escape the water's surface. 8. Heat needs to be supplied to cause a change of phase: a) To melt a solid (ice, wax, iron) we must supply the "latent heat of fusion" LF for unit mass of the solid. For mass m we must supply an amount of heat Q  mLF to overcome the attraction between molecules. No temperature change occurs in the melting. b) To vapourize a substance (convert to gas or vapour), we must supply the "latent heat of vapourization" LV for unit mass of the substance. For mass m we must supply an amount of heat Q  mLV. No temperature change occurs in the vapourization. c) To raise the temperature of a substance we must supply the "specific heat" C for unit mass of the substance by 1 degree Kelvin. To change the temperature of mass m by T , we must supply heat Q  mC T. Q  ci mT Q  mL F Q  c w m  T Q  mLV Q  cs mT no increase in no increase in temperature temperature added heat added heat melts ice vaporizes water In the above diagram you see what happens as ice is heated. The amount of heat needed is indicated as well. Note that no change of temperature happens until the phase change is completed. The same physical situation is represented to the right as well where the temperature is plotted against Q. © Copyright Virtual University of Pakistan 129 PHYSICS –PHY101 VU 9. A very important concept is that of entropy. It was first introduced in the context of thermodynamics. Then, when the statistical nature of heat became clear a century later, it was understood in very different terms. Let us begin with thermodynamics: suppose a system is at temperature T and a small amount of heat dQ is added to it. Then, the small dQ increase in the entropy of the system is dS . What if we keep adding little bits of T heat? Then the temperature of the system will change by a finite amount, and the change T2 dQ in entropy is got by adding up all the small changes: S   T1 T. Now let's work out how much the entropy changes when we add heat dQ to an ideal gas. dV From the First Law, dQ  dU  dW  dU  PdV  CV dT  Nk BT. The entropy change V dQ dT dV is dS   CV  Nk B , and for a finite change we simply integrate, T T V dQ T V 3 T V S    CV ln 2  Nk B ln 2  Nk B ln 2 + Nk B ln 2. Note that the following: T T1 V1 2 T1 V1 a)The entropy increases if we heat a gas (T2  T1 ). b)The entropy increases if the volume increases (V2  V1 ). 10. In statistical mechanics, we interpret entropy as the degree of disorder. A gas with all atoms at rest is considered ordered, while a hot gas having atoms buzzing around in all directions is more disordered and has greater entropy. Similarly, as in the above example, when a gas expands and occupies greater volume, it becomes even more disordered and the entropy increases. 11. We always talk about averages, but how do you define them mathematically? Take the example of cars moving along a road at different speeds. Number of cars per 5-mph interval Call n  vi  the number of cars with speed vi , then N   n  vi  is the total number of cars and i  v nv  i i the average speed is defined as v  i. nv  i i We define the probability of finding a car with n  vi  n  vi  Speed speed vi as P  vi   . We can  n  vi  N ©i Copyright Virtual University of Pakistan 130 also write the average speed as, v   vi P  v i . i