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Topic 10 Thermal Physics 2012

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Topic 10 Thermal Physics 2012

1. 1. Thermal Physics AHL Topic 10
2. 2. Thermodynamics Thermodynamics is the study of heat and its transformation into mechanical energy, as heat and work. The word is derived from the Greek meaning ‘movement of heat’. It was developed in the mid 1800’s before atomic and molecular theory was developed.
3. 3. Thermodynamics Work is defined as: The quantity of energy transferred from one system to another by ordinary mechanical processes. Heat is defined as: A transfer of energy from one body to another body at a lower temperature.
4. 4. Thermodynamics From this we can see that thermodynamics describes the relationship between heat and work. To distinguish the two: Heat is the transfer of energy due to a temperature difference. Work is the transfer of energy that is not due to a temperature difference.
5. 5. Thermodynamics The foundation of this area of study is The law of conservation of energy and the fact that heat flows from hot to cold. In discussing thermodynamics, we will refer to different systems. A system is just a group of objects we wish to consider. Everything else in the universe will be called the environment.
6. 6. Thermodynamics Consider a hot gas separated from a cold gas by a glass wall. In macroscopic terms, we know that the hot gas gets cooler and the cool gas gets hotter. The molecules in the hot gas hit the glass and set those molecules in faster motion.
7. 7. Thermodynamics This then sets in train a set of collisions which sees the energy being transferred to the cold gas. If we were to observe a single collision, we could analyse the energy transfer using the laws of mechanics.
8. 8. Thermodynamics We could say that one molecule has transferred energy by doing work on another. Heat is therefore the work done on a molecular level.
9. 9. Thermodynamics This is not the complete story. Although the cool gas contains, on average, slower molecules than in the hot gas, it does contain some fast moving molecules. Likewise, the hot gas contains slow moving molecules.
10. 10. Thermodynamics From above, it should be possible to for the cold gas to transfer energy to the hot gas So the cool gas would get cooler and the hot gas hotter. This does not disobey any classical theory of mechanics. We do know however that this cannot occur.
11. 11. Thermodynamics To explain this, we cannot look at this the effects of single molecules or even a few molecules. We must, when discussing heat, look at the overall effects of a large number of molecules and the average energies and distribution of energies and velocities.
12. 12. Thermodynamics This is what is meant by a system of particles in thermodynamics. A system could be any group of atoms, molecule or particles we wish to deal with. It may be the steam in a steam engine, the earth’s atmosphere or the body of a living creature.
13. 13. Thermodynamics The operation of changing the system from its initial state to a final state is called the, Thermodynamic process.
14. 14. Thermodynamics During this process, heat may be transferred into or out of the system, and work may be done on or by the system. We assume all processes are carried out very slowly, so that the system remains in thermal equilibrium at all stages.
15. 15. Isothermal & Adiabatic Processes Previously, we discussed the relationship between pressure and volume and found that: P 1/V
16. 16. Isothermal & Adiabatic Processes We also stated that this was true, the temperature was constant. A graph of P vs V is shown below.
17. 17. Isothermal & Adiabatic Processes The volume has increased from Vi to Vf while the pressure has decreased. The solid line is an isotherm, a curve giving the relationship between V and P at a constant temperature. This is known as isothermic expansion.
18. 18. Isothermal & Adiabatic Processes The process of compression or expansion of a gas; so that no heat enters or leaves the system, is said to be adiabatic. This comes from the Greek which means ‘impassable’.
19. 19. Isothermal & Adiabatic Processes Adiabatic changes of volume can be achieved by; performing the process so rapidly that, heat has little time to enter or leave the system, like a bicycle pump. Or by thermally insulating a system; from its surroundings, with Styrofoam.
20. 20. Isothermal & Adiabatic Processes A common example of a near adiabatic system; is the compression and expansion of gases, in the cylinders of a car engine. Compression and expansion occur too rapidly; for heat to leave the system.
21. 21. Isothermal & Adiabatic Processes When work is done on a gas by; adiabatically compressing it, the gas gains internal energy, and becomes warmer. When the gas adiabatically expands; it does work on the surroundings, and gives up its internal energy, and becomes cooler.
22. 22. Isothermal & Adiabatic Processes Adiabatic processes occur in the atmosphere in large masses of air. Due to their large size, mixing of different pressures and temperatures only occur at the edges of these large masses and do little to change the composition of these air masses.
23. 23. Isothermal & Adiabatic Processes As it flows up the side of a mountain; its pressure reduces, allowing it to expand and cool. The reduced pressure results in a reduced temperature.
24. 24. Isothermal & Adiabatic Processes It has been shown that dry air will drop by 10oC for every kilometre it rises. Air can flow over high mountains or rise in thunderstorms or cyclones many kilometres.
25. 25. Isothermal & Adiabatic Processes If a mass was 25oC at sea level and was lifted 6 kilometres, its temperature would become -35oC. An air mass that was -20OC at 6 km, would be 40oC at sea level.
26. 26. Isothermal & Adiabatic Processes An example of this is when cold air is blown over the Mt Lofty Ranges. Warm moist air is cooled as it rises over the ranges starts to rain. On the other side, the air begins to warm as it flows down the other side causing a warm wind.
27. 27. Isothermal & Adiabatic Processes As the Mt Lofty ranges are not very high; the change in temperature is not as great, compared to the Rocky Mountains in the USA..
28. 28. P – V DiagramsThermodynamic processes can berepresented by pressure - volume graphs.
29. 29. P – V Diagrams In the, an ideal gas is expanding isothermally, absorbing heat Q, and doing work W. The system has not been restored; to its original state, at the end of the process.
30. 30. P – V Diagrams
31. 31. P – V Diagrams The previous diagram is from a reversible heat engine. Process 1-2 takes place at a constant volume; Isochoric process 2-3 is adiabatic, process 3-1 is at a constant pressure; Isobaric.
32. 32. P – V Diagrams In the next case; the volume of an ideal gas is decreased, by adding weight to the piston. The process is adiabatic ( Q = 0).
33. 33. P – V Diagrams
34. 34. P – V DiagramsThe process is as shown below on a graph.
35. 35. P – V Diagrams In the next case, the temperature of an ideal gas is raised from T; to T + T, by a constant pressure process. Heat is added; and work is done, in lifting the loaded piston.
36. 36. P – V Diagrams
37. 37. P – V DiagramsThe process is shown below on a P-V diagram
38. 38. P – V Diagrams The work; P V, is the shaded area under the line, connecting the initial and final states.
39. 39. Work Done By a Gas To calculate the work done in a process, some Year 10 knowledge is important. Imagine the pressure is kept constant during a process.
40. 40. Work Done By a Gas
41. 41. Work Done By a Gas If the gas expands slowly against the piston; the work done to raise the piston is the force F multiplied by the distance d. But the force is just the pressure P of the gas; multiplied the area A of the piston, F = PA.
42. 42. Work Done By a Gas W = Fd = PAd W = P V = p(V2 – V1) The sign of the work done depends on whether the gas expands or is compressed.
43. 43. Work Done By a Gas If the gas expands, V is +ive and work is +ive. The equation also is valid if the gas is compressed.
44. 44. 1st Law of Thermodynamics A long, long time ago; heat was thought to be an invisible fluid, called a caloric, which flowed like water, from hot objects, to cold objects.
45. 45. 1st Law of Thermodynamics Caloric was conserved in its interactions which led to the discovery of the conservation of energy. Within any system, the less heat energy it has, the more ordered is the motion of its molecules.
46. 46. 1st Law of Thermodynamics This can be seen in solids; where the molecules all vibrate, about a mean position. As heat is added; the more disorderly the motion until in a gas, we say that all molecules, move in random motion.
47. 47. 1st Law of Thermodynamics In a sense then, heat is the disordered energy of molecules. There can be no heat in a single molecule. Heat is a statistical concept; applies only to a large number of molecules.
48. 48. 1st Law of Thermodynamics It is only when there is a great number of molecules does the concept of random or disorderly movement have meaning.
49. 49. 1st Law of Thermodynamics The discussion of heat, internal energy and temperature. Has given rise to the law of conservation of energy, and when applied to thermal systems, is often referred to as the, First law of thermodynamics.
50. 50. 1st Law of Thermodynamics In a general form it is: Whenever heat is added to a system, it transforms to an equal amount of some other form of energy.
51. 51. 1st Law of Thermodynamics The added energy does one or both of two things to the system:1. It increases the internal energy of the system if it remains in the system.2. It does external work if it leaves the system.
52. 52. 1st Law of Thermodynamics Heat added = increase in internal energy + external work done by the system. It can also be described mathematically: Q= U+ W Q = heat energy U = internal energy W = work For an isolated system; W = Q = 0 and U = 0
53. 53. 1st Law of Thermodynamics This can apply to a number of cases:1.Adiabatic Processes. In this case, no heat enters or leaves the system, ie Q = 0. Substituting this into the 1st Law;
54. 54. 1st Law of Thermodynamics 0 = U + W or, U = - W. This means that if work is done; there must be a decrease, in the internal energy of the system.
55. 55. 1st Law of Thermodynamics2. Constant Volume Processes. (Isovolumetric or Isochoric Process) If the volume of a system is held constant; the system can do no work, ie W =0. Substituting this into the 1st Law;
56. 56. 1st Law of Thermodynamics Q = U. If heat is added to the system; Q is +ive, the internal energy of the system increases. The converse is also true.
57. 57. 1st Law of Thermodynamics3.Cyclical Processes. There are processes in which; after certain interchanges of heat and work, the system is returned to its initial state. No property of the system can change, including the internal energy, ie U =0.
58. 58. 1st Law of Thermodynamics Substituting this into the 1st Law; Q = W. The net work done must exactly; equal the net amount of heat transferred.
59. 59. 1st Law of Thermodynamics4.Free Expansion. This is an adiabatic process; no work is done on or by the system, ie Q = W = 0.
60. 60. 1st Law of Thermodynamics Substituting this into the 1st Law; U =0. An example of this is given below.
61. 61. 1st Law of Thermodynamics A gas confined in an insulated container; is released into another container,that originally was a vacuum and then waiting until an equilibrium is established. as shown below.
62. 62. 1st Law of Thermodynamics No heat is transferred because; of the insulation No work is done because; the expanding gas rushes into an evacuated space, its motion unopposed by any counteracting pressure.
63. 63. 1st Law of Thermodynamics A summary is given: Process Restriction Consequence Adiabatic Q=0 U=- W Const V W=0 U= Q Closed Cycle U=0 Q= WFree Expansion Q= W=0 U=0
64. 64. Thermodynamic Cycles A thermodynamic cycle is; where heat may be transferred into (or out of), a system, or work may be done on or by the system. It is assumed that all transfers are done; very slowly so, the system remains essentially, in thermodynamic equilibrium at all stages.
65. 65. Thermodynamic Cycles An engine is a device that changes heat into mechanical work. Examples include: The steam engine external combustion Petrol & diesel internal combustion engines. It is impossible to convert all the heat energy into mechanical work.
66. 66. Thermodynamic Cycles Consider the internal combustion engine. Once the fuel is injected into the cylinder; the piston moves up, compresses the gas, Q = 0.
67. 67. Thermodynamic Cycles Spark plug fires, temperature increases. Adiabatic expansion pushes the piston down, burnt gases are pushed out.
68. 68. Thermodynamic Cycles A heat engine is a device that changes internal energy into mechanical work. Examples include; Steam engine, Internal combustion engine, Jet engine.
69. 69. Thermodynamic Cycles The mechanical work can only be obtained when; heat flows from a high to low temperature, only some of the heat is transferred into work.
70. 70. Thermodynamic Cycles Every heat engine will:1. absorb internal energy from a reservoir of higher temperature.2. convert some of this energy into mechanical work expel the remaining energy to some lower temperature reservoir, often called a sink.
71. 71. Thermodynamic Cycles
72. 72. Thermodynamic Cycles This section is for interest, not assessment Sadi Carnot; a French engineer, In 1924, Analysed the compression and expansion of in a heat engine Made a fascinating discovery when he examined the ideal engine, now called a Carnot engine.
73. 73. Thermodynamic Cycles The upper fraction of heat that can be converted to useful work; even under ideal conditions, depends on the temperature difference between, • the hot reservoir and the cold sink. The cycle starts at a in the diagram below.
74. 74. Thermodynamic Cycles
75. 75. Thermodynamic Cycles a b The gas expands isothermally by adding heat QH at temp TH. b c Gas then expands adiabatically no heat is exchanged but temp drops to TL
76. 76. Thermodynamic Cycles c d Compressed at const temp TL and heat QL flows out. d a Gas then compressed adiabatically to its original state.
77. 77. Thermodynamic CyclesHis equation gives the ideal or Carnot efficiency of aheat engine. TH TL TH TH TL TH TH TL 1 TH
78. 78. Thermodynamic Cycles The efficiency of the cycle only depends on; the absolute temperature of the high and, low temperature reservoirs. The greater the difference between them; the greater the efficiency.
79. 79. Entropy Back to assessment Heat flows naturally from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object. Clausius introduced the word entropy; from the Greek words meaning, transformation content.
80. 80. 2nd Law of Thermodynamics Entropy is a function of; the state of the system. Entropy can be interpreted as; a measure of of the order or disorder of a system.
81. 81. 2nd Law of Thermodynamics A coin, when put flat on a table; cannot spontaneously rise into the air, suddenly get too hot to touch, flatten out to something twice its diameter. These phenomena can easily be explained.
82. 82. 2nd Law of Thermodynamics Each of these situations requires energy to be added to the system and so violate the conservation of energy.
83. 83. 2nd Law of Thermodynamics We also know that coffee in a cup cannot; spontaneously cool down and start to swirl around, one end of a spoon gets hot while the other end cools down. molecules of air in the room do not move to one corner of the room and stay there.
84. 84. 2nd Law of Thermodynamics These events however do obey the conservation of energy and the first law of thermodynamics.
85. 85. 2nd Law of Thermodynamics The coffee could get its energy from the cooling process, the hot end of the spoon could get its energy from the cool end molecules of air do not need to change their kinetic energy, only their position.
86. 86. 2nd Law of Thermodynamics These events, however, do not happen although the reverse does happen. There are many other cases where an event will happen in one direction but not the other.
87. 87. 2nd Law of Thermodynamics The direction in which natural events happen is Determined by the Second Law of Thermodynamics. It can be described on a macroscopic and microscopic base:
88. 88. 2nd Law of Thermodynamics No device is possible whose sole effect is to transform a given amount of heat completely into work. This is the Kelvin-Planck formulation of the Second Law.
89. 89. 2nd Law of Thermodynamics The second law suggests that everything; is tending to disorder. Heat is a lower form of energy; so when heat is given off, it suggests that the system is tending to disorder. When your parents ask you to clean your room, you might like to suggest that you are only obeying the second law of thermodynamics.
90. 90. 2nd Law of Thermodynamics According to Clausius, the change in entropy S of a system; when an amount of heat Q is added to it, by a reversible process, at constant temperature, is given by: S = Q/T (not required for assessment) The units of entropy are J K-1
91. 91. 2nd Law of Thermodynamics In the example, although one part of the system decreased in entropy; the total entropy for the system increased. The second law stated in terms of entropy becomes: The entropy of an isolated system never decreases. It can only stay the same or increase.
92. 92. 2nd Law of Thermodynamics Entropy can only remain the same; for an idealised (reversible) process. For any real process; the change in entropy is greater than zero. The general statement of the second law of thermodynamics becomes:
93. 93. 2nd Law of Thermodynamics The total entropy of any system plus that of its environment increases as the result of any natural process.
94. 94. 2nd Law of Thermodynamics Although the entropy of one part of the universe may decease in any process; the entropy is some other part of the universe increases by a greater amount, the total entropy always increases.
95. 95. 2nd Law of Thermodynamics In the process of heat conduction from a hot body to a cold one; entropy increases and order goes to disorder. Useful work can be obtained while there is a temperature difference but; when the two heat reservoirs reach the Vale temperature, no work can be obtained from them.
96. 96. 2nd Law of Thermodynamics No energy is lost, it instead becomes less useful; the energy becomes degraded. The natural outcome of this is that as time goes on; the universe will reach a state of maximum disorder.
97. 97. 2nd Law of Thermodynamics The whole universe will be at one temperature; no work can be done. All the energy will have become degraded; to thermal energy. All change will cease. This is known as heat death.
98. 98. Refrigerators & Heat Pumps This section is not required for assessment Heat flows from the inside of warm houses in winter to the cold outside. The reverse can happen, but only by imposing external effort as do heat pumps. Air conditioners or refrigerators use these.
99. 99. Refrigerators & Heat Pumps
100. 100. Refrigerators & Heat Pumps The second form of the 2nd law of thermodynamics states: It is not possible for heat to flow from one body to another body at a higher temperature, with no other change taking place.
101. 101. Refrigerators & Heat Pumps A device that causes heat to move from a cold place to a hot place is called a refrigerator.
102. 102. Refrigerators & Heat Pumps
103. 103. Refrigerators & Heat Pumps In the diagram on the left, heat Qc is extracted from a low temperature reservoir the food storage area Some work W is done on the system by an external agent.
104. 104. Refrigerators & Heat Pumps The heat and work are combined and discharged as heat QH; to a high temperature reservoir, • the kitchen. The work shows up on the quarterly electricity bill; done by the motor that drives the unit.
105. 105. Refrigerators & Heat Pumps The diagram on the right shows a perfect refrigerator where no work is required. This fridge is yet to be built.
106. 106. Refrigerators & Heat Pumps In an air-conditioner, the low temperature reservoir is the room to be cooled the high temperature reservoir is the outside air • where the condenser coils are located. Again, the motor does the work.
107. 107. Refrigerators & Heat Pumps Both the fridge and the air-conditioner are rated by the amount of work they have to do. The ratings are by the coefficient of performance K.
108. 108. Refrigerators & Heat Pumps This is defined from: Qc QcK W QH Qc Design engineers want the performance of a fridge to be high as possible.
109. 109. Refrigerators & Heat Pumps A value of 5 is typical for a household fridge, while a room air-conditioner range 2-3. If there was a perfect fridge, the value of K = .
110. 110. The 2nd Law & Technology Back to assessment The second law indicates limits for technology. Heat engines and refrigerators cannot be perfect.. It is not possible for heat to flow from one body to another; at a higher temperature, with no other changes.
111. 111. The 2nd Law & Technology As the world is full of low-grade thermal energy from concept of entropy, Why can’t we concentrate and harvest that energy? Why not lower the temperature of the oceans by 1 oC and use that enormous amount of energy?
112. 112. The 2nd Law & Technology It can be done but it requires; work be put into the system which requires, energy to drive a fridge like machine. This energy input would make it unfeasible from an energy perspective.
113. 113. The 2nd Law & Technology Every living creature from bacteria to higher life forms such as you Extract energy from their surroundings to increase their own organisation.
114. 114. The 2nd Law & Technology This tends to indicate that all life; Including you, plus their waste products have a net increase in entropy. The 1st law is a universal law for which no exceptions have been observed.
115. 115. The 2nd Law & Technology The 2nd law is a probability statement. Given enough time, even the most improbable states could exist. The 2nd law tells us the most probable event, not the only possible event.
116. 116. The 2nd Law & Technology The laws of thermodynamics are often put this way: You can’t win, because you can’t get more energy out of a system than you put in,
117. 117. The 2nd Law & Technology You can’t break even, because you can’t even get as much energy as you put in,
118. 118. The 2nd Law & Technology You can’t get out of the game, entropy in the universe is always increasing.
119. 119. TRY EXAMPLE 1 1.00 kg of water is to be converted to steam at standard atmospheric pressure. The volume changes from an initial value of 1.00 x 10-3 m3 as a liquid to 1.671 m3 as steam.a) How much work is done by the system during this process?b) How much heat must be added to the system during the process? (L = 2260 kJ kg-1)c) What is the change in the internal energy of the system during the boiling process?
120. 120. Solution – Part (a) W=P V W = 1.01 x 105 (1.671 - 1 x 10-3) W = 1.69 x 105 J
121. 121. Solution – Part (b) Q = mL Q = 1 x 2260 Q = 2260 kJ Q = 2.26 x 106 J
122. 122. Solution – Part (c) Q= U+ W U= Q- W U = 2.26 x 106 - 1.69 x 105 U = 2.090 x 106 J As the result is positive; the internal energy of the system, has increased during the boiling process.
123. 123. TRY EXAMPLE 2 A steam turbine uses steam at 127 oC which then is cooled to 27 oC. What is the ideal efficiency of the turbine?
124. 124. Solution Convert all temperatures to SI units 127 oC = 273 + 127 = 400 K 27 oC = 273 + 27 = 300 K 400 - 300Efficiency 400 Efficiency = ¼ This means only 25% of the energy is converted into work while 75% is expelled as waste.
125. 125. TRY EXAMPLE 3 A household fridge, whose coefficient of performance is 4.7, extracts heat from a cooling chamber at the rate of 250 J per cycle.a) How much work per cycle is required to operate the fridge?b) How much heat per cycle is discharged to the room, which forms the high temperature reservoir of the fridge?
126. 126. Solution – Part (a) From W = Qc /K W = 250/4.7 W = 53 J As work is done on the system, W = -53 J
127. 127. Solution – Part (b) The net work done per cycle must equal the heat transferred per cycle. W = QH - Q c QH = W + Q c QH = 53 + 250 = 303 J
128. 128. TRY EXAMPLE 4 An ice cube of mass 60 g is taken from a storage compartment at 0 oC and placed in a paper cup. After a few minutes, exactly half of the mass of the ice cube has melted, becoming water at 0 oC. Find the change in entropy.
129. 129. Solution Q = mL Q = 0.06 x 3.34 x 105 Q = 1.002 x 104 J S = Q/T S = 1.002 x 104 /273 S = 36.7 J K-1
130. 130. TRY EXAMPLE 5 A sample of 50.0 kg of water at 20.0 oC is mixed with 50.0 kg of water at 24 oC. Estimate the change in entropy.
131. 131. Solution The final temperature of the mixture will be 22 oC as we started with equal amounts of water. A quantity of heat, Q = mc T Q = 50.0 x 4.18 x 103 x 2 Q = 4.18 x 105 J
132. 132. Solution flows out of the hot water and flows into the cold water. The total change in entropy will be the sum of changes in entropy of the hot and cold water. S = SH + SC To calculate the entropy, we must use the average temperature of 23 oC for the hot water and 21oC for the cold water.
133. 133. Solution SH = 4.18 x 105/296 SH = -1.41 x 103 J K-1 (-ive as heat flows out of the hot water) SC = 4.18 x 105/294 SC = 1.42 x 103 J K-1 S = SH + SC S = -1.41 x 103 + 1.42 x 103 S = 10 J K-1