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Topic 10 Thermal Physics

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  • 1. Thermal Physics AHL Topic 10
  • 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. Thermodynamics
    • Work is defined as :
    • T he 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. Thermodynamics
    • From this we can see that thermodynamics describes the relation ship 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. Thermodynamics
    • The foundation of this area of study is
    • T he 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. 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. 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. 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. 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.
    • L ikewise, the hot gas contains slow moving molecules.
  • 10. Thermodynamics
    • From above, it should be possible to for the cold gas to transfer
      • energy to the hot gas
    • S o 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. 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. 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. Thermodynamics
    • The operation of changing the system from its initial state to a final state is called the,
    • T hermodynamic process .
  • 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. Isothermal & Adiabatic Processes
    • Previously, we discussed the relationship between
      • pressure and volume and found that:
    •   P  1/V
  • 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. Isothermal & Adiabatic Processes
    • The volume has increased from V i to V f 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. 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. 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 .
    • O r by thermally insulating a system ;
      • from its surroundings ,
      • with Styrofoam.
  • 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. 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. 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. 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. Isothermal & Adiabatic Processes
    • It has been shown that dry air will drop
      • by 10 o C
      • for every kilometre it rises.
    • Air can flow over high mountains
      • or rise in thunderstorms
      • or cyclones
      • many kilometres.
  • 25. Isothermal & Adiabatic Processes
    • If a mass was 25 o C at sea level and was lifted 6 kilometres,
      • its temperature would become -35 o C .
    • A n air mass that was -20 O C at 6 km ,
      • would be 40 o C at sea level.
  • 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. Isothermal & Adiabatic Processes
    • As the Mt Lofty ranges are not very high ;
      • the change in temperature is not as great ,
      • c ompare d to the Rocky Mountains in the USA. .
  • 28. P – V Diagrams
    • Thermodynamic processes can be represented by pressure - volume graphs.
  • 29. P – V Diagrams
    • In the, an ideal gas is expanding isothermally,
    • absorbing heat  Q ,
    • and doing work  W .
    • T he system has not been restored ;
      • to its original state ,
      • at the end of the process.
  • 30. P – V Diagrams
  • 31. P – V Diagrams
    • Th e 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. 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. P – V Diagrams
  • 34. P – V Diagrams
    • The process is as shown below on a graph .
  • 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. P – V Diagrams
  • 37. P – V Diagrams
    • The process is shown below on a P-V diagram
  • 38. P – V Diagrams
    • The work ;
      • P  V ,
      • is the shaded area under the line ,
      • connecting the initial and final states.
  • 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. Work Done By a Gas
  • 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. Work Done By a Gas
    • W = Fd = PAd
    • W = P  V = p(V 2 – V 1 )
    • The sign of the work done depends on whether the gas expands or is compressed.
  • 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. 1 st 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. 1 st 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. 1 st 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. 1 st 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. 1 st Law of Thermodynamics
    • I t is only when there is a great number of molecules
      • does the concept of random
      • or disorderly movement have meaning.
  • 49. 1 st Law of Thermodynamics
    • The discussion of heat,
      • internal energy and temperature .
    • H as given rise to the law of conservation of energy,
      • and when applied to thermal systems ,
      • is often referred to as the ,
    • F irst law of thermodynamics .
  • 50. 1 st 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. 1 st 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. 1 st 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. 1 st 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 1 st Law;
  • 54. 1 st 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. 1 st Law of Thermodynamics
    • 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 1 st Law;
  • 56. 1 st 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. 1 st Law of Thermodynamics
    • 3 . Cyclical Processes .
    • There are processes in which ;
      • after certain interchanges of heat and work ,
      • the system is returned to its initial state.
    • N o property of the system can change,
      • including the internal energy,
      • ie  U =0.
  • 58. 1 st Law of Thermodynamics
    • Substituting this into the 1 st Law;
    •  Q =  W .
    • The net work done must exactly ;
      • equal the net amount of heat transferred.
  • 59. 1 st Law of Thermodynamics
    • 4. Free Expansion .
    • This is an adiabatic process ;
      • no work is done on or by the system,
      • ie Q = W = 0.
  • 60. 1 st Law of Thermodynamics
    • Substituting this into the 1 st Law;
    •  U =0.
    • An example of this is given below.
  • 61. 1 st 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. 1 st Law of Thermodynamics
    • No heat is transferred because ;
      • of the insulation
    • N o work is done because ;
      • the expanding gas rushes into an evacuated space,
      • its motion unopposed by any counteracting pressure.
  • 63. 1 st Law of Thermodynamics
    • A summary is given:
     U = 0  Q =  W = 0 Free Expansion  Q =  W  U = 0 Closed Cycle  U =  Q  W = 0 Const V  U = -  W  Q = 0 Adiabatic Consequence Restriction Process
  • 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. Thermodynamic Cycles
    • An engine is a device that changes heat into mechanical work.
    • Example s include :
    • T he steam engine
      • external combustion
    • P etrol & diesel
      • internal combustion engines.
    • I t is impossible to convert all the heat energy into mechanical work.
  • 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. Thermodynamic Cycles
    • S park plug fires ,
      • temperature increases.
    • Adiabatic expansion pushes the piston down ,
      • burnt gases are pushed out.
  • 68. Thermodynamic Cycles
    • A heat engine is a device that changes internal energy into mechanical work.
    • Examples include ;
      • S team engine ,
      • I nternal combustion engine ,
      • J et engine.
  • 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. Thermodynamic Cycles
    • Every heat engine will:
    • absorb internal energy from a reservoir of higher temperature .
    • convert some of this energy into mechanical work expel the remaining energy to some lower temperature reservoir ,
      • often called a sink.
  • 71. Thermodynamic Cycles
  • 72. Thermodynamic Cycles
    • Sadi Carnot ;
      • a French engineer ,
      • In 1924 ,
    • A nalysed the compression and expansion of in a heat engine
    • M a de a fascinating discovery
      • when he examined the ideal engine ,
      • now called a Carnot engine.
  • 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. Thermodynamic Cycles
  • 75. Thermodynamic Cycles
    • a  b
      • The gas expands isothermally by
      • adding heat Q H at temp T H .
    • b  c
      • Gas then expands adiabatically
      • no heat is exchanged but temp drops to T L
  • 76. Thermodynamic Cycles
    • c  d
      • Compressed at const temp T L and heat Q L flows out.
    • d  a
      • Gas then compressed adiabatically to its original sta t e.
  • 77. Thermodynamic Cycles
    • His equation gives the ideal or Carnot efficiency of a heat engine.
  • 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. Entropy
    • 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. Entropy
    • 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. Entropy
    • 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.
  • 82. Entropy
    • 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 entropy
  • 83. Entropy
    • 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 :
  • 84. Entropy
    • 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
    • The units of entropy are J K -1
  • 85. Entropy
    • 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.
  • 86. Entropy
    • 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.
  • 87. Entropy
    • The total entropy of any system plus that of its environment increases as the result of any natural process.
  • 88. 2 nd 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.
  • 89. 2 nd Law of Thermodynamics
    • Each of these situations requires energy to be added to the system and so violate the conservation of energy.
  • 90. 2 nd Law of Thermodynamics
    • We also know that coffee in a cup cannot ;
      • s pontaneously 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.
  • 91. 2 nd Law of Thermodynamics
    • These events however do obey the conservation of energy
      • and the first law of thermodynamics.
  • 92. 2 nd 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.
  • 93. 2 nd 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.
  • 94. 2 nd Law of Thermodynamics
    • The direction in which natural events happen is
    • D etermined by the Second Law of Thermodynamics .
    • It can be described on a macroscopic and microscopic base:
  • 95. 2 nd 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. 2 nd 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. 2 nd 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. Refrigerators & Heat Pumps
    • 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. Refrigerators & Heat Pumps
  • 100. Refrigerators & Heat Pumps
    • The second form of the 2 nd 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. Refrigerators & Heat Pumps
    • A device that causes heat to move from a cold place to a hot place is called a refrigerator.
  • 102. Refrigerators & Heat Pumps
  • 103. Refrigerators & Heat Pumps
    • In the diagram on the left, heat Q c is extracted from a low temperature reservoir
      • the food storage area
    • S ome work W is done on the system
      • by an external agent.
  • 104. Refrigerators & Heat Pumps
    • The heat and work are combined and discharged as heat Q H ;
      • 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. 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. 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. Refrigerators & Heat Pumps
    • For interest
      • not in syllabus
    • 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. Refrigerators & Heat Pumps
    • This is defined from:
    • Design engineers want the performance of a fridge to be high as possible.
  • 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. The 2 nd Law & Technology
    • 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. The 2 nd Law & Technology
    • As the world is full of low - grade thermal energy from concept of entropy ,
    • W hy can’t we concentrate and harvest that energy?
    • Why not lower the temperature of the oceans by 1 o C
      • and use that enormous amount of energy?
  • 112. The 2 nd 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. The 2 nd Law & Technology
    • E very living creature from bacteria
      • to higher life forms such as
      • you
    • E xtract energy from their surroundings
      • to increase their own organisation.
  • 114. The 2 nd Law & Technology
    • This tends to indicate that all life ;
      • Including you ,
      • plus their waste products
      • have a net increase in entropy.
    • The 1 st law is a universal law for which no exceptions have been observed.
  • 115. The 2 nd Law & Technology
    • The 2 nd law is a probability statement.
    • Given enough time,
      • even the most improbable states could exist.
    • The 2 nd law tells us the most probable event,
      • not the only possible event.
  • 116. The 2 nd 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. The 2 nd Law & Technology
    • Y ou can’t break even ,
      • because you can’t even get as much energy as you put in,
  • 118. The 2 nd Law & Technology
    • Y ou can’t get out of the game ,
      • entropy in the universe is always increasing.
  • 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 m 3 as a liquid to 1.671 m 3 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. Solution – Part (a)
    • W = P  V
    • W = 1.01 x 10 5 (1.671 - 1 x 10 -3 )
    • W = 1.69 x 10 5 J
  • 121. Solution – Part (b)
    •  Q = mL
    •  Q = 1 x 2260
    •  Q = 2260 kJ
    •  Q = 2.26 x 10 6 J
  • 122. Solution – Part (c)
    •  Q =  U +  W
    •  U =  Q -  W
    •  U = 2.26 x 10 6 - 1.69 x 10 5
    •  U = 2.090 x 10 6 J
    • As the result is positive ;
      • the internal energy of the system ,
      • has increased during the boiling process.
  • 123. TRY EXAMPLE 2
    • A steam turbine uses steam at 127 o C which then is cooled to 27 o C. What is the ideal efficiency of the turbine?
  • 124. Solution
    • Convert all temperatures to SI units
    • 127 o C = 273 + 127 = 400 K
    • 27 o C = 273 + 27 = 300 K
    • Efficiency = ¼
    • This means only 25% of the energy is converted into work while 75% is expelled as waste.
  • 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. Solution – Part (a)
    • From  W  =  Q c  / K
    •  W  = 250/4.7
    •  W  = 53 J
    • As work is done on the system, W = -53 J
  • 127. Solution – Part (b)
    • The net work done per cycle must equal the heat transferred per cycle.
    •  W  =  Q H  -  Q c 
    •  Q H  =  W  +  Q c 
    •    Q H  = 53 + 250 = 303 J
  • 128. TRY EXAMPLE 4
    • An ice cube of mass 60 g is taken from a storage compartment at 0 o C 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 o C. Find the change in entropy.
  • 129. Solution
    • Q = mL
    • Q = 0.06 x 3.34 x 10 5
    • Q = 1.002 x 10 4 J
    •  S = Q/T
    •  S = 1.002 x 10 4 /273
    •  S = 36.7 J K -1
  • 130. TRY EXAMPLE 5
    • A sample of 50.0 kg of water at 20.0 o C is mixed with 50.0 kg of water at 24 o C. Estimate the change in entropy.
  • 131. Solution
    • The final temperature of the mixture will be 22 o C as we started with equal amounts of water. A quantity of heat,
    • Q = mc  T
    • Q = 50.0 x 4.18 x 10 3 x 2
    • Q = 4.18 x 10 5 J
  • 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 =  S H +  S C
    • To calculate the entropy, we must use the average temperature of 23 o C for the hot water and 21 o C for the cold water.
  • 133. Solution
    •  S H = 4.18 x 10 5 /296
    •  S H = -1.41 x 10 3 J K -1 (-ive as heat flows out of the hot water)
    •    S C = 4.18 x 10 5 /294
    •  S C = 1.42 x 10 3 J K -1
    •  S =  S H +  S C
    •  S = -1.41 x 10 3 + 1.42 x 10 3
    •  S = 10 J K -1