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    Topic 10   Thermal Physics Topic 10 Thermal Physics Presentation Transcript

    • Thermal Physics AHL Topic 10
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • Thermodynamics
      • The operation of changing the system from its initial state to a final state is called the,
      • T hermodynamic process .
    • 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.
    • Isothermal & Adiabatic Processes
      • Previously, we discussed the relationship between
        • pressure and volume and found that:
      •   P  1/V
    • Isothermal & Adiabatic Processes
      • We also stated that this was true,
        • the temperature was constant.
      • A graph of P vs V is shown below.
    • 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.
    • 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’.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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. .
    • P – V Diagrams
      • Thermodynamic processes can be represented by pressure - volume graphs.
    • 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.
    • P – V Diagrams
    • 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.
    • 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).
    • P – V Diagrams
    • P – V Diagrams
      • The process is as shown below on a graph .
    • 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.
    • P – V Diagrams
    • P – V Diagrams
      • The process is shown below on a P-V diagram
    • P – V Diagrams
      • The work ;
        • P  V ,
        • is the shaded area under the line ,
        • connecting the initial and final states.
    • 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.
    • Work Done By a Gas
    • 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 .
    • 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.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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 .
    • 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.
    • 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 .
    • 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.
    • 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.
    • 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
    • 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;
    • 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.
    • 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;
    • 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.
    • 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.
    • 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.
    • 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.
    • 1 st Law of Thermodynamics
      • Substituting this into the 1 st Law;
      •  U =0.
      • An example of this is given below.
    • 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.
    • 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.
    • 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
    • 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.
    • 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.
    • Thermodynamic Cycles
      • Consider the internal combustion engine.
      • Once the fuel is injected into the cylinder ;
        • the piston moves up ,
        • compresses the gas ,
        •  Q = 0 .
    • Thermodynamic Cycles
      • S park plug fires ,
        • temperature increases.
      • Adiabatic expansion pushes the piston down ,
        • burnt gases are pushed out.
    • 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.
    • 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.
    • 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.
    • Thermodynamic Cycles
    • 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.
    • 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.
    • Thermodynamic Cycles
    • 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
    • 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.
    • Thermodynamic Cycles
      • His equation gives the ideal or Carnot efficiency of a heat engine.
    • 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.
    • 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.
    • 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.
    • 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.
    • 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
    • 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 :
    • 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
    • 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.
    • 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.
    • Entropy
      • The total entropy of any system plus that of its environment increases as the result of any natural process.
    • 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.
    • 2 nd Law of Thermodynamics
      • Each of these situations requires energy to be added to the system and so violate the conservation of energy.
    • 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.
    • 2 nd Law of Thermodynamics
      • These events however do obey the conservation of energy
        • and the first law of thermodynamics.
    • 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.
    • 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.
    • 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:
    • 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.
    • 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.
    • 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 .
    • 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.
    • Refrigerators & Heat Pumps
    • 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.
    • Refrigerators & Heat Pumps
      • A device that causes heat to move from a cold place to a hot place is called a refrigerator.
    • Refrigerators & Heat Pumps
    • 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.
    • 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.
    • Refrigerators & Heat Pumps
      • The diagram on the right shows a perfect refrigerator where no work is required.
      • This fridge is yet to be built.
    • 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.
    • 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 .
    • Refrigerators & Heat Pumps
      • This is defined from:
      • Design engineers want the performance of a fridge to be high as possible.
    • 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 =  .
    • 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.
    • 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?
    • 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.
    • 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.
    • 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.
    • 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.
    • 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,
    • The 2 nd Law & Technology
      • Y ou can’t break even ,
        • because you can’t even get as much energy as you put in,
    • The 2 nd Law & Technology
      • Y ou can’t get out of the game ,
        • entropy in the universe is always increasing.
    • 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?
    • 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
    • Solution – Part (b)
      •  Q = mL
      •  Q = 1 x 2260
      •  Q = 2260 kJ
      •  Q = 2.26 x 10 6 J
    • 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.
    • 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?
    • 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.
    • 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?
    • 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
    • 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
    • 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.
    • 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
    • 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.
    • 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
    • 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.
    • 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