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Thermodynamics

… the study of how thermal energy can do work. Thermal energy … can produce useful work. Thermodynamics. work can produce … Thermal energy. Materials have internal energy U. Internal Energy is KE of random motions of atoms + PE due to forces between atoms. Internal energy.

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Thermodynamics

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  1. … the study of how thermal energy can do work Thermal energy … can produce useful work Thermodynamics work can produce … Thermal energy

  2. Materials have internal energy U Internal Energy is KE of random motions of atoms + PE due to forces between atoms Internal energy Can be modeled as vibrating springs joining atoms to each other in solids or within molecules Energy is also stored as vibrational, rotational and translational motions

  3. Materials have internal energy U, (thermal energy + potential energy in bonds) U is the sum of all KE and PE of atoms/molecules in the material Internal energy U is the change of internal energy If U > 0 then internal energy has increased If U < 0 then internal energy has decreased

  4. For each “degree of freedom” (different direction in 3D space) an atom (or molecule) can store energy:  kT k is Boltzmann’s constant, 1.38 (10-23) J/K, T is absolute temperature Internal Energy of Ideal gases  U =  NkT for a monatomic gas with N molecules, since there are 3 dimensions (directions) In an Ideal Gas U  T (in K) In polyatomic gases the molecules can store energy in rotations, and vibrations, as well as translations and this adds more degrees of freedom increasing the internal energy at a given temperature so that complex gases are slower to warm up

  5. Heat is not Internal Energy Heat is the flow of Thermal Energy from one object to another and will increase the Internal Energy of the receiver and decrease the Internal Energy of the donor Heat and Internal Energy HEAT more random motion (<KE>) higher temperature HEAT stretched bonds  higher PE, without changing T (Like Work is not Mechanical Energy Work is the transfer of Mechanical Energy from one object to another)

  6. The reasons for studying thermodynamics were mainly practical – engines and the Industrial Revolution Efficient engines were needed which meant analyzing how fuel (thermal energy) may be harnessed to do useful work. Heat Engines and Refrigerators The earliest known engine is Hero’s – a Greek from Alexandria 2000 years ago. Engines use a working fluid, often a gas, to create motion and drive equipment Engines (and refrigerators) must repeat their cycles over and over to continue to do work

  7. (this is the 0th law because it was added after 1,2, and 3) Temperature exists and can be measured 0th Law of Thermodynamics When 2 objects are in thermal equilibrium separately with a 3rd object then they are in thermal equilibrium with each other T1 = T2 and T2 = T3 T1 = T3 Thermal equilibrium means there is no net thermal energy flow between the objects

  8. Energy is conserved – it is neither created not destroyed Energy may be transferred from one object to another, or changed in form (KE to PE for example) 1st Law of Thermodynamics U = Q – W For thermodynamic systems The energy change of a system is the heat in less the work done by the system

  9. EXAMPLE 1000 J of thermal energy flows into a system (Q = 1000 J).  At the same time, 400 J of work is done by the system (W = 400 J).What is the change in the system's internal energy U? U = Q –W = 1000 - 400 = 600 J

  10. EXAMPLE 800 J of work is done by a system (W = 800 J) as 500 J of thermal energy is removed from the system (Q = -500 J).What is the change in the system's internal energy U? U = Q –W = -500 – 800 = -1300 J NB: work done on the system is +, work done by the system is - & heat into the system is +, heat out of the system is -

  11. A system can change its state A state is a unique set of values for P, V, n, & T (so PV = nRT is also called a “State Equation”) Thermodynamic Processes When you know the state of a system you know U since U =  NkT =  nRT =  PV, for a monatomic gas A “process” is a means of going from 1 state to another There are 4 basic processes with n constant Isobaric, a change at constant pressure Isochoric or isovolumetric, a change at constant volume, W = 0 Isothermal, a change at constant temperature (U = 0, Q = W) Adiabatic, a change at no heat (Q = 0) “iso” means “same”

  12. Isobar P (P1,V1) T1 (P2,V2) T2 1 2 Thermodynamic Processes Isochore Adiabat 4 (P4,V4) T4 Q = 0 T3 = T4 3 (P3,V3) T3 Isotherm The trip from 12341 is call a “thermodynamic cycle” Each part of the cycle is a process V All state changes can be broken down into the 4 basic processes

  13. Isobar, expansion at constant pressure, work is done P 1 2 Thermodynamic Processes Isochoric pressure change, W = 0 Adiabatic expansion; no heat, Q = 0 4 3 Isothermal compression W = Q, U is constant The area enclosed by the cycle is the total work done, W The work done, W, in a cycle is + if you travel clockwise V

  14. Engines use a working fluid, often a gas, to create motion and drive equipment; the gas moves from 1 state (P, V, n, & T define a state) to another in a cycle Heat Engines and Refrigerators The Stirling Cycle: 2 isotherms 2 isochores Stirling designed this engine in the early 18th century – simple and effective The Stirling Engine

  15. Isobaric expansion of a piston in a cylinder The work done W = Fd = PAd = PV The work done is the area under the process W = PV 4 stroke engine

  16. Isochoric expansion of a piston in a cylinder The work done W = 0 since there is no change in volume Thus U = Q – W = Q

  17. Adiabatic expansion of an ideal gas The work done W = 0 here because chamber B is empty and P = 0 Thus U = Q – W = 0, that is adiabatic expansion against no resistance does not change the internal energy of a system

  18. EXAMPLE How much work is done by the system when the system is taken from: (a)  A to B  (900 J)(b)  B to C  (0 J)(c)  C to A  (-1500 J) Each rectangle on the graph represents 100 Pa-m³ = 100 J (a) From A B the area is 900 J, isobaric expansion (b) From B  C, 0, isovolumetric change of pressure (c) From C A the area is -1500 J

  19. EXAMPLE 10 grams of steam at 100 C at constant pressure rises to 110 C: P   = 4 x 105 Pa             T = 10 C   V = 30.0 x 10-6 m3        c = 2.01 J/g What is the change in internal energy? U = Q – W U = mcT – PV U = 189 J So heating the steam produces a higher internal energy and expansion

  20. EXAMPLE Aluminum cube of side L is heated in a chamber at atmospheric  pressure. What is the change in the cube's internal energy if L = 10 cm and T = 5 °C? Q = mcT m = V0V0 = L3W = PVV = V0T U = Q – W U = mcT – PV cAl = 0.90 J/g°C Al = 72(10-6) °C-1 U = V0cT – PV0T Patm = 101.5 kPa U = V0T (c – P) Al = 2.7 g/cm³ U = L³T (c – P) U = 0.10³(5)((2700)(900) – 101.5(10³)(72(10-6)) U = 12,150 J NB: P is neglible

  21. EXAMPLE Find the work done for a cycle if P1 = 1000 kPa, V1 = 0.01 m³, V2 = 0.025 m³, V3 = V4 = 0.04 m³, T1 = 400 K, T2 = 600K, n = 2 mol W = Area enclosed = P1V12 +  (P2+P3)V23 –  (P1+P4)V41 P 1, (P1,V1) T1 2, (P2,V2) T2 Isobar Isotherm 3, (P3,V3) T3 Isotherm Isochore 4, (P4,V4) T4 V 1. P2 = P1 = 1000 kPa W = Area enclosed = P1V12 +  (P2+P3)V23 +  (P1+P4)V41 = (15 + 12.188 – 18.75)(10³) = 8.44 kJ 2. T4 = T1 = 400 K 3. T3 = T2 = 600 K 4. P3 = P2V2/V3 = 625 kPa 5. P4 = P1V1/V4 = 250 kPa

  22. EXAMPLE Find the internal energy for each state if P1 = 1000 kPa, V1 = 0.01 m³, V2 = 0.025 m³, V3 = V4 = 0.04 m³, T1 = 400 K, T2 = 600K, n = 2 mol 1. P2 = P1 = 1000 kPa 2. T4 = T1 = 400 K P 3. T3 = T2 = 600 K 1, (P1,V1) T1 2, (P2,V2) T2 Isobar 4. P3 = P2V2/V3 = 625 kPa 5. P4 = P1V1/V4 = 250 kPa Isotherm 3, (P3,V3) T3 Isotherm Isochore 4, (P4,V4) T4 V 6. U1 =  nRT1 = 9972 J 7. U4 = U1 = 9972 J 8. U2 =  nRT2 = 14958 J 9. U3 = U2 = 14958 J

  23. EXAMPLE Find the thermal energy change Q for each state if P1 = 1000 kPa, V1 = 0.01 m³, V2 = 0.025 m³, V3 = V4 = 0.04 m³, T1 = 400 K, T2 = 600K, n = 2 mol 1. P2 = P1 = 1000 kPa Q12 2. T4 = T1 = 400 K P 3. T3 = T2 = 600 K 1, (P1,V1) T1 2, (P2,V2) T2 Isobar 4. P3 = P2V2/V3 = 625 kPa 5. P4 = P1V1/V4 = 250 kPa Q34 Q41 Isotherm Q34 3, (P3,V3) T3 Isotherm Isochore 4, (P4,V4) T4 V 10. Q12 = U12 + W12 = 34986 J 6. U1 =  nRT1 = 9972 J 11. Q23 = W23 (U23 = 0) W23 =  (P2+P3)V23 = 12.188 kJ 7. U4 = U1 = 9972 J 12. Q34 = U34 = -4986 J 8. U2 =  nRT2 = 14958 J 13. Q41 = W41 (U41 = 0) W41 =  (P4+P1)V41 = - 18.75 kJ 9. U3 = U2 = 14958 J

  24. The Wankel Rotary engine is a powerful and simple alternative to the piston engine used by Nissan and invented by the German, Wankel in the 1920s Heat Engines and Refrigerators The Wankel Cycle: 2 adiabats 2 isochores The Wankel Engine

  25. Work done by system Heat flow into system Increase in internalenergy of system Recap • 1st Law of Thermodynamics • energy conservation Q= DU + W P • U depends only on T (U = nRT = PV) • point on PV plot completely specifies state of system (PV = nRT) • work done is area under curve • for complete cycle DU = 0  Q = W V

  26. HEAT ENGINE REFRIGERATOR system TH TH QH QH W W QC QC TC TC What do the cycles apply to? • system taken in closed cycle  Usystem = 0 • therefore, net heat absorbed = work done QH - QC = W (engine) QC - QH = -W (refrigerator) energy into blue blob = energy leaving bluegreen blob

  27. HEAT ENGINE TH QH W QC TC Heat Engine: Efficiency Goal: Get work from thermal energy in the hot reservoir 1st Law: QH - QC = W, (U = 0 for cycle) Define efficiency as work done per thermal energy used e  What is the best we can do? Solved by Sadi Carnot in 1824 with the Carnot Cycle W QH

  28. Designed by Sadi Carnot in 1824, maximally efficient QH enters from 1-2 at constant THand QC leaves from 3-4 at constant TL P 1 QH Work done Wnet = WH – WC = QH – QC = W Carnot Cycle Isotherm QH = WH Efficiency is W / QH = ( QH – QC ) / QH 2 Adiabat Q = 0 Since U  T then Q – W is also proportional to T but from (1-2) and (3-4) Q = W so Q  T Adiabat Q = 0 4 Efficiency is W / TH = ( TH – TC ) / TH Isotherm QC = WC 3 TC QC / TH V emax = 1 –

  29. HEAT ENGINE TH QH W QC TC Heat Engine: Entropy We can define a useful new quantity Entropy, S Entropy measures the disorderof a system Only changes in S matter to us S = Change in entropy depends on thermal energy flow (heat) at temperature T Q T

  30. HEAT ENGINE TH QH W QC TC Heat Engine: Entropy Entropy, S measures the disorderof a system changes in S matter S = If = as in the Carnot Cycle … then there is no net change in entropy for the cycle and efficiency is a maximum, … because we do as much work as is possible QC QH Q TH TC T

  31. Heat flows from hot to cold naturally “One cannot convert a quantity of thermal energy entirely to useful work” (Kelvin) 2nd Law of Thermodynamics “One cannot transfer thermal energy from a cold reservoir to hot reservoir without doing work” (Clausius) In closed systems, S > 0 for all real processes The entropy, disorder, always increases in closed systems Only in the ideal case of maximum efficiency would S = 0

  32. EXAMPLE Does the apparent order of life on Earth imply the 2nd law is wrong or that some supernatural being is directing things? No. The second law applies to closed systems, those with no energy coming in or going out. As long as the Sun shines more energy falls on the Earth, and more work can be done by the plants to build new mass, release oxygen, grow, metabolize.

  33. EXAMPLE What is happening to the Universe? The universe is slowly coming to an end. When the entire universe is at the same temperature, then no work will be possible, and no life and no change … billions and billions and billions of years from now … Heat Death

  34. EXAMPLE Consider a hypothetical device that takes 1000 J of heat from a hot reservoir at 300K, ejects 200 J of heat to a cold reservoir at 100K, and produces 800 J of work. Is this possible? The maximum efficiency is emax = 1 – TL/TH = 67%, but the proposed efficiency is eprop = W/QH = 80%. This violates the 2nd law – do not buy shares in the company designing this engine!

  35. EXAMPLE Consider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K. Is this possible? The entropy of the cold reservoir decreases by SC = 1000 J / 100 K = 10 J/K The entropy of the heat reservoir increases by SH = 1200 J / 300 K = 4 J/K There would be a net decrease in entropy which would violate the 2nd Law, so this refrigerator is not possible What is the minimum work needed?  2000 J, so that SH becomes at least 10 J/K

  36. Air Conditioners Uses a “working fluid” (freon or other nicer gas) to carry heat from cool room to hot surroundings – same as a refrigerator, moving Q from inside fridge to your kitchen, which you must then air condition!

  37. Air Conditioners

  38. Air Conditioners • Evaporator located in room air transfers heat from room air to fluid • Compressor located in outside air does work on fluid and heats it further • Condenser located in outside air transfers heat from fluid to outside air • Then the fluid reenters room for next cycle

  39. Evaporator Heat exchanger made from a long metal pipe Fluid nears evaporator as a high pressure liquid near room temperature A constriction reduces the fluid pressure Fluid enters evaporator as a low pressure liquid near room temperature Working fluid evaporates in the evaporator – requires energy LV to separate molecules, so fluid cools & Q flows from room to fluid • Fluid leaves evaporator as a low pressure gas near room temperature, taking thermal energy with it, leaving the room cooler!

  40. Compressor Working fluid enters compressor as a low pressure gas near room temperature Gas is compressed (PV work) so gas T rises (1st Law, T U & U ↑ when PV work is done) Compressing gas forces Q out of it into surroundings (open air) Fluid leaves compressor as hot, high pressure gas

  41. Condenser Fluid enters condenser (heat exchanger made from long metal pipe) as a hot, high pressure gas Q flows from fluid to outside air Gas releases energy across heat exchanger to air and condenses forming bonds releases energy LV – thermal energy & fluid becomes hotter liquid so even more heat flows from fluid into outside air • Fluid leaves condenser as high pressure • liquid near room temperature to repeat the cycle

  42. Summary Evaporator – in room transfers heat from room air to working fluid Condenser – in outside air transfers heat from fluid to outside air, including thermal energy extracted from inside air and thermal energy added by compressor Compressor – outside does work on fluid, so fluid gets hotter Entropy of room has decreased but entropy of outside has increased by more than enough to compensate – order to disorder

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