3. Heat Engines Coal fired steam engines. Petrol engines Diesel engines Jet engines Power station turbines

8. DECChttp://www.decc.gov.uk/asset s/decc/statistics/publications/flow/1 93-energy-flow-chart-2009.pdfhttp://www.decc.gov.uk/asset s/decc/statistics/publications/flow/1 93-energy-flow-chart-2009.pdf

12. Combined Cycle

14. THE LAWS OF THERMODYNAMICS 1. You cannot win you can only break even. 2. You can only break even at absolute zero. 3. You can never achieve absolute zero.

15. S = k log W

17. Atoms don’t care. What happens most ways happens most often

18. p Boyle’s Law p  1/V 1/V

19. T V Charles’s Law V  T

20. T p Pressure Law p  T

21. Number of molecules, N p Common sense Law p  N

22. Isotherms (constant temperature) 1/V p T V Isobars (constant pressure) Isochors (constant volume) T p

23. p  1/V V  T p  T In summary… pV T = constant For ideal gases only A gas that obeys Boyles law

24. Ideal gas? Most gases approximate ideal behaviour Ideal gases assume:- No intermolecular forces Volume of molecules is negligible Not true - gases form liquids then solids as temperature decreases Not true - do have a size

25. p1V1p1V1 T1T1 p2V2p2V2 T2T2 = pV T = constant Only useful if dealing with same gas before (1) and after (2) an event

26. Ideal Gas Law pV = nRT p = pressure, Pa V = volume, m 3 n = number of moles R = Molar Gas constant (8.31 J K -1 mol -1 ) T = temperature, K Macroscopic model of gases

27. pV = NkT N = number of molecules k = Boltzmann’s constant (1.38 x 10 -23 J K -1 ) Which can also be written as …

28. First there was a box and one molecule… Molecule:- mass = m velocity = v x y z v Kinetic Theory

29. Molecule hits side of box…(elastic collision) v -v  p mol Molecule  p box = -  p mol = 2mv Box mv - mu = -mv - mv = -2mv 2mv -2mv Remember  p = F so a force is felt by the box t

30. Molecule collides with side of box, rebounds, hits other side and rebounds back again. Time between hitting same side, t v s = v = 2x x y z

31. Average force, exerted by 1 molecule on box F = pp t =  p v 2x = 2mv v 2x = mv 2 x Force exerted on box Time Average Force Actual force during collision

32. x y z v1v1 Consider more molecules v4v4 v2v2 v5v5 vNvN -v 6 -v 7 All molecules travelling at slightly different velocities so v 2 varies - take mean - v 2 v3v3 -v 8

33. Pressure = Force Area Force created by N molecules hitting the box… F = Nmv 2 x = Nmv 2 xyz = Nmv 2 V But, molecules move in 3D p =p = Nmv 2 V 1 3 Mean square velocity Kinetic Theory equation

34. Brownian Motion Why does it support the Kinetic Theory? confirms pressure of a gas is the result of randomly moving molecules bombarding container walls rate of movement of molecules increases with temperature confirms a range of speeds of molecules continual motion - justifies elastic collision

35. MicroscopicMacroscopic pV = Nmv 2 1 3 pV= NkT (In terms of molecules) (In terms of physical observations) =Nmv 2 1 3 NkT

36. Already commented that looks a bit like K.E. K.E. = ½mv 2 Rearrange (and remove N) Substitute into (1) = 3kT mv 2 (1) K.E. = 3 2 kT Average K.E. of one molecule

37. Total K.E. of gas (with N molecules) K.E. Total = 3 2 NkT This is translational energy only - not rotational, or vibrational And generally referred to as internal energy, U U = 3 2 NkT

38. U = 3 2 NkT Internal Energy of a gas Sum of the K.E. of all molecules How can the internal energy (K.E.) of a gas be increased? 1)Heat it - K.E.  T 2)Do work on the gas Physically hit molecules Energy and gases

39. Change in Internal Energy Work done on material Energy transferred thermally =+  U = W + Q Basically conservation of energy Also known as the First Law of Thermodynamics

40. Heat, Q – energy transferred between two areas because of a temperature difference Work, W – energy transferred by means that is independent of temperature i.e. change in volume +ve when energy added -ve when energy removed +ve when work done on gas - compression -ve when work done by gas - expansion

41. Einstein’s Model of a solid Bonds between atoms Atom requires energy to break them U  kT Jiggling around (vibrational energy)

42. Mechanical properties change with temperature T = high can break and make bonds quickly – atoms slide easily over each other T = low difficult to break bonds – atoms don’t slide over each other easily Liquid: less viscousSolid: more ductile Liquid: more viscous Solid: more brittle

43. Activation energy,  - energy required for an event to happen i.e. get out of a potential well Activation energy,  Can think of bonds as potential wells in which atoms live

44. The magic  /kT ratio  - energy needed to do something kT - average energy of a molecule  /kT = 1  /kT = 10 - 30  /kT > 100 Already happened Probably will happen Won’t happen

45. Probability of molecule having a specific energy Exponential Energy Probability 1 0

46. Boltzmann Factor e -  /kT Probability of molecules achieving an event characterised by activation energy,  1 10 - 30 > 100 0.37 4.5 x 10 -6 - 9.36 x 10 -14 3.7 x 10 -44 e -  /kT  /kT Nb. 10 9 to 10 13 opportunities per second to gain energy

47. Entropy Number of ways quanta of energy can be distributed in a system Lots of energy – lots of ways Not much energy – very few ways An “event” will only happen if entropy increases or remains constant Amongst particles

48. S = k ln W 2 nd law of thermodynamics S = entropy k = Boltzmann’s constant W = number of ways

50. ΔS = ΔQ T

51. Energy will go from hot to cold At a thermal boundary Hot Cold Number of ways decreases – a bit Number of ways increases – significantly Result - entropy increase

54. Efficiency = W/Q H = (Q H – Q C ) / Q H BUT Δ S = Q/T SO Efficiency = (T H – T C )/ T H = 1 – T C /T H

55. Atoms don’t care. What happens most ways happens most often

56. Specific Thermal Capacity Energy required to raise 1kg of a material by 1K Symbol = cUnit = J kg -1 K -1 Energy and solids (& liquids) Supplying energy to a material causes an increase in temperature

57.  E = mc    E = Energy needed to change temperature of substance / J m =Mass of substance / kg c =Specific thermal capacity of substance / J kg -1 K -1  = Change in temperature / K

58. Energy gained by an electron when accelerated by a 1V potential difference E = 1.6 x 10 -19 x 1= 1.6 x 10 -19 J= 1eV From E = qV Energy Units From E = N A kT Energy of 1 mole’s worth of particles kJ mol -1

59. Latent Heat Extra energy required to change phase Solid  liquid Latent Heat of vaporisation Liquid  gas At a phase boundary there is no change in temperature - energy used just to break bonds Latent Heat of fusion

60. SLHV - water Calculate 1) Number of molecules of water lost 2) Energy used per molecule to evaporate 3) Energy used to vaporise 1kg of water mass evaporated molar mass NANA  energy used n o of molecules evaporated 1kg molar mass  N A  Energy to vaporise one molecule N A = 6.02 x 10 23 Molar mass water = 18g