1. Electricity production Generally (except for solar cells) a turbine is turned, which turns a generator, which makes electricity.

2. Fossil fuels

3. In electricity production they are burned, the heat is used to heat water to make steam, the moving steam turns a turbine etc.

4. Fossil fuels - Advantages Relatively cheap High energy density Variety of engines and devices use them directly and easily Extensive distribution network in place

5. Fossil fuels - Disadvantages Will run out Pollute the environment (during mining sulphur and heavy metal content can be washed by rain into the environment) Oil spillages etc. Contribute to the greenhouse effect by releasing greenhouse gases

6. Example question A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35%

7. Calculate the rate at which thermal energy is provided by the coal

8. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35% Calculate the rate at which thermal energy is provided by the coal Efficiency = useful power output/power input Power input = output/efficiency Power input = 400/0.35 = 1.1 x 10 3 MW

9. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35% Calculate the rate at which coal is burned (Coal energy density = 30 MJ.kg-1)

10. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35% Calculate the rate at which coal is burned (Coal energy density = 30 MJ.kg-1) 1 kg of coal burned per second would produce 30 MJ. The power station needs 1.1 x 10 3 MJ per second. So Mass burned per second = 1.1 x 10 3 /30 = 37 kg.s -1 Mass per year = 37x60x60x24x365 = 1.2 x 10 9 kg.yr -1

11. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35% The thermal energy produced by the power plant is removed by water. The temperature of the water must not increase by more than 5 °C. Calculate the rate of flow of water.

12. A coal powered power plant has a power output of 400 MW and operates with an overall efficiency of 35% The thermal energy produced by the power plant is removed by water. The temperature of the water must not increase by moe than 5 °C. Calculate the rate of flow of water. Rate of heat loss = 1.1 x 10 3 – 0.400 x 10 3 = 740 MW In one second, Q = mcΔT 740 x 10 6 = m x 4200 x 5 m = 35 x 10 3 kg So flow needs to be 35 x 10 3 kg.s -1

13. Nuclear Fission

14. Uranium Uranium 235 has a large unstable nucleus.

15. Capture A lone neutron hitting the nucleus can be captured by the nucleus, forming Uranium 236.

16. Capture A lone neutron hitting the nucleus can be captured by the nucleus, forming Uranium 236.

17. Fission The Uranium 236 is very unstable and splits into two smaller nuclei (this is called nuclear fission)

18. Fission The Uranium 236 is very unstable and splits into two smaller nuclei (this is called nuclear fission)

19. Free neutrons As well as the two smaller nuclei (called daughter nuclei), three neutrons are released (with lots of kinetic energy)

20. Fission These free neutrons can strike more uranium nuclei, causing them to split.

21. Chain Reaction If there is enough uranium (critical mass) a chain reaction occurs. Huge amounts of energy are released very quickly.

22. Bang! This can result in a nuclear explosion! YouTube - nuclear bomb 4 YouTube - nuclear bomb 4

23. Controlled fission The chain reaction can be controlled using control rods and a moderator. The energy can then be used (normally to generate electricity).

24. Fuel rods In a Uranium reactor these contain Enriched Uranium (the percentage of U-235 has been increased – usually by centrifuging)

25. Moderator This slows the free neutrons down, making them easier to absorb by the uranium 235 nuclei. Graphite or water is normally used. 1 eV neutrons are ideal)

26. Control rods These absorb excess neutrons,making sure that the reaction does not get out of control. Boron is normally used.

27. Heat The moderator gets hot from the energy it absorbs from the neutrons.

28. Heat This heat is used to heat water (via a heat exchanger), to make steam, which turns a turbine, which turns a generator, which makes electricity.

29. Useful by-products Uranium 238 in the fuel rods can also absorb neutrons to produce plutonium 239 which is itself is highly useful as a nuclear fuel (hence breeder reactors) It makes more fuel!!!

30. Nuclear Power That’s how a nuclear power station works!

31. Nuclear power - Advantages High power output Large reserves of nuclear fuels No greenhouse gases

32. Nuclear power - disadvantages Waste products dangerous and difficult to dispose of Major health hazard if there is an accident Problems associated with uranium mining Nuclear weapons

33. Solar power

34. The solar constant

35. The sun’s total power output is 3.9 x 10 26 W!

36. The solar constant The sun’s total power output is 3.9 x 10 26 W! Only a fraction of this power actually reaches the earth, given by the formula I (Power per unit area) = P/4πr 2 For the earth this is 1400 W.m -2 and is called the solar constant

37. The solar constant For the earth this is 1400 W.m -2 and is called the solar constant This varies according to the power output of the sun (± 1.5%), distance from sun (± 4%), and angle of earth’s surface (tilt)

38. Solar power - advantages “Free” Renewable Clean

39. Solar power - disadvantages Only works during the day Affected by cloudy weather Low power output Requires large areas Initial costs are high

40. Hydroelectric power

41. Water storage in lakes “High” water has GPE. AS it falls this urns to KE, turns a turbine etc.

42. Pumped storage Excess electricity can be used to pump water up into a reservoir. It acts like a giant battery.

43. Tidal water storage Tide trapped behind a tidal barrage. Water turns turbine etc. YouTube - TheUniversityofMaine's ChannelYouTube - TheUniversityofMaine's Channel

44. Hydroelectric - Advantages “Free” Renewable Clean

45. Hydroelectric - disadvantages Very dependent on location Drastic changes to environment (flooding) Initial costs very high

46. Wind power

47. Calculating power

48. Wind moving at speed v, cross sectional area of turbines = A V A

49. V A Volume of air going through per second = Av Mass of air per second = Density x volume Mass of air per second = ρAv

50. Wind moving at speed v, cross sectional area of turbines = A V A Mass of air per second = ρAv If all kinetic energy of air is transformed by the turbine, the amount of energy produced per second = ½mv 2 = ½ρAv 3

51. Wind power - advantages “Free” Renewable Clean Ideal for remote locations

52. Wind power - disadvantages Works only if there is wind! Low power output Unsightly (?) and noisy Best located far from cities High maintainance costs

53. Wave power

54. OWC Oscillating water column

55. Modeling waves We can simplfy the mathematics by modeling square waves. λ L 2A

56. Modeling waves If the shaded part is moved down, the sea becomes flat. λ L 2A

57. Modeling waves The mass of water in the shaded part = Volume x density = Ax(λ/2)xLxρ = AλLρ/2 λ L 2A

58. Modeling waves Loss of E p of this water = mgh = = (AλLρ)/2 x g x A = A 2 gLρ(λ/2) λ L 2A

59. Modeling waves Loss of E p of this water = mgh= A 2 gLρ(λ/2) # of waves passing per unit time = f = v/λ λ L 2A

60. Modeling waves Loss of E p per unit time = A 2 gLρ(λ/2) x v/λ = (1/2)A 2 Lρgv λ L 2A

61. Modeling waves The maximum power then available per unit length is then equal to = (1/2)A 2 ρgv λ L 2A

62. Power per unit length A water wave of amplitude A carries an amount of power per unit length of its wavefront equal to P/L = (ρgA 2 v)/2 where ρ is the density of water and v stands for the speed of energy transfer of the wave

63. Wave power - Advantages “Free” Reasonable energy density Renewable Clean

64. Wave power - disadvantages Only in areas with large waves Waves are irregular Low frequency waves with high frequency turbine motion Maintainance and installation costs high Transporting power Must withstand storms/hurricanes

65. Radiation from the Sun http://www.youtube.com/watch?NR=1&v=1pfqIcSydgE

66. Black-body radiation Black Body - any object that is a perfect emitter and a perfect absorber of radiation object does not have to appear "black" sun and earth's surface behave approximately as black bodies

67. Black-body radiation http://phet.colorado.edu/sims/blackbody- spectrum/blackbody-spectrum_en.htmlhttp://phet.colorado.edu/sims/blackbody- spectrum/blackbody-spectrum_en.html Need to “learn” this!

68. Wien’s law λ max T = constant (2.9 x 10 -3 mK)

69. Example The sun has an approximate black-body spectrum and most of its energy is radiated at a wavelength of 5.0 x 10 -7 m. Find the surface temperature of the sun. From Wien’s law 5.0 x 10 -7 x T = 2.9 x 10 -3 T = 5800 K

70. Spectral ClassColourTemperature/K OBlue25 000 – 50 000 BBlue - white12 000 – 25 000 AWhite7 500 – 12 000 FYellow - white6 000 – 7 500 GYellow4 500 – 6 000 KYellow - red3 000 – 4 500 MRed2 000 – 3 000 In the astrophysics option you need to remember the classes and their order. How will you do this?

71. Spectral classes Oh be a fine girl….kiss me!

72. Stefan-Boltzmann law The amount of energy per second (power) radiated from a body depends on its surface area and absolute temperature according to P = eσAT 4 where σ is the Stefan-Boltzmann constant (5.67 x 10 -8 W.m -2.K -4 ) and e is the emissivity of the surface ( e = 1 for a black object)

73. Example By what factor does the power emitted by a body increase when its temperature is increased from 100ºC to 200ºC?

74. Example By what factor does the power emitted by a body increase when its temperature is increased from 100ºC to 200ºC? Emitted power is proportional to the fourth power of the Kelvin temperature, so will increase by a factor of 473 4 /373 4 = 2.59

75. Graph sketching

76. Global Warming

77. The Sun The sun emits electromagnetic waves (gamma X-rays, ultra-violet, visible light, infra-red, microwaves and radio waves) in all directions.

78. The earth Some of these waves will reach the earth

79. Reflected Around 30% will be reflected by the earth and the atmosphere. This is called the earth’s albedo (0.30). (The moon’s albedo is 0.12) Albedo is the ratio of reflected light to incident light. 30%

80. Albedo The Albedo of a body is defined as the ratio of the power of radiation reflected or scattered from the body to the total power incident on the body.

81. Albedo The albedo depends on the ground covering (ice = high, ocean = low), cloud cover etc.

82. Absorbed by the earth Around 70% reaches the ground and is absorbed by the earth’s surface. 70%

83. Absorbed by the earth Infrared This absorbed solar energy is re-radiated at longer wavelengths (in the infrared region of the spectrum)

84. Temperature of the earth with no atmosphere? Remember the solar constant is around 1360 W.m -2. This can only shine on one side of the Earth at a time, and since the silhouette of the earth is a circle, the power incident = 1360 x πr 2 = 1360 x π x (6.4 x 10 6 ) 2 = 1.75 x 10 17 W

85. Temperature of the earth with no atmosphere? Power incident on earth = 1.75 x 10 17 W Since the albedo is 30%, 70% of the incident power will be absorbed by the Earth 70% of 1.75 x 10 17 W = 1.23 x 10 17 W

86. Temperature of the earth with no atmosphere? Power absorbed by Earth = 1.23 x 10 17 W At equilibrium, the Power absorbed = Power emitted Using the Stefan Boltzmann law; 1.23 x 10 17 = eσAT 4

87. Temperature of the earth with no atmosphere? Using the Stefan Boltzmann law; 1.23 x 10 17 = eσAT 4 1.23 x 10 17 = 1 x 5.67 x 10 -8 x 4πr 2 x T 4 This gives T = 255 K (-18°C)

88. Temperature of the earth with no atmosphere? T = 255 K (-18°C) This is obviously much colder than the earth actual temperature. WHY?

89. Absorbed by the earth Infrared This absorbed solar energy is re-radiated at longer wavelengths (in the infrared region of the spectrum) http://phet.colorado.edu/en/simulation/green house http://phet.colorado.edu/en/simulation/green house

90. Absorbed Various gases in the atmosphere can absorb radiation at this longer wavelength (resonance) C O O C H H H H They vibrate more (become hotter) HH O

91. Greenhouse gases These gases are known as “Greenhouse” gases. They include carbon dioxide, methane, water and N 2 O. C O O C H H H H HH O

92. Transmittance curves

93. Re-radiated These gases in the atmosphere absorb the infra-red radiation and re-emit it, half goes into space but half returns to the earth.

94. It’s complex!!!

95. Balance There exists a balance between the energy absorbed by the earth (and its atmosphere) and the energy emitted. Energy in Energy out

96. Balance This means that normally the earth has a fairly constant average temperature (although there have been big changes over thousands of years) Energy in Energy out

97. Balance Without this normal “greenhouse effect” the earth would be too cold to live on. Energy in Energy out

98. Greenhouse gases Most scientists believe that we are producing more of the gases that absorb the infra-red radiation, thus upsetting the balance and producing a higher equilibrium earth temperature. This is called the enhanced greenhouse effect.

99. What might happen?

100. Polar ice caps melt

101. What might happen? Higher sea levels and flooding of low lying areas as a result of non-sea ice melting and expansion of water

102. Coefficient of volume expansion Coefficient of volume expansion is defined as the fractional change in volume per unit temperature change

103. Coefficient of volume expansion Given a volume V 0 at temperature θ 0, the volume after temperature increase of Δθ will increase by ΔV given by ΔV = γV 0 Δθ

104. Definition Coefficient of volume expansion is the fractional change in volume per unit temperature change. ΔV = γV 0 Δθ

105. Example The area of the earth’s oceans is about 3.6 x 10 8 km 2 and the average depth is 3.7 km. Using γ = 2 x 10 -4 K -1, estimate the rise in sea level for a temperature increase of 2K. Comment on your answer.

106. Example The area of the earth’s oceans is about 3.6 x 10 8 km 2 and the average depth is 3.7 km. Using γ = 2 x 10 -4 K -1, estimate the rise in sea level for a temperature increase of 2K. Comment on your answer. Volume of water = approx depth x area = 3.6 x 10 8 x 3.7 = 1.33 x 10 9 km 3 = 1.33 x 10 18 m 3 ΔV = γV 0 Δθ ΔV = 2 x 10 -4 x 1.33 x 10 18 x 2 = 5.3 x 10 14 m 3 Δh = ΔV/A = 5.3 x 10 14 /3.6 x 10 14 = 1.5 m Evaporation? Greater area cos of flooding? Uniform expansion?

107. What else might happen? More extreme weather (heatwaves, droughts, hurricanes, torrential rain)

108. What might happen? Long term climate change

109. What might happen? Associated social problems (??)

110. Evidence?

111. Ice core research Weather records Remote sensing by satellites Measurement! How do ice cores allow researchers to see climate change? | GrrlScientist | Science | guardian.co.ukHow do ice cores allow researchers to see climate change? | GrrlScientist | Science | guardian.co.uk

112. Surface heat capacitance C s Surface heat capacitance is defined as the energy required to increase the temperature of 1 m 2 of a surface by 1 K. Cs is measured in J.m -2.K -1. Q = AC s ΔT

113. Example Radiation of intensity 340 W.m -2 is incident on the surace of a lake of surface heat capacitance Cs = 4.2 x 10 8 J.m -2.K -1. Calculate the time to increase the temperature by 2 K. Comment on your answer.

114. Example Radiation of intensity 340 W.m -2 is incident on the surface of a lake of surface heat capacitance Cs = 4.2 x 10 8 J.m -2.K -1. Calculate the time to increase the temperature by 2 K. Comment on your answer. Each 1m 2 of lake receives 340 J.s -1 Energy needed to raise 1m 2 by 2 K = Q = AC s ΔT = 1 x 4.2 x 10 8 x 2 = 8.4 x 10 8 J Time = Energy/power = 8.4 x 10 8 /340 = 2500000 seconds = 29 days Sun only shines approx 12 hours a day so would take at least twice as long

115. Let’s read! Pages 198 to 211 of SL Physics by Hamper and Ord. Pages 434 to 450 of Physics for the IB Diploma by Tsokos

116. Homework Page 450 Qs 1, 2a, 5, 7, 9, 20, 30.