1. Types of Energy HeatChemical LightGravitational SoundElastic/strain KineticNuclear Electric Stored/potential

2. The Law of Conservation of Energy Energy can be changed (transformed) from one type to another, but it can never be made or destroyed.

3. This means that the total amount of energy in the Universe stays the same!

4. Energy Flow diagrams We can write energy flow diagrams to show the energy changes that occur in a given situation. For example, when a car brakes, its kinetic energy is transformed into heat energy in the brakes. Kineticheat sound

5. Other examples When a rocket launches. Chemicalkinetic gravitational sound heat

6. Energy degradation! In any process that involves energy transformations, the energy that is transferred to the surroundings (thermal energy) is no longer available to perform useful work.

7. Energy transfer (change) A lamp turns electrical energy into heat and light energy

8. Sankey Diagram A Sankey diagram helps to show how much light and heat energy is produced

9. Sankey Diagram The thickness of each arrow is drawn to scale to show the amount of energy

10. Sankey Diagram Notice that the total amount of energy before is equal to the total amount of energy after (conservation of energy)

11. Efficiency Although the total energy out is the same, not all of it is useful.

12. Efficiency Efficiency is defined as Efficiency = useful energy output total energy input

13. Example Efficiency = 75 = 0.15 500

14. Energy efficient light bulb Efficiency = 75 = 0.75 100 That’s much better!

15. Energy Density The energy that can be obtained from a unit mass of the fuel J.kg -1 If the fuel is burnt the energy density is simply the heat of combustion

16. Energy density Coal - 30 MJ.kg -1 Wood - 16 MJ.kg -1 Gasoline – 47 MJ.kg -1 Uranium – 7 x 10 4 GJ.kg -1 (70000000 MJ.kg -1 )

17. Hydroelectric energy density? Imagine 1 kg falling 100m. Energy loss = mgh = 1x10x100 = 10 3 J If all of this is turned into electrical energy it gives an “energy density” of the “fuel” of 10 3 J.kg -1

18. Electromagnetic induction If a magnet is moved inside a coil an electric current is induced (produced)

19. Electromagnetic induction A electric current is induced because the magnetic field around the coil is changing.

20. Generator/dynamo A generator works in this way by rotating a coil in a magnetic field (or rotating a magnet in a coil)

21. Non-renewable Finite (being depleted – will run out) In general from a form of potential energy released by human action

22. Fossil fuels – Coal, oil, gas

23. Nuclear fuels

24. Renewable Mostly directly or indirectly linked with the sun The exception is tidal energy

25. Photovoltaic cells (photoelectric effect)

26. Active solar devices

27. Wind

28. Wave

29. Tidal

30. Biomass

31. World energy production Fuel% total energy production CO 2 emission g.MJ -1 Oil4070 Natural gas2350 Coal2390 Nuclear7- Hydroelectric7- Others< 1-

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

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

34. Fossil fuels - Advantages Relatively cheap High energy density Variety of engines and devices use them directly and easily Extensive distribution network in place Gas power stations are about 50% efficient

35. Fossil fuels - Disadvantages Will run out (finite) Burning coal can cause acid rain Oil spillages etc. Contribute to the greenhouse effect by releasing carbon dioxide

36. 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

37. 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

38. 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

39. Nuclear Fission

40. Uranium Uranium 235 has a large unstable nucleus.

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

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

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

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

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

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

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

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

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

50. 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)

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

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

53. 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.

54. 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!!!

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

56. 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 Expensive to build and maintain

57. Solar power

58. Photovoltaic cells (photoelectric effect)

59. Active solar devices

60. The solar constant

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

62. 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

63. 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)

64. The solar constant This 1400 W.m -2 can only shine on the cross sectional area of the earth as seen from the sun. Area = πr e 2

65. The solar constant This 1400 W.m -2 can only shine on the cross sectional area of the earth as seen from the sun. Area = πr e 2 However, as the earth turns this is spread over the TOTAL surface area of the earth = 4πr e 2

66. The solar constant Therefore the average intensity of the sun falling on the earth = (πr e 2 /4πr e 2 ) 1400 W.m -2 = 350 W.m -2

67. Solar power - advantages “Free” once built Renewable Clean

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

69. Hydroelectric power

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

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

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

73. Hydroelectric - Advantages “Free” once built Renewable Clean

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

75. Wind power Calculating power

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

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

78. 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

79. Wind power - advantages “Free” once built Renewable Clean Ideal for remote locations

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

81. Wave power

82. OWC Oscillating water column

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

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

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

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

87. 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

88. 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

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

90. 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

91. Wave power - Advantages “Free” once built Reasonable energy density Renewable Clean

92. 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

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

94. 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

95. 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!

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

97. 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

98. 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)

99. 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

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

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

102. 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%

103. 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.

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

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

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

107. 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

108. 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

109. 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

110. 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)

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

112. 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

113. 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

114. 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

115. 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.

116. It’s complex!!!

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

118. 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

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

120. 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.

121. What might happen? Polar ice caps melt

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

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

124. 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

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

126. 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.

127. 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?

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

129. What might happen? Long term climate change

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

131. Evidence? 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

132. Other possible causes of global warming? Increase in solar activity Volcanic activity increasing CO2 concentrations Earth orbitting closer to sun?!

133. 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

134. 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