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Most Earth’s E comes from the sun.

Most Earth’s E comes from the sun. Solar Intensity E received (any planet). I = power = P I = intensity W/m 2 , area 4 p r 2 . P = power of sun W. (sphere) r = distance.

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Most Earth’s E comes from the sun.

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  1. Most Earth’s E comes from the sun.

  2. Solar Intensity E received (any planet) I = power = P I = intensity W/m2, area 4pr2. P = power of sun W. (sphere) r = distance.

  3. 1. The sun radiates ~3.9 x 10 26 J /s, if the distance btw earth – sun is 1.5 x 1011 m, What is the intensity of radiation reaching Earth? 3.9 x 10 26 Js-1 4p(1.5 x 1011 m)2, = 1380 W/m2 solar “constant” ~ 1400 W m2.

  4. Energy from the Sun - Insolation Sun radiates 43% visible, 49 % IR, 8 % UV. Earth receives very small fraction of total solar power ~ 1400 W/m2 - most does not reach surface.

  5. Earth’s day/night cycle, tilt, & varying orbital distance affect the insolation hitting surface. Accounting for day/night & seasons could average to ~ 170 W/m2 or less.

  6. To find the exact E reaching an object, we need to know how much is absorbed & reflected by atmosphere & surface.

  7. Albedo - Ability of planet to reflect or scatter radiation. It’s a ratio. Albedo a = total reflected/scattered power total incident poweralways 0-1 1 = high reflectionSee tables.

  8. Albedo % Snow - 80. Ground - 10. Ice - 90 Ocean – 10. M ean Earth Albedo = 30%

  9. What factors Affect Albedo? • Season (cloud cover) • Cloud type • Surface type (water, land, forest, snow, ice).

  10. Natural Greenhouse EffectNatural warming effect due to atmosphere. • The moons av T is -18oC. • Earth is +16 oC. • Same dist fr sun. • Atmospheric greenhouse gasses absorb outgoing IR radiation from Earth, re-radiate some back to Earth.

  11. Sun emits Visible, some IR, &UV. • Visible light gets through atmosphere to Earth. • (UV & IR mostly absorbed in atmosphere) • Earth surface either reflects, or absorbs & later emits E as IR radiation. • Greenhouse gasses absorb, re-radiate IR back to Earth.

  12. How does the solar E interact with matter on Earth?

  13. Individual Atoms (gasses): • Can model photon absorption with Bohr • Excite e- to different orbits by difl of photons • Ionization of atom

  14. Indv. Molecules More Complexlike to vibrate at specific resonant f. Photons w/ E at the resonant f, are absorbed. KE increases.T increase.Usu absorb IR f.

  15. E Interaction with polyatomic Solidsdif than single atoms or molecules. • Solids absorb large range of f • over broader spectrum. • Molc’s vibrate, • Emit low f E IR.

  16. 5 Greenhouse Gases • Natural Resonant f of greenhouse gasses is in IR region-the emission f of solids. Visible light f too high. • When molc absorbs proper IR l / photon resonance occurs. • Molecular KE increase so T increases. • CO2 • H20 • CH4 • N2O • O3.

  17. CO2 absorbs specific IR f’s l, molecular resonance occurs. wavelength

  18. IR Spectrum Methanemore vibrational modes, more l absorbed. wavelength

  19. Gasses absorb specific f, solids absorb a range of f.

  20. Black Bodies & Absorption & Emission of EM Solids absorb & wide range f ‘s When hot emit same f range: Black bodies – absorb & emit all l.

  21. Black Body emissionWhat do you notice as T

  22. Black Body SpectrumTotal intensity goes up.The shorter l more intense as T increases.

  23. Equations of Climate

  24. Peak l is calculated by Wein’s displacement law: b = 2.89 x 10-3 m K

  25. 1. What is the peak wavelength for a lamp that glows at 1800o C? • 1800o C = 2073 K • 2.89 x 10-3 m K 2073 K • 1.39 x 10-6 m.

  26. Stefan-Boltzmann Law - Relates emitted power & to object’s T & area (m2). s is a constant for black body. Emitted P = power Watts. A = surface area m2 (Area sphere = 4 p r2) sun, Earth. T = Kelvin T s = 5.67x 10 – 8 Wm-2 K-4

  27. 2. A tungsten filament has a length of 0.5 m and a radius of 5.0 x 10-5 m. The power rating is 60 W. Estimate the temperature of the filament if it acts as a black body. (Use surface A = 2prh) (ignore ends). 60 J/s = A s T4. A = (2)(p)(5.0 x 10-5 m)(0.5m) = 1.57 x 10-4m2 T4 = (1.57 x 10-4m2) (5.67x 10 – 8 ) 60 J/s T = 1611 K = 1600 K.

  28. 3. If the Sun behaves as a perfect black body with T = 6000 K, what is the energy radiated per second? The radius sun is 7 x 108 m. • Area sphere = 4 p r2. P = 4(p)(7 x 108 m)2(5.67x 10 – 8 )(6000 K)4. P = 4.53 x 10 26 W.

  29. Most objects are not as emissive as a black body.

  30. Emissivity (e) • Is a number from 0 – 1 telling how an object’s emitted radiation compares w/ perfect black body. From Stefan’s law: • e is emissivity = ratio energy emitted/black body energy at a T. • Shiny objects have low e, dark objects have high e. • Emissivity + albedo = 1.

  31. 4. An object at 500 K with a surface area of 5 m2, emits 5300 W of power. What is its emissivity? • P = eAsT4. • 5300 W = e (5m2) (5.67x 10 – 8) (500 K )4. • e = 0.3

  32. SurfaceHeatCapacity (Cs) – amount E required to heat 1m2 of a surface by 1oC or 1K. Cs = Q ADT Q –Joules DT temp dif.

  33. 5. It takes 2 x 10 11 J of E to heat 50 m2 of Earth by 10 K. Find the surface heat capacity for Earth. Cs = Q 2 x 10 11 J ADT (50 m2)(10 K) 4 x 108 Jm-2K-1

  34. 6. Find the approximate radiation power of the sun & the earth, given the following data: • Sun radius 7 x 108 m • Earth radius 6.4 x 106 m • Surface T sun 5800 K • Surface T Earth 25oC. • Earth e 0.7 • Sun e 0.95

  35. Psun = (0.95) (4p)(7x108)2(5.67x10-8) (5800)4. Psun = 3.8 x 1026 W. • Pearth = = 1.6 x 10 17W.

  36. Hamper Read 8.9 Do pg 201 #18-21and Handout Greenhouse Effect 1and mult choice question in packet.

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