most earth’s e comes from the sun
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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. - PowerPoint PPT PresentationTRANSCRIPT
Most Earth’s E comes from the sun.
Solar Intensity E received (any planet)
I = power = P I = intensity W/m2,
area 4r2. P = power of sun W.
(sphere) r = distance.
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 4(1.5 x 1011 m)2, = 1380 W/m2
solar “constant” ~ 1400 W m2.
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.
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.
To find the exact E reaching an object, we need to know how much is absorbed & reflected by atmosphere & surface.
Albedo - Ability of planet to reflect or scatter radiation. It’s a ratio.
Albedo = total reflected/scattered power total incident power
always 0-11 = high reflectionSee tables.
Albedo %
Snow - 80.Ground - 10.
Ice - 90Ocean – 10.
Mean Earth Albedo = 30%
What factors Affect Albedo?
• Season (cloud cover)
• Cloud type
• Surface type (water, land, forest, snow, ice).
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.
• 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.
How does the solar E interact with matter on Earth?
• Individual Atoms (gasses):
• Can model photon absorption with Bohr
• Excite e- to different orbits by dif of photons
• Ionization of atom
Indv. Molecules More Complex
like to vibrate at specific resonant f.
Photons w/ E at the resonant f, are absorbed. KE increases.
T increase.Usu absorb IR f.
• E Interaction with polyatomic Solids dif than single atoms or molecules.
• Solids absorb large range of fover broader spectrum.
Molc’s vibrate,Emit low f E IR.
5 Greenhouse Gases
• CO2
• H20
• CH4
• N2O
• O3.
• 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 / photon resonance occurs.
• Molecular KE increase so T increases.
CO2 absorbs specific IR f’s , molecular resonance occurs.
wavelength
IR Spectrum Methanemore vibrational modes, more absorbed.
wavelength
Gasses absorb specific f, solids absorb a range of f.
Black Bodies & Absorption & Emission of EM
Solids absorb & wide range f ‘s
When hot emit same f range:
Black bodies – absorb & emit all .
Black Body emissionWhat do you notice as T
Black Body SpectrumTotal intensity goes up.
The shorter more intense as T increases.
Equations of Climate
Peak is calculated by Wein’s
displacement law:
b = 2.89 x 10-3 m K
T
b
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.
Stefan-Boltzmann Law - Relates emitted power & to object’s T & area (m2). is a constant for black body.
Emitted P = power Watts.
A = surface area m2 (Area sphere = 4 r2) sun, Earth.
T = Kelvin T
= 5.67x 10 – 8 Wm-2 K-4
PTA 4
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 = 2rh) (ignore ends).
60 J/s = A T4.
A = (2)()(5.0 x 10-5 m)(0.5m) = 1.57 x 10-4m2
T = 1611 K = 1600 K.
PTA 4
T4 = (1.57 x 10-4m2) (5.67x 10 – 8 )
60 J/s
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 r2.
PTA 4 424 TrP
P = 4()(7 x 108 m)2(5.67x 10 – 8 )(6000 K)4.
P = 4.53 x 10 26 W.
Most objects are not as emissive as a black body.
Emissivity (e)• Is a number from 0 – 1 telling how an object’s
emitted radiation compares w/ perfect black body. From Stefan’s law:
4TeAP • 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.
4. An object at 500 K with a surface area of 5 m2, emits 5300 W of power. What is its emissivity?
• P = eAT4.
• 5300 W = e (5m2) (5.67x 10 – 8) (500 K )4.
• e = 0.3
Surface Heat Capacity (Cs) – amount E required to heat 1m2 of a surface by 1oC or 1K.
Cs = Q AT
Q –JoulesT temp dif.
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
AT (50 m2)(10 K)
4 x 108 Jm-2K-1
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
• Psun = (0.95) (4)(7x108)2(5.67x10-8) (5800)4.
Psun = 3.8 x 1026 W.
4TeAP
• Pearth = = 1.6 x 10 17W.
Hamper Read 8.9 Do pg 201 #18-21and Handout Greenhouse Effect 1and mult choice question in packet.