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.