Do Now. The sun radiates ~3.9 x J /s, Earth av. orbit = 1.5 x m, calculate intensity of radiation reaching Earth. 3.9 x Js -1 4  (1.5 x m) 2, = 1380 W/m 2 solar “constant” ~ 1400 W m 2.

Solar Intensity E received (any planet) I = power = P I = intensity W/m 2, area 4  r 2. P = radiated power W. (sphere) r = distance.

Energy Transfer

Name the 3 types of E transfer Conduction. Convection Radiation. All important in climate study. Only way for Earth to lose E to space is….

Radiation All matter can both absorb and emit EM radiation. Radiation is absorbed by matter at specific frequencies,. Solids absorb many more different  than the atmospheric gases.

Black-body radiation  A solid’s color determines the wavelengths it cannot absorb.  A green object reflects (does not absorb) green light.  A black object absorbs and emits all wavelengths. When bathed in white light a solid is the color of the light that it cannot absorb.

EM Emission Matter above 0 K emits EM radiation. EM generated by accelerating charges. Electrons, protons, nuclei, ions. Emitted radiation is related to the T, and type of surface.  As a black body heats up it emits all wavelengths, called black-body radiation.

Demo What happens as T increases?

Black Body Spectrum Total intensity goes up. The shorter more intense as T increases. Sun = 5800 K Earth = 288 K Vis, IR IR only

Peak is calculated by Wein’s displacement law: b = 2.89 x m K

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

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

2. 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 10 8 m. Area sphere = 4  r 2. P = 4(  )(7 x 10 8 m) 2 (5.67x 10 – 8 )(6000 K) 4. P = 4.53 x W.

3. A tungsten filament has a length of 0.5 m and a radius of 5.0 x m. The power rating is 60 W. Estimate the temperature of the filament if it acts as a black body. (Use surface A = 2  rh). 60 J/s = A  T 4. A = (2)(  )(5.0 x m)(0.5m) = 1.57 x m 2 T = 1611 K = 1600 K. T 4 = 60 J/s (1.57 x m 2 ) (5.67x 10 – 8 )

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: e is emissivity = ratio energy emitted/black body energy at a T. Shiny objects have low e, dark objects have high e.

4. An object at 500 K with a surface area of 5 m 2, emits 5300 W of power. What is its emissivity? P = eA  T W = e (5m 2 ) ( 5.67x 10 – 8 ) (500 K ) 4. e = 0.3

5. The sun’s surface has a temperature of 4500K. What is the prevalent wavelength of light? SOLUTION: max = 2.90  / 4500 max = 6.44  m = 644 nm. Intensity Wavelength (nm) Wavelength / nm Visible Light

FYI  Since no body is at absolute zero (K = 0) it follows from the Stefan-Boltzmann law that all bodies radiate. 6. The planet Mercury has a radius of 2.50  10 6 m. Its sunny side has a T 400°C and its shady side -200°C. Treat it as a black-body, find its average power. SOLUTION:  A sphere = 4  (2.50  10 6 ) 2 = 7.85  m 2.  For T use T AVG = ( ) / 2 = 373 K P =  AT 4 = (5.67  )(7.85  )373 4 = 8.62  W.

Energy from the Sun - Insolation Sun radiates 43% visible, 49 % IR, 8 % UV. Earth receives very small fraction of total solar power ~ 1400 W/m 2 - most does not reach surface.

Solar Constant at top of Atmosphere average ~ 1390 W/m 2 11 year sunspot cycle ~ 0.1 % Elliptical orbit ~ 7 % Longer term cycles (Milankovitch)

Absorbed by Earth’s surface? Some radiation reflected or scattered before absorption.

Albedo - Ability of planet to reflect or scatter radiation. It’s a ratio. Albedo  = total reflected/scattered I total incident I always = high reflection See tables.

7. Given the following values, find the albedo: Incident power = 340 W/m 2 Reflected power = 100 W/m 2. Re-radiated power = 2 W/m %

8. Match the surface materials to their correct albedo. Snow Ground Ice charcoal

Albedo % Mean Earth Albedo = 30%

Kerboo sheet 8.2 solar constant albedo. End here Wed

Earth’s day/night cycle, tilt, & varying orbital distance affect the insolation hitting surface. Accounting for day/night & seasons could average to ~ 170 W/m 2 or less.

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

Natural Greenhouse Effect Natural warming effect due to atmosphere. The moons av T is -18 o C. Earth is +16 o C. Same dist fr sun but no atmosphere Atmospheric greenhouse gasses absorb outgoing IR radiation from Earth, re-radiate some back to Earth.

What happens to the radiation? 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 in all directions, some back to Earth.

Interaction of solar E 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

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 gas molecules. Solids absorb large range of f over broader spectrum. Molc’s vibrate, Emit low f E IR.

5 Greenhouse Gases CO 2 H 2 0 CH 4 N 2 O O 3. 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.

CO 2 absorbs specific IR f’s, molecular resonance occurs. wavelength

IR Spectrum Methane more vibrational modes, more absorbed. wavelength

Gasses absorb & emit specific f, Have absorption and emission spectrum.

Solids Absorb & emit wide range f ‘s Black bodies – absorb & emit all.

Film Clip How do greenhouse gasses work? 3.1 min qvTghttps:// qvTg hi

Phet black bodies

Equations of Climate

All incident radiation is either reflected or absorbed. Fraction absorbed = 1 – .

Surface Heat Capacity (C s ) – amount E required to heat 1m 2 of a surface by 1 o C or 1K. C s = Q A  T Q –Joules  T temp dif.

5. It takes 2 x J of E to heat 50 m 2 of Earth by 10 K. Find the surface heat capacity for Earth. C s = Q 2 x J A  T (50 m 2 )(10 K) 4 x 10 8 Jm -2 K -1

6. Find the approximate radiation power of the sun & maybe the earth, given the following data: Sun radius7 x 10 8 m Earth radius6.4 x 10 6 m Surface T sun5800 K Surface T Earth25 o C. Earth e0.7 Sun e0.95

P sun = (0.95) (4  )(7x10 8 ) 2 (5.67x10 -8 ) (5800) 4. P sun = 3.8 x W. P earth = = 1.6 x W.

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