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Unformatted text preview: Physics 112 Fall 2010 Professor William Holzapfel Homework 4 Solutions Problem 1: Temperature of the Earth and the Greenhouse Effect (a) In a steadystate the Earth’s temperature is constant and so it must absorb as much power as it emits: Treating the earth as a blackbody, we have P emit E = Area × Flux = ( 4 πR 2 E ) σ b T 4 E = P abs E . The power absorbed by the earth P abs E will be the solar flux at the Earth’s surface Flux sun times the Earth’s cross sectional area A cs e = πR 2 E . To find Flux sun we treat the sun as a blackbody, emitting total power 4 πR 2 S σ b T 4 s . This power spreads out isotropically over a sphere as it travels radially away from the sun; the solar flux at a given distance from the sun d is then Flux sun = Power / Area = σ b T 4 S × 4 πR 2 S 4 πd 2 . Thus the total power absorbed by the earth is P abs E = A cs E × Flux sun = ( πR 2 E ) σ b T 4 S × 4 πR 2 S 4 πd 2 = P emit E = ( 4 πR 2 E ) σ b T 4 E where d is the distance from the earth to the sun. We can solve for the temperature of the earth in terms of the temperature of the sun and the ratio (2 R s ) /d = tan( θ ) ≈ θ where θ ≈ 1 / 2 o = 1 2 2 π rad 360 o ≈ 9 * 10 3 rad is the angle the sun subtends as viewed from the earth. (Remember that the angle that the Sun subtends is given by the ratio of the diameter of the sun to the distance to the sun; that’s where the factor of 2 comes from). Solving for T E then gives T E = T S r 1 4 2 R S d = T S r 1 4 θ. To figure out the temperature of the sun T S we can use the given fact that the spectral intensity as a function of wavelength I λ peaks at λ = 480 nm. The only hiccup is that we don’t know I λ , we know I ν , and to relate the two we CANNOT simply just substitute in the relationship between frequency and wavelength, because I ν is the energy density per unit frequency range and a range of unit frequency does not correspond to a range of unit wavelength. To take this into account we must relate I ν and I λ not as functions but as densities : I ν dν = I λ dλ = ⇒ 2 h ( c λ ) 3 c 2 e h ( c λ ) τ 1 d c λ = I λ dλ = ⇒ I λ = 2 hc 2 λ 5 e hc λτ 1 . By defining x ≡ hc λτ , we can find the maximum of I λ : 1 d dλ I λ = = ⇒ d dx x 5 e x 1 = = ⇒ 5 x 4 e x 1 + ( 1) x 5 e x ( e x 1) 2 = = ⇒ 5 ( e x 1) xe x = = ⇒ e x (5 x ) 5 = . This can be solved numerically, giving x max = 4 . 97 = hc λk b T = ⇒ T S = hc 4 . 97 λk b ....
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 Fall '06
 StevenG.Louie
 Physics, Power, Work, Statistical Mechanics, Entropy, TS, σb tu

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