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Unformatted text preview: 1 ATOC/ASEN 5235 Homework #4 Due Thursday, 30 September Please staple all pages together !!! All problems worth 10 points 1. Petty problem 5.2. A field of snow that is just starting to melt is a special case, in that absorbed solar radiation will contribute primarily to a further phase change (solid to liquid) at a constant temperature of 0°C. Assume that the shortwave albedo of wet snow is 60% and that the solar flux reaching the surface on a particular sunny day is 500 W m-2 . The latent heat of fusion of ice L f is 3.3 × 10 5 J kg-1 and you can assume that the density ρ ws of wet snow stays constant (assuming all excess meltwater drains away) at around 200 kg m-3 . At what rate (cm/hr) is the snow pack depleted by direct solar heating? Answer: A = 0.6 F o = 500 W m-2 L f = 3.3 × 10 5 J/kg ρ = 200 kg m-3 Rate = F (1-A) / (L f × ρ ) = (500 J s-1 m-2 × 0.4) / (3.3e5 J kg-1 × 200 kg m-3 ) Rate = 1.09 cm/hr 2. Use Kirchhoff’s law to answer the following: If ozone in the upper atmosphere is a “good absorber” of solar ultraviolet radiation, why isn’t it also a “good emitter” of such radiation, thereby undoing the good it is said to do? (This is actually problem 1.9 from Bohren and Clothiaux). Answer : Kirchhoff’s law says that absorptivity = emissivity, so any molecule that has a high absorptivity at a given wavelength will also have a high emissivity at that wavelength. It must be realized, however, that absorptivity refers to the fraction of incident radiation that is absorbed, whereas emissivity refers to the fraction of the Planck function that is emitted. The Planck function at stratospheric temperatures has very little contribution from the UV, so even a molecule with a high emissivity cannot emit very much UV radiation. a molecule with a high emissivity cannot emit very much UV radiation....
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This note was uploaded on 01/19/2011 for the course ATOC 5235 taught by Professor Randell during the Fall '10 term at Colorado.
- Fall '10