the net evaporation, thermodynamic

timer Asked: Dec 3rd, 2014

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If you did Problem 2 correctly, you learned that (net) evaporation would rapidly case if water did not interact with its surroundings. The total amount of radiant energy incident on a surface outside Earth's atmosphere and pointed towards the sun is about 1365 W/m^2. This is maximum flux of solar radiation. The actual flux at the surface is less because of attenuation by the atmosphere and because the direction of the sun is not, in general, perpendicular to the surface. For sake of argument, let's say that a representative value. Further assume that all this solar radiation is absorbed by water and its only source of heating(i.e, neglect emission and absorption of infrared radiation and also convective energy transfer from the atmosphere). given these assumptions, estimate the rate of net evaporation of water exposed to sunshine on a clear day. Express this evaporation rate as a change of depth per unit time(e.g mm/h or cm/h). After you have done so, compare your estimate with measurement of net evaporation. the net 

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