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sci.geo.meteorology (Meteorology) (sci.geo.meteorology) For the discussion of meteorology and related topics. |
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On Feb 19, 8:49*pm, wrote:
It is a most basic fact that objects in a room will all reach the same temperature as the air in a room. OK. So how does the air transfer the energy to the solid substances such as metal? Conduction and radiation. It is not denied that a piece of metal radiates according to Planck's Blackbody Radiation Law, which describes the distribution of the energy of the radiation. The total energy radiated from the surface conforms with the Boltzman Stefan equation. If the air brings the metal to it's temperature at equilibrium, the metal will be radiating from it's surface ... Sir... I am sorry to point out that it's (it is) _its_ temperature and _its_ surface. _No_ apostrophe. The contrary is incorrect and creates a bad impression on the reader. the infrared energy according to the Boltzman Stefan equation, which for 300K, (81F) is 460Wm-2. So a steel ball with 1 meter surface area will be radiating this quantity of energy per second. If this quantity of energy is leaving the surface of the metal at c and in this quantity per second, the metal must be absorbing this quantity of energy at this rate from the gases in the air. Actually, the ball must be absorbing this power from the surrounding EM field -- aka "thermal radiation bath" with which it is in equilibrium. A situation could be devised wherein it could be clearly shown that the energy is not received as radiation from other solid surfaces. That would be a non-equilibrium situation, and would violate the principle of detailed balance. At equilibrium radiation out equals radiation in, on any surface. This energy cannot be transmitted to the metal merely by the collisions or conduction through the air molecules. No, it's transmitted via radiation, at equilibrium. A quantification of the energy of the gas molecules and the transfer of energy to the metal must be achieved in order to have valid theoretical physics of this most basic of phenomena of physics. True. Something we are lacking! This means the gas molecules are not transparent to the infrared as is often believed, but are absorbing and radiating the full continous spectrum of the infrared in the distribution similar to the Planck Radiation Law. That doesn't follow. At equilibrium the radiation in the room is in equilibrium with the gas molecules, however weakly coupled (transparent). We can make no deductions about the degree of transparency or opacity from the mere fact of thermal equilibrium, and equilibrium with the radiation bath. Your argument breaks down here. ... |
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