Sentences with phrase «equilibrium of a blackbody»

Not exact matches

The emissivity of the surface in the infrared is unimportant because it behaves as though it were in a blackbody cavity in equilibrium with the lowest layers of the atmosphere.

At equilibrium, it will have a temperature at which the blackbody flux at 15 microns would be half of the actual OLR at 15 microns.

Scattering may also drive the distribution over polarizations toward an equilibrium (which would be, at any given frequency and direction, constant over polarizations so long as the real component of the index of refraction is independent of polarization) Interactions wherein photons are scattered by matter with some exchange of energy will eventually redistribute photons toward a Planck - function distribution — a blackbody spectrum — characteristic of some temperature, and because the exchange involves some other type of matter, the photon gas temperature (brightness temperature) will approach the temperature of the material it is interacting with -LRB-?

I.absorbed / I.incident = absorptivity; I.absorbed = I.emitted; I.incident = B.emitted (because they have the same brightness temperature, where B.emitted is what would be emitted by a blackbody, and is what would be in equilibrium with matter at that temperature), emissivity = I.emitted / B.emitted; therefore, given that absorptivity is independent of incident intensity but is fixed for that material at that temperature at LTE, and the emitted intensity is also independent of incident intensity but is fixed for that material at that temperature, emissivity (into a direction) = absorptivity (from a direction).

So while, in the isothermal blackbody surface approximation, if the starting surface temperature is 288 K and we know the OLR is reduced from surface emission by 150 W / m2 via GHE, we know that removing all greenhouse agents will have a TOA forcing of -150 W / m2, (and some forcing at the tropopause, etc.) which will cool the surface temperature to about 255 K at equilibrium, absent non-Planck feedbacks.

This was the blackbody radiator fed with 240 W, that need to transfer 240W at equilibrium to a reflective blanket that only absorbs 10 % of the incident energy.

If the radiaitve thermal equilibrium is both with the sun (a blackbody) and space (a blackbody), then no matter what kind of non-blackbody the earth is, absorptivity must still equal emissivity at every wavelength, unless someone is arguing that Kirchoff's law no longer holds.

At equilibrium with cell contents and source at the same temperature, the spectrophotometer will see the same blackbody curve and total energy flux for T whether the cell is evacuated or filled with any gas or mixture of gases.

I'm not going to review the various arguments that indicate that this is indeed the equilibrium — they are straightforward consideration of the integrals over the blackbody spectra from the two bodies that shows that the hotter one loses heat (on average) and the colder one gains heat (on average) until they are at the same temperature and have identical spectra, where the (time / frequency averaged integral of the) flux of the Poynting vector vanishes within microscopic thermal fluctuations of the sort that are routinely ignored in thermodynamics.

Given that the blackbody equilibrium temperature of earth as seen from space is a function of solar irradiance arriving and earth albedo and not much of anything else apart from factors that change those two, anyone claiming earth's temperature isn't affected by solar output better have a pretty good theory and data to support that.

For example, a flat, perfect, blackbody, receiving 965 W / m ^ 2 would, at equilibrium, have a temperature of 361.2 K (88 ºC, 190.5 ºF)

the blackbody is already in radiative thermal equilibrium with a hotter source of energy That is simply wrong — That might be true if the blackbody was surrounded by a perfect insulator but is definitely not true for planets in space.