Sentences with phrase «equals emissivity»

Two infinite planes separated by a distance in equilibrium (the earth being one and some height in the atmosphere being the other)-- the radiosity of the one equals the emissivity of the other.

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.

Not exact matches

Applying aperture photometry on the azimuthally averaged deconvolved PACS images and using a modified blackbody of the form Bν · λ − β, as expected for a grain emissivity Qabs ~ λ − β with β equal to 1.2 (representing amorphous carbon, Mennella et al. 1998), we derived a dust temperature between 108 ± 5 K at 20 ′ ′ and 40 ± 5 K at 180 ′ ′.

UWLWIR is proportional to T ^ 4, (2) with emissivity constant, so the increase in UWLWIR, assuming that the global mean surface temperature is equal 288K, works out to delta U = (288.5 / 288) ^ 4 × 398 — 398 = 2.8 W / m ^ 2.

... For example, the formula given by JBS that T = (e + f - ef) / (e * (2 - e)-RRB--RRB- **.25 * TB can't really be correct since setting the solar input fraction f to zero does not reproduce temperature of the earth TB unless the emissivity e is equal to unity.

For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.

And I # is emitted and absorbed by emission and absorption cross sections and emissivities and absorptivities that are equal for emission into a direction and absorption from that direction at the same location or over the same path length for the same frequency and polarization.

By Kirchhoff's radiation law, the layer emissivity is also equal to the layer absorptivity.

Now if you were an Astrophysicist that has never really don't anything productive in the real world, you would consider that the top of the atmosphere emissivity of 0.69 and that 5.35 is the correct multiplier for the surface because through your Web Hubble Telescope all you can see is albedo and emissivity which has to equal absorption.

Pekka, if absorptivity = emissivity for every wavelength separately, then they must also be equal for every possible mixture of wavelengths.

For the equal temperature surface (emissivity ~ 1)-- atmosphere (emissivity ~ 0.6 in temperate regions), the surface operational emissivity is ~ (1 — 0.6) = 0.4.

Notice that when the absorptivity and emissivity are equal the temperature T1 is pretty much independent of the actual value of emissivity / absorptivity — why is that?

It will not rise at all if the absorption is balanced by an equal amount of emission (as would occur if its emissivity would be increased from a change in its molecular composition — e.g. the formation of ozone from UV radiation or mixing a little CO2 within it).

The emissivity at a given wavelength is generally equal to the absorptivity at that same wavelength.

With some generality, you can say that at a given wavelength, the emissivity and absorptivity are equal.

I believe at equilibrium emissivity and absorption must be equal.

And since the solar input is equal in both cases, then the initial terrestrial output would also be equal, relying only on surface temperature (and emissivity).

The cause of the BB radiation is obviously the energy from the source mediated by the BB Absorbtivity and Emissivity (which at certain wavelengths will be equal once radiative equilibrium is reached)

The earth is almost a perfect blackbody emitter in the mid & far - IR and since Kirchkoff's Law imply that the emissivity and absorptivity must be equal at each wavelength, the means that essentially all of the radiation that is incident on these objects is absorbed.

It is not physically reasonable to assume the true value of Earth's average emissivity -LCB- ε -RCB- to be equal to 1.

As early as 1859, Gustav Kirchhoff proposed that «At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity» and as far as I can understand, nobody objected and his proposition was accepted as part of «Kirchhoff's Law», and, to me, it seems logical and should be unavoidable as it is based on «energy conservation».

The albedo considering only the Surface is closer to 12 % which would make emissivity (if it is equal to absorptivity) about 0.88 rather than 0.7.

Therefore, to assume that the absorptivity of the earth's surface for solar radiation is equal to the emissivity of the earth's surface is a huge mistake.

It is important to understand that absorptivity for longwave radiation will be equal to emissivity for longwave radiation (see Planck, Stefan - Boltzmann, Kirchhoff and LTE), therefore, if the surface and the atmosphere are at the same temperature then the exchange of radiation will be equal.

If we break the atmosphere into 100 layers, the optical depth of each layer is 0.03 making the emissivity equal 0.029554.

whereF is radiant - energy flux at the emitting surface; εis emissivity, set at 1 for a blackbody that absorbs and emits all irradiance reaching its emitting surface (by Kirchhoff's law of radiative transfer, absorption and emission are equal and simultaneous), 0 for a whitebody that reflects all irradiance, and (0, 1) for a graybody that partly absorbs / emits and partly reflects; and σ ≈ 5.67 x 10 — 8 is the Stefan - Boltzmann constant.

At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.

The value of emissivity is equal to the value of absorptivity, but the concepts are opposed.

So Kirchoff's Law NEVER applies to earth atmospheric situations; EMISSIVITY NEVER EQUALS ABSORPTANCE.