Sentences with phrase «emissivity of»

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.

An emissivity 1 is only for ideal blackbodies and, as you have accepted, blackbodies don't exist in real nature, consequently, the emissivity of the Earth CAN NOT be the emissivity of a black body.

However, you assigned a total emissivity to the Sun of 1.0, which is the emissivity of the non-existent blackbodies when you wrote:

«One possible (repeat possible) answer to your question about the cotton field example is that the calibration of the instruments have been set to «assume» that CO2 has an emissivity of 1»

The emissivity of the wall is defined as the ratio of emitted energy to the amount that would be radiated if the wall were a perfect black body.

Putting aside the fact that the above represents how much a surface at 293K would be emitting (not absorbing) with an emissivity of 0.82 for a second there is another potentially more serious problem.

The emissivity of the atmosphere is going to be prefferentialy concerntrated in the H2O and CO2 bands because these are several orders of magnitude «better» at absorbing and emitting at these wavelengths than the N2 AND O2 are at any wavelength.

This is only accomplished by calculating temperature using an emissivity of 1 for the 443.17 W / m ^ 2 he says the surface is emitting towards the atmosphere.

I referred you back to my original calculation because the end result -LRB--18.3 C) was for an earth absorbing with absorptivity of 0.7 across solar wavelengths and an emissivity of 1.0 across terrestrial wavelengths.

If the emissivity of the gray body remains constant during the temperature increase, the % increase in the gray body radiation is the same as for a black body.

If the atmosphere is radiating 202W / m ^ 2 towards the surface and the surface has an emissivity of near 1 then that is all absorbed.

luke; Nasif is right about the reduced emissivity of H2O in the overlapping spectrum when CO2 is added; that shreds the positive feedback basis of AGW.

The Sun has not an emissivity of 1.0, but 0.9875, as I have been saying along this thread.

The only way anyone can get a temperature of 24C (297.34 K) from a flux of 443.17 W / m ^ 2 is to use an emissivity of 1.

Luke, Could that possibly be because the measuring instruments and resultant computer modelling are «calibrated» or «assuming» that the emissivity of CO2 is 1?

This sphere can, for the purposes of this demonstration, be approximated by a Grey Body with emissivity of 0.7 for solar wavelengths (λ 4μm) and an emissivity of 1.0 for terrestrial wavelengths (λ > 4μm)-- REGARDLESS of whether these numbers are correct.

If an emissivity of about 0.87 is assigned to that Earth, the radiative - equilibrium temperature increases to about 288.7 K. Each of these base temperatures gives a different value for the zero - feedback sensitivity; 0.75 and 0.78, respectively.

The emissivity of the system of blanket + person is significantly lower (0.1) than the system of just the person (near 1.0).

The 5.4 C result (line 20) is for an earth with uniform emissivity of 0.7 across both spectrums.

This sphere can, for the purposes of this demonstration, be approximated by a Grey Body with emissivity of 0.7 for solar wavelengths (λ 4μm)-- REGARDLESS of whether these numbers are correct.

But to be an apples to apples comparison keep the parameters the same: Solar power 3.8415 * 10 ^ 26W (Which is the same as saying the sun is at 5778K and emissivity 1) Earth emissivity of 0.7 The reason in keeping the parameters the same is to see clearly the difference in the math.

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

The «flaw» was that the emissivity of the earth was pegged at 0.7 across all wavelengths.

If we were to use the 329W / m ^ 2 and calculate the temperature of a body emitting that much we would be at 276K via the Stefan - Boltzmann equation (using emissivity of 1).

To set the lower bound assume the other 30 % of the earth has an emissivity of 0.0 the total emissivity of te earth would be 0.95 * 0.7 +0.0 * 0.3 = 0.665.

No one accurately knows emissivity of the sun and its actual surface area (the present area is based on a visual projection).

Across the wavelengths where the earth is absorbing solar radiation (0.2μm - 4μm) the earth has approximately an emissivity of 0.7.

The earth is about 70 % oceans, the emissivity of water (measured) is 0.95 - 0.98.

The emissivity of the Sun is not one, but 0.9875.

The latter is coincident with its total emissivity of almos zero, as experiments have demonstrated (Hottel, Leckner, etc., etc.).

The emissivity of water is > 0.99, so if it is dead calm the surface can lose a lot of heat in a single night.

The emissivity of CO2 and H2O and other GHGs is NOT always close to zero, but instead there are broad bands where the emissivity is not close to 1 (ie where these molecules can and do absorb and emit IR effectively given sufficient path length thru the molecules).

«The new definition of effective emissivity of non-isothermal rough surface and its approximate expression for continuous canopy vegetation» Chen IEEE explore

(Their calculation used: Absorptivity of 0.69, Emissivity of 1.0, Incident flux at earth of 1368W / m ^ 2 and absorption as the product of Absorptivity and Incident.)

How do we know that the radiation absorption and emissivity of the oceans is not changing over multi-decadal scales which would affect the heat content of the oceans?

It also gives you 238K when calculating the temperature change of the earth when you use it with your figure of 181.64 W / m ^ 2 if you assume emissivity of 1 (tarmac, for example, would be pretty close)-RRB-.

However, I give you a clue: find Hottel's and Leckner's charts on the total emissivity of the carbon dioxide; look for the total emissivity of the latter at 290 K, a partial pressure of 0.00034 m atm, and a total pressure of 1 m atm.

As the emissivity of the atmosphere reduces then the equation won't stay exactly proportional to the 4th power of temperature.

The water vapor has a total emissivity of 0.5, while the carbon dioxide has a total emissivity of 0.0017; therefore, the warming of the atmosphere is due to its content of water vapor.

It depends on the emissivity of the atmosphere.

It is the fundamental formula for calculating the Total Emissivity of any absorbent gas.

Additionally, total emissivity doesn't apply to earth emissions since most of the emissivity of CO2 resides in just these wavelengths, the fact that a CO2 molecule can not absorb visible light has no effect when there is no visible light photons present: every single such photon in the flux is absorbed.

The outgoing radiation 240 W / m2 can also be explained as coming from a surface at 288K, with an emissivity of 0,62.

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.

Using mean beam lengths and CO2 partial pressures, I use PcL charts by Hoyt C hottel to obtain the emissivity of CO2 in the atmosphere.

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».

Indeed, it is physically impossible from usual known radiative processes for any opaque body to have a surface albedo (reflectivity) as high as 0.30 yet have an emissivity of 1.

So the conclusion (furthermore the equation has a term for the emissivity of the object and no term for the emissivity of the surroundings or the absorptiveness of the object, both of which would be required if the surroundings were transferring energy to the object.)

If the atmosphere contained no IR - absorbing substances, then all the IR emitted by the earth's surface would escape into space and radiative balance would dictate that the earth's average surface temperature (or really the average of emissivity * T ^ 4 where T is the absolute temperature and the emissivity of most terrestrial materials in the wavelength range of interest is very close to 1) is set by the condition that the earth must radiate as much energy as it absorbs from the sun.

The Stefan - Boltzmann Law requires heat flux, the Stefan - Boltzmann constant, and the emissivity of the surface (or effective emissivity of a gas).