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Black body radiation is given by Planck's formula (see link for variables) Here is the measured irradiance of the sun and the attempt to fit it with the black body formula: The effective temperature, or black body temperature, of the Sun (5,777 K) is the temperature a black body of the same size must have to yield the same total emissive power. ...


8

Maybe the simplest way to think about this is that the Sun is in approximate thermal equilibrium and would absorb any photon, of any frequency, that is incident upon it. This is essentially the definition of a BB. There are many radiative processes that can absorb (and hence emit) radiation at all frequencies, not just those corresponding to atomic ...


0

Since Cavity 1 has "presumably" higher energy density $p_t(v)$ than cavity 2 then: a) energy would flow from cavity 1 to cavity 2 b)"presumably" this happens, then energy of cavity 1 will drop and cavity 2 energy will rise. c)This would violate second law of thermodynamics, the two cavities are at Thermal Equilibrium, the energy exchange between them is ...


0

There is no much Physics involved. The two graphs below ($\sin(x)$ and $\sin(x^{-1})/x^2$) summarize it. The largest areas are the green and the blue. In the first graph, the areas have equal widths. In the second graph, the areas are reversed in order and the widths are not the same.


7

I'll try to point by point address the question. So we have $E(\omega)$ the energy radiated at a given frequency. Actually, no, we have an energy per unit frequency. It's like a density. For a mass density $\rho$ you have a mass per unit volume, and the density can vary from place to place, so to get the total mass you break up space into a bunch of ...


2

The physics is simple: the wavelengh is not proportional to the frequency, but a "distorted" function of the latter, so the density functions (factors at $d\nu$ and $d\lambda$) are different.


5

I will not be able to feel the heat form the fire as a result of convection as there is no atmosphere True, but no atmosphere also means no oxygen for the fire, so this example is kind'a off. Now my question is will I be able to feel the heat from the fire as a result of radiation? Yes. If the fire was at Earth (in an atmosphere with oxygen) the ...


0

At 300K, radiation from your ball of Nitrogen should be pretty well approximated by a black body spectrum. The power density emitted by a black body is given by the Stefan-Boltzmann law: $$j = \sigma T^4$$ Where $j$ is the power radiated per unit surface area. Therefore, we'll need to know the size of your sphere. Let's keep it general, and say it has a ...


0

The body will lose heat according to the Stefan-Boltman law, that is the heat loss per unit surface area goes as $$\epsilon \sigma T^4$$ where $\epsilon$ is the emissivity (which is a function of wavelength). It will be hard to judge how quickly the sphere will lose heat - but it is absolutely certain that it will lose heat, given that the surroundings are ...



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