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The question operates under a false premise. It is not the case that fitting a blackbody spectrum to the Sun gives you its "surface temperature". The Sun does not have a black body spectrum, although sometimes that approximation is made. A star also does not have a single "surface temperature". What you can do is divide the luminosity of the Sun by $4\pi ... 4 There are some issues with the experimental setup you proposed (apart from the fact that when its temperature is lowered the gas would become a liquid and then a solid - if it's not$^4$He: in that case it will stay a liquid). Let's see why. In the picture above, I've sketched your experimental setup. The black box must be impermeable to matter in order ... 4 It won't work because your perfect vacuum is permeated by the cosmic background radiation, which itself is only asymptotically reducing to zero with the expansion of the universe. Trying to exclude the cosmic background radiation backs you into the infinite steps that forms the basis of the third law again. Also, using a container results in quantum ... 10 You seem to make the implicit assumption that your vessel is placed in an environment that does not emit any thermal radiation, i.e. is already at 0 K temperature. The temperature of your container will asymptotically decrease to 0 K but will never actually reach it. Assuming black-body radiation, fixed heat capacity$c$, and sufficient thermal ... 0 How does the matter-radiation system, on its own, goes over to the Blackbody distribution? Evolution towards equilibrium (in macroscopic sense) happens when the system matter + radiation is isolated, for example if a piece of matter is inside a cavity that slows down leakage of energy out of the system. For example, a piece of coal in a well reflecting ... 3 If the absorptivity of a medium really was discrete, then there would be no way it could emit blackbody radiation. The defining characteristic of a blackbody is that it absorbs light of all frequencies that are incident upon it (and that it is in thermal equilibrium). There is a close relationship (a direct proportionality) between the Einstein absorption ... 1 As you said, vibrational and rotational transitions are also possible, and I believe that the energy differences involved there are enough to have a quasi-continuous absorption spectrum in most real-life scenarios (of course, in real life you will never have perfect absorption at all wavelengths). In the following picture, you can see two energy wells ... 0 A black-body can absorb all the radiation falling on it(light at all wavelengths) and appears black when cold. When it gets heated it can emit radiation at all wavelengths like a heated piece of metal. The hotter it gets the higher the photon frequency (energy) and so a shorter wavelength. Hotter objects emit more total radiation per unit surface area. This ... 0 The answer lies in your previous question. Use the heat loss function for radiative cooling and plug it into the differential equation I set up for you in the answer. Then integrate. 1 If you assume that the object radiates as a blackbody, then the Stefan-Boltzmann Law tells you that the power radiated by the object will be$P = A \sigma T^4$where$A$is the surface area,$\sigma$is the Stefan-Boltzmann constant, and$T$is the temperature. If your object is not a blackbody, you can parameterize the above equation with an additional ... 2 High pressure gases do not produce a true continuum, and they do not become closer to the blackbody case with increasing pressure. The individual emission lines only become broader with pressure and make the emission spectrum appear continuous because the lines start to overlap. The number of molecular transitions does not change with pressure. There is a ... -1 According to wikipedia, it can reach 3500-4000 °C You can increase the temperature by choosing material which is black at visible light range and white at IR range. 0 The spectral radiation rate depends on the material in question and the temperature of the material. You can start to learn about this by Googling "Black Body Radiation" and "Planck" . Your terms "glossy black" and "dull white" are far too vague (in a scientific or engineering sense) to be able to answer. Further, the visual color is not necessarily ... 0 Yes, the amount of radiation (i.e. power) would decrease$N^4$times, since the whole geometry of the system does not change, only the power emitted by the sun. But the temperature of the earth would decrease only$N$times, since the ratio of the temperatures remains equal - again because of the unchanged geometry. The earth radiates a power proportional ... 1 Background The specific intensity or brightness,$I_{\nu}$, is defined as: $$I_{\nu} = \frac{ dE }{ dA \ dt \ d\Omega \ d\nu } \tag{1}$$ where$\nu$is the frequency,$dE$the differential energy,$dA$the differential area,$dt$the differential time,$d\Omega$the differential solid angle, and$d\nu\$ the differential frequency. We can define a net flux ...