thermal-radiation
I'm Having trouble replicating the Black Body model for sun shown on this plot
To my understanding I should only use Planck's formula:
$$I(\lambda) = \frac{2\pi hc^2}{\lambda^5}\dfrac{1}{exp\left(\frac{hc}{k_bT\lambda}\right)}\cdot 10^{-9}$$
Being $\lambda$ in [m] and having the last factor, $10^{-9}$, added for the units presented. What i get is
Can anyone see why?
Ok, so here is my answer to my own question:
Of the three options I presented in the question, the answer is 1: the emission of radiation into space is actually an irreversible process. At first I couldn't see how this could be the case, because the transfer of energy from the hot body to the outgoing radiation field doesn't involve a change in temperature, so it seemed like it should be reversible. What I had failed to consider is that the body interacts not only with the outgoing radiation field, but also with the incoming one (i.e. the cosmic microwave background in the case of a star.)
A useful metaphor is a heat exchanger, in which a pipe enters carrying cold water (temperature $T_C$), which is placed in contact with a body at a higher temperature $T_H$ until it equilibriates. Another pipe carries the warm water out.
Although the outgoing fluid is at the same temperature as the solid body, it is clear that this is an irreversible process. The entropy is produced not in the transporting away of warm water at $T_H$ but in the heating of water from $T_C$ to $T_H$.
A star may be seen as an analogous kind of "cosmic heat exchanger". In this case we have to imagine a small volume of space containing a small amount of cosmic microwave background energy at $3\:\mathrm{K}$ coming into contact with the star and being heated up to $6000\:\mathrm{K}$, and then moving away again.
These volumes of space should be thought of as moving at the speed of light. As with the heat exchanger, the transporting of the $6000\:\mathrm{K}$ radiation away from the star is a reversible process, but the "heating up of space" from $3\:\mathrm{K}$ to $6000\:\mathrm{K}$ is irreversible and produces entropy.
In my question I asked what would happen if this radiation were absorbed by a colder body, since it seems as if the entropy can decrease. For example, if $U$ Joules of thermal radiation at $4000\:\mathrm{K}$ (with entropy $\frac{4}{3}\cdot\frac{U}{4000} = \frac{U}{3000}\:\mathrm{JK^{-1}}$) were absorbed by a body at $3500\:\mathrm{K}$ then the body's entropy would increase by only $\frac{U}{3500}\:\mathrm{JK^{-1}}$ and it seems like the total entropy must have decreased. But if the second body can absorb all the radiation then it must be a black body, and hence it must emit black body radiation of its own according to the Stefan-Boltzmann law at a rate $A\sigma T^4$ (with A its surface area and $\sigma$ the Stefan-Boltzmann constant). It turns out that when you take the entropy of this outgoing radiation into account, the total entropy production is always positive. (Unless the absorbing/emitting body is at the same temperature as the radiation field, in which case it is zero, as we should expect in the case of thermal equilibrium.) This is like a heat exchanger operating the other way, with warm water coming into contact with a cold body, and cold water running out. If you forgot to take into account the entropy of the cold water it might seem as if the total entropy was decreasing.
Finally I should say why my argument about the outgoing radiation not doing work against a piston doesn't work. Imagine the following thought experiment. First we reversibly fill a piston with thermal radiation as described in the question. Then we make a small hole in the cylinder and let that radiation escape into space. This escaping radiation is exactly the same as black body radiation. The entropy of the radiation in the cylinder cannot decrease as it goes through the hole, and hence its entropy flux must be at least $\frac{4}{3}\frac{u}{T}$.
Interesting and complicated question. The things to consider:
"Black body radiation" assumes perfect absorption / radiation at all wavelengths. The greenhouse effect comes about from having absorption in the IR: the hot (short wavelength) radiation from the sun can penetrate the atmosphere, but the cooler earth radiates at a lower temperature - longer wavelength. And that longer wavelength light is reflected by the atmosphere (water, carbon dioxide, methane, etc). Think of the story of Winnie the Pooh visiting Rabbit's burrow. He goes in through the hole, has a "little smackerel" of honey (read: the whole pot), and then is too fat to get out again - commemorated on a postage stamp:
That's your photon. It had no difficulty penetrating the atmosphere as a short wavelength photon - but as a long wavelength photon it gets stuck as it tries to leave earth...
If you worry about your spaceship getting too cold (how big is it?) you should probably consider lowering its reflectivity - this directly scales with the heat loss. Notice how space ships are often "shiny metal". This isn't just because paint is expensive to lift into orbit (it is), but also to keep the emitted power down - keep the people inside protected from too much heat loss when not in the sun, and too much heat gain when they are. If you want to simulate the "Venus" effect you would want to create your own greenhouse effect - add a film that is transparent in the visible and opaque in the near IR.
Either way, your black body model needs to take account of the reflectivity as a function of wavelength - and instead of using the simple Stefan-Boltzmann law (which deals with total power per unit area), use the wavelength formulation (Planck law):
$$S_\lambda=\frac{2\pi hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$$
But yes - the amount of heat that a large object loses through radiation is substantial, even when it is at room temperature. There's a handy calculation on wolframalpha.com - it shows that the heat loss for an emissivity of 0.1 is still over $40 W/m^2$ at 298 K. The best thing you can do to insulate yourself is not let the outer shell get so hot in the first place - if you used a double shell, with the outer being thermally insulated from the inner, then you can see how this will lower the power emitted at the outer shell comes to equilibrium at some temperature $T_o$.
Assuming that the outer shell reflects half its power back to the inner shell, and half to the universe (which is so close to absolute zero that we ignore the difference), you can write
$$\epsilon \sigma T_i^4 = 2 \epsilon \sigma T_o^4$$
since the outer shell loses heat from both surfaces; thus if the inner shell is at $298 K$ the outer shell temperature will be at 250 K, but the inner shield is now losing heat at
$$\epsilon \sigma (T_i^4-T_o^4) = \frac12 \epsilon \sigma T_i^2$$
In other words - you halved it. If you add additional skins, the heat loss will be further reduced.
I must admit that I did that last bit of analysis "by the seat of my pants". It makes sense, intuitively, that the heat loss is reduced by a radiation shield; I have never attempted to come up with a number before, nor do I remember seeing this analysis. There could be a blooper in here - in which case I would be happy to have someone point it out.
I did find an online book that seemed to follow a similar approach but had a cylindrical geometry and uses different reflectivity on the inner and outer faces, which complicated matter further. But they show that multiple layers of shielding can significantly reduce these heat loads - which was really what I tried to say.
Best Answer
The way you have expressed the power density or irradiance is not correct. There is a (-1) in the denominator which is subtracted from the exponential term. Look at the following link to correctly express your result: black body radiation curves