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In space, the sun transfers heat via radiation to equipment and astronauts. Although the sun’s peak emission is in the visible region (about 500 nm), you can see that there is also a fair amount IR (infrared) and UV (ultraviolet) emitted as well at the top of the atmosphere. To control the surface temperature of an object that is exposed to IR (heat waves)...

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The formula you want is called Planck's Law. Copying Wikipedia: The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. $$... 38 No, it is not possible to hide a person's heat signature indefinitely. Even with the best suit imaginable, you will eventually either begin leaking the heat, overheating the person, or both. One problem is that there are no perfect thermal insulators. This means that you must either use the best available and keep emissions below some threshold of detection,... 38 Typically, satellites use radiative cooling to maintain thermal equilibrium at a desired temperature. How they do this depends greatly on the specifics of the satellite's orbit around Earth. For instance, sun-synchronous satellites typically always have one side in sunlight and one side in darkness. These are particularly easy to keep cool because you can ... 29 If you heat a metal (or anything else) up to a temperature T then the average energy of any degree of freedom of the metal will be of order kT. At 600ºC this is about 0.075eV, and as you say the energy of visible light is around 2 - 3eV, which is a factor of 30 or so higher. The reason that visible light can be produced is because the thermal energy is ... 27 Is darkness really light? No. But not all light is visible to your eyes. You cannot see infrared light, because it is so "red" that your eye can't pick it up. But warm objects radiate it, which is why infrared cameras see warm objects. As objects get warmer, not only do they radiate more, but the light they radiate moves up the spectrum. First they glow ... 24 Broadly speaking, fire is a fast exothermic oxidation reaction. The flame is composed of hot, glowing gases, much like a metal that is heated sufficiently that it begins to glow. The atoms in the flame are a vapor, which is why it has the characteristic wispy quality we associate with fire, as opposed to the more rigid structure we associate with hot metal. ... 24 It's exactly the point of thermodynamics – and statistical physics – that one doesn't have to know the microscopic origin of similar processes if he is only interested in thermodynamic and/or statistical properties. The black body radiation arises from all conceivable interactions between the electromagnetic field and the "black body" – from the electric ... 23 This is a fantastic question, and a topic I was very confused about when I first took a class on Radiative Processes. The ultimate answer, as hinted at by @LubošMotl, is anything---if you start with a 'white-noise' of radiation (i.e. equal amounts of every frequency), it will equilibrate with the medium/material into a black-body distribution because of its ... 21 If it's an incandescent bulb, it's because the whole operating principle of the bulb is based on getting the filament really hot, hot enough to glow. When you cut off the current, it stops heating the filament, so it cools down fairly rapidly, but there may be enough residual heat for a faint glow lasting a little while afterwards. If it's a CFL bulb or an ... 21 Firstly, 'Fire', according to numerous comments and answers [here][1] is a 'process', in which case, the answer to the question will be 'no', since plasma is a state of matter. It would be unfair to leave it there by blaming the semantics, and given the abundant references to 'flame' region, I am going to assume that that is what the question meant to ask. I ... 20 The color of a surface doesn't reliably indicate the emissivity at non-visible wavelengths. The color in the visible spectrum is more of a side effect than anything. Most thermal radiation around body temperature or room temperature happens in the infrared region, not the visible, and that's not reliably indicated by visible color: The visibly ... 19 A NASA team recently (july 2011) doubled the amount of available data related to the pioneer anomaly, and they saw an exponentially decreasing acceleration with a 27 year half life, consistent with the thermal scenario. More precisely they say : The rationale for an exponential model is based on the possibility that the acceleration may be due to thermal ... 19 Such technology is in its infancy, but it definitely exists. The images below are produced by several companies promoting their thermal/IR camouflage clothes. Obviously the applications are well-suited for the military, so who knows what more the military has developed. This last image is made by a company called Blucher Systems. The link provides much ... 19 In practice, no. In theory, also no. The Universe is filled with photons with an energy distribution corresponding to 2.73 K. Every cm^3 of space holds around 400-500 of them. That means that if you place your "stable body" in an ever-so-isolated box, the box itself will never come below 2.73 K, and neither will the body inside. It will asymptotically go ... 18 I'll stick my neck out and say that the answer to your question is simply "yes." First off, these detailed thermal models are complex and hard to do, so we want confirmation from independent groups. We have that: Rievers and Lämmerzahl, "High precision thermal modeling of complex systems with application to the flyby and Pioneer anomaly," gr-qc/1104.3985 ... 17 You are looking at this incorrectly. Pale skin allows the UV to penetrate more deeply than dark skin (that has the melanin in the dead skin cells). Since dark skin individuals absorb the UV in the dead skin layer, it make no difference if it causes DNA damage. 16 Expanding on Ron's comment:$$I(\nu ,T)d\nu =\frac{2h\nu ^3}{c^2}\frac{d\nu }{e^{\frac{h\nu }{kT}}-1}\nu \to \frac{c}{\lambda },\quad d\nu \to c\frac{d\lambda }{\lambda ^2}I(\lambda ,T)d\lambda =\frac{2h}{c^2}\left(\frac{c}{\lambda }\right)^3\frac{1}{e^{\frac{hc}{\lambda kT}}-1}c\frac{d\lambda }{\lambda ^2}=\frac{2hc^2}{\lambda ^5}\frac{d\lambda }{...

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perhaps the degree of quantization is so small the radiation curves look continuous Yes, this is the reason. The correspondence principle says that quantum mechanics has to become classical in the appropriate limit. One way to obtain an appropriate limit is with large numbers of particles. As you increase the number of particles in a material many-body ...

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The electromagnetic spectrum covers a wide range of values: (source) What we see is visible light (just right of the middle of the page), which is a very small section of the spectrum. If the light source emits brightly in the infrared, our eyes will not be able to see it. As an example, consider our own Milky Way galaxy: (full size, source) The ...

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Let's start by assuming you're in the shade, so you're not receiving any radiation (apart from the cosmic microwave background, which I think we can ignore). The amount of heat per unit area that you radiate is given by Stefan's law: $$J = \varepsilon \sigma T^4 \tag{1}$$ The emissivity of human skin is allegedly 0.98, and the area of skin of an adult ...

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You probably are under the misconception that heat travels only via molecular interactions. (i.e, heat transfer by conduction, which needs a medium of sorts). Heat also transfers by radiation, which the sun is an enormous source of. Electromagnetic radiation does not need a 'medium' to travel through. All types of electromagnetic radiation carry energy, ...

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According to Wikipedia, Pure hydrogen-oxygen flames emit ultraviolet light and with high oxygen mix are nearly invisible to the naked eye, as illustrated by the faint plume of the Space Shuttle Main Engine (The picture they provide is the same or very similar to that in the question). So, maybe Crazy Buddy is right.

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At the radius of the earth, the solar irradiance is approximately $1.412\;\mathrm{kW/m^2}$, giving a total power hitting the foil sheet (assuming normal incidence) of $\sim7.06\;\mathrm{MW}$. The average human in America is around $1.7\;\mathrm{m}$ tall, and somewhere around $0.5\;\mathrm{m}$ wide, making his cross sectional area around $0.85\;\mathrm{m^2}$. ...

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Not all the radiation from the outer shell reaches the inner shell. When you take into account the intensity distribution of radiation from the outer shell (Lambertian distribution, i.e. $\propto\cos\theta$) you will see that the amount of radiation for the inner to the outer shell is the same as in the other direction. No violation of the second law.

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Dark skin absorbs UV better than lighter skin. More specifically, melanin absorbs most of the UV radiation so that your skin cells don't have to.

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This problem can be solved with noise-shaping. Since the shape of the spectrum is known, it can be used as a base for the power spectral density: $$P(f,T)=\frac{ 2 h f^3}{c^2} \frac{1}{e^\frac{h f}{k_\mathrm{B}T} - 1}$$ where $k_\mathrm{B}$ is the Boltzmann constant, $h$ is the Planck constant, and $c$ is the speed of light. This outputs the relative ...

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Your formula is wrong, is the short answer. It works fine as an order of mangnitude approximation, but it's only an approximation. What you need is Wien's displacement law: $\lambda=\frac{b}{T}$ where Wien's displacement constant $b=2.8977729(17)×10^{−3} m K$. Put $T=288K,308K$ into that and you get: $\lambda=9.4-10.0\mu m$, which as you'd expect is at ...

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Worrying about the walls can be misleading. See A blackbody is not a blackbox for an illuminating account of the derivation of the Planck spectrum without enclosing the field in a box. If you cant get the published version, see the arxiv version. EDIT (25 March 2012) Planck's Radiation Law: A Many Body Theory Perspective discusses blackbody radiation ...

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On your figures, it's going to depend on conditions, and also critically on what the referee was wearing; but in general it would certainly put the hapless referee in very dangerous position. It probably wouldn't be quite so dramatic as in the tale. On a typical day, let's say the Sun delivers $750{\rm\;W\;m^{-2}}$ intensity when straight overhead. Scale it ...

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