# Tag Info

171

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 ...

37

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 ...

36

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 ...

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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 ...

21

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. ...

19

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 ...

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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 ...

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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

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 ...

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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

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 ...

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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.

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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 ...

16

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 ...

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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 ...

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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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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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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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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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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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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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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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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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As an example, the International Space Station (ISS) has external thermal radiators. They looks similar to solar panels, but instead of pointing the flat side towards the sun, they point towards empty space. An ammonia loop carries heat from various parts of the space station to the radiators. This is a picture of a radiator: (source) External Active ...

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Oh, but the edge of the atmopheres of Jupiter and Saturn (and the others) are fuzzy! Look at these Cassini images from a few years ago, at the CICLOPS website: "Adrift at Saturn" (PIA 07667), "Beyond the Limb" (PIA 10426), "Off Saturn's Shoulder" (PIA 09791) There is so much gas, such strong gravity, that it gets thicker and thicker quite rapidly as you ...

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It is due to thermal radiation. Bodies with temperature above absolute zero emits radiation. If frequency of the radiation is in the visual range the body "glows". When the electrons in the atom are excited, for example by being heated, the additional energy pushes the electrons to higher energy orbits. When the electrons fall back down and leave the ...

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You would be unlikely to see green. The problem is that in order to see green, you would need the spectrum of emitted light to peak at green and have relatively little contribution from other frequencies. This, for example, is the reason you do not see green stars (There is a cute Feynman story about this). An explanation of color temperature is given on ...

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You're pretty much right and the principle - that many hot bodies are "nearly" black bodies and therefore the colour of their radiation is related to their temperature through the Planck law (or, more succinctly, the Wien displacement law) - is the basis for the optical pyrometer (see this page on howstuffworks.com). There are some approximations though. Hot ...

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