# Tag Info

140

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 ...

19

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. ...

16

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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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 ...

14

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 ...

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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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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 ...

11

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.

10

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 ...

10

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 ...

9

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 ...

8

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 ...

8

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 ... 8 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 ... 8 If you can measure the input energy (Sun's spectrum) and compare it to the reflected energy ~30% (Earth's reflected spectrum) you can compute the absorbed energy ~70% (planet's absorption). Isolating that absorbed energy to just the surface requires some approximations and a lot of instrument calibration. The sun (in space) has a spectrum of a black body ... 8 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 ... 7 The thermal radiation associated with some object is typically described in terms of the "black-body" spectrum for a given temperature, given by the Planck formula. This formula is based on an idealization of an object that absorbs all frequencies of radiation equally, but it works fairly well provided that the object whose thermal spectrum you're interested ... 7 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 transparent ... 7 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 ... 7 If the majority of the radiation emitted by a star is infrared, the majority of the visible light emitted will be red. We don't see the lower half of the spectrum the star emmits If the majority of the radiation emitted by a star is infrared but some is visible, the average visible light emitted will be yellow. Little blue is emmitted and any green ... 6 Blackbody radiation is an idealized description of thermal radiation of a substance which is in thermal equilibrium with the photon field. Your description in the question, which equates thermal radiation and blackbody radiation is therefore not quite accurate. Indeed, blackbody radiation is quite simple compared to thermal radiation in general -- ... 6 You are confusing two very different concepts (but you are certainly not the first) whose biggest connection is that they are both named after Max Planck. First, there is the energy of a single photon:$$ E = h\nu = \frac{hc}{\lambda}.  If you plotted $E$ as a function of $\lambda$, you would indeed get the monotonic relation you seek. This is the Planck ...

6

This is the second time in only a few days that I've cited Luboš Motl's excellent answer to What are the various physical mechanisms for energy transfer to the photon during blackbody emission?. As Luboš points out, the precise microscopic mechanisms of the radiation are unimportant because the statistical properties ensure that it follows Planck's law. To ...

6

As I understand, you are asking for Planck's motivation for his law and thus quantization of energy. Well, there is a letter written in 1931 by Planck to Robert Williams Wood. I haven't found it anywhere on the web, so I'll quote part of it here. I'm taking it from Theoretical Concepts in Physics by Malcolm S. Longair (as Longair, I also find it rather ...

6

Generally speaking solids absorb light by converting the EM radiation to lattice vibrations (i.e. heat). The incident light causes electrons in the solid to oscillate, but if there is no way for electrons to dissipate the energy then electrons will simply reradiate the light and the light is reflected. In metals the transfer of energy from oscillations of ...

5

Your second equation, $P(\nu,T) = \frac{2 h {\nu}^3}{c^2}$ $\frac{1}{\exp\bigl(\frac{h \nu}{kT}\bigr) - 1}$ is what is commonly referred to as Planck's law for radiation, although a more standard symbol used is $B_\nu(T)$. This is the energy radiated per time, per area, per frequency interval, per steradian. It is a formula for the 'specific intensity' of a ...

5

Assuming continuous emission generated results at odds with experiment (the UV catastrophe): that is the theory was false to fact. Those are the circumstances under which theorists start trying all kinds of crazy stuff just to see if it will work, and sometimes the justification comes after the math.

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