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A star with a temperature of 50.000 °K would have to be much larger than our sun to sustain that rate of fusion, and our Earth as we know it would not exist. It would have to be much further out and it would be unlikely to form a similar atmosphere. There would probably be more ozone as a direct result of more UV to break stuff, though all emission at UVB ...

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The best representative of a black body curve is the cosmic microwave background radiation. Graph of cosmic microwave background spectrum measured by the FIRAS instrument on the COBE, the most precisely measured black body spectrum in nature. The error bars are too small to be seen even in an enlarged image, and it is impossible to distinguish the ...

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Is the Sun absorbing energy from it's surroundings? No, of course not in a net sense. The Sun loses far more energy than it absorbs from its surroundings. It is not in thermal equilibrium. The Sun is also not a blackbody at a single temperature, even though it most definitely absorbs nearly all radiation that is incident upon it. That is because the Sun is ...

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There is no direct relation between melting point and colour of light produced, it's just that some heat energy is used in breaking intermolecular forces and a part of it is transferred to atoms. So for a higher melting point, a bigger part of energy is used in breaking intermolecular attraction and to change its state of matter. The rest is explained by ...

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Just take the volume as your system and integrate the Plank distribution over the wavelengths/frequencies you are interested in (tip: don't use the wave-length but the angular frequency version, it is more concise). This works, because the Planck law describes an equilibrium setting, whether your volume is in equilibrium with the walls or the surrounding ...

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It is true that the most contribution to heat comes from compressing the air. The temperature of a falling meteor was in fact in my aerodynamics II exam where I had to predict its temperature using shockwaves. According to my estimation it was about 10,000 K. You need a proper understanding of compressible air flows in order to answer this question. And I ...

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Remembering that entanglement is a property of the global quantum state, the statement "one of the entangled objects is at some finite temperature" makes no sense without some additional information. Either the local (marginal) states of $A_1$ and $A_2$ are thermal, or the global state is thermal, but not both. If the two objects are called $A$ and $B$, ...

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I guess wearing dark coloured clothes just makes us feel warmer. They actually have nothing to do with the body temperature of a person. They retain our temperature but do not increase it, do they? So "dark clothes for winter" might not be relevant after all. Often by wearing light clothes and by doing heavy exercise, one can increase one's body temperature ...

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If you are cooling your object that you wish to hear, then the exact sound will depend on the exact temperature (as given by yuki96's answer at 17nK). However, any temperature above the nanoKelvin temperature scale will sound the same, but the volume will increase with temperature (according to the Stefan-Boltzmann law). The sound of a warm blackbody (such ...

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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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OK, I asked my Physics TA today. The best explanation from him, up to now, is that this method is nowhere an exact solution, but an approximation. Generally, when the black body is heated up enough, there are many, many different oscillator modes happening at the same time, so the m in the m-space is very large, and thus approximates a continuous space. ...

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A hot object will glow like a light bulb, even in the absence of an electric current. As such, I think that the glowing is a "heat effect".

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I don't think either is right. You can find by integrating over solid angle that the spectral power per unit area from the surface is $\pi I_{\nu}$ in Watts per square metre per Hertz. The luminosity spectrum (assuming isotropy) is this times $4\pi R^2$. The flux at Earth is this divided by $4\pi d^2$ and then you integrate over whatever spectral range you ...

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