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Internal Energy = Heat - Work, or U = Q - W. By definition an adiabatic process has constant heat, so when finding the rate of change of each in a process by taking the time derivative: dU = dQ(0) - dW Q goes to zero, and U = W.


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There's no violation of the second law here. You have a system that is out of thermal equilibrium. That black bodies absorb and radiate is the driving mechanism that tries to move this system toward thermal equilibrium. By way of analogy, suppose you are from a southern clime and take a trip at this time of year to a northern clime. You, as a southerner, ...


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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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Your function $I(\lambda,T)$ is the spectral radiance. It's the intensity per steradian per square meter per unit wavelength. To get the total intensity radiated per steradian per square meter over all wavelengths between $\lambda$ and $\lambda + d\lambda$ you multiply it by $d\lambda$, as you've done in your second equation (strictly speaking you integrate ...


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Resolution of the ellipsoid paradox in thermodynamics: the finite sizes of the black bodies lead to a resolution of the apparent paradox


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I have discovered the source of the discrepancy with help from the researcher (thanks Dr. Lipinski!). It seems that many sources (including Wikipedia, the undergraduate heat transfer texts by Incropera & DeWitt and Lienhard & Lienhard) mention only breifly, or not at all, the dependence of the speed of light in the medium on the refractive index, ...


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In addition to the other answers, "bright light can never hurt your eyes" has to be false -- you up the intensity of the light enough, and the energy density can get arbitrarily high. In principle, it's possible to have light so bright that it collapses to a black hole. Before then, you'll get pair production in the light beam.


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You're question is one of the canonical questions that inevitably led to Quantum-Mechanics. It is true that in Classical Mechanics the electron is rotating -> radiating, but if that were true the system would be losing energy all the time, until the electron collapses into the proton. by the same token a proton would be radiating since it also rotates ...


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Let us try to clear up terminology here: A black body is classically defined as a perfect absorber of radiation. No, it is not. The classical description of the perfect absorber also includes an emitter black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. A ...


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It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ...


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1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body). 2) It will be reflected back and forth, but only a finite amount of time. This is ...


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Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone ...


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It's tempting to think that heat will flow more readily across a cavity from a high emissivity surface to a low emissivity surface than in the opposite direction when the surface temperatures are interchanged, but this is fallacious. The reason is that repeated inter-reflection between the surfaces restores symmetry. Call the radiant flux across the cavity ...


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If we assume that both the aluminum foil and the paint have scalar heat conductivities then their composite will also be scalar, and thus symmetrical. Being both material either poly-crystalline or amorphous this is probably reasonable assumption. @Floris suggested that I expand on this, but not being my area I can only summarize a few ideas from ...


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The second law of thermodynamics forbids materials that conduct better in one (forward) direction than the reverse direction - such a material placed between two containers at thermal equilibrium would drive the temperature away from equilibrium, decreasing the entropy of the whole system and paving the way for a perpetuum mobile... Reflectance and ...



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