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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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The heat that makes a filament lamp glow is derived from electrons bashing into the lattice of atoms in the filament and transferring energy to them. The kinetic energy of the electrons becomes vibrational energy of the lattice, and this is exactly what heat is. However the interaction of electrons with the lattice is also what resistance is, and ...

2

The concept of the storage heater is very common in Britain: they usually include a phase change medium which means that a relatively small mass of material can contain a lot of heat without becoming very hot (the latent heat of fusion provides a "thermal cushion" where the medium can give off a lot of heat at a constant temperature, keeping the rate of ...

2

Water does not, in general, help extinguish a fire. Typical fires, however, can be successfully attacked using water alone, as it can cool the fuel at the base of the fire or generate a vapor barrier between atmospheric oxygen and the hot fuel. Water can accelerate liquid hydrocarbon fires by dispersing fuel. Water can generate explosive gaseous mixtures ...

2

To turn thermal energy into useful work completely one would need a thermal bath at the temperature of absolute zero. This is explicitly forbidden by the third law of thermodynamics. The best one can do is given by the efficiency of the (theoretical) Carnot cycle: http://en.wikipedia.org/wiki/Carnot_cycle. Th efficiency of the Carnot cycle only depends on ...

2

The cold air does flow down, but instead of flowing out of the fridge it is sucked into a channel, and pumped back out at the top of the fridge.

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In thermodynamic equilibrium, the solidification process can be tracked using the phase diagram of water and salt. One example (from wikipedia) is: It is fairly straightforward as a binary phase diagram. Above 0C, adding NaCl to water results in complete dissolution until somewhere above 26.3 wt.%. At that point, trying to stir more in will result in ...

1

Interesting question. I would have thought that if you were aware of the exact number of energy states and the populations thereof, you could apply boltzmann statistics to each of the levels in order to fit an appropriate temperature to the population in each state. This temperature, if comparable amongst the included levels, would therefore require that the ...

1

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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The inherent idea is, from that equation $$I = \exp (\frac{eU}{k_B T})$$ if you plot $(\ln I)$ versus $U$, that would be a straight line (of the form $y=mx$), with a slope $$\alpha = \frac{e}{k_B T}$$ That's all you have to do in the experiment, use least square fitting to find an accurate value of $\alpha$ and then find the Boltzmann constant using the ...

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To answer your second question, yes, condensation typically requires nucleation. Supersaturated vapours like the one you describe are the basis of the cloud chambers that used to be used as particle detectors. The energetic particles passing through the vapor would ionize molecules, and these ions would act as nucleation sites.

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I don't think you need to overthink this so much. Mechanical equilibrium in this context basically means that from a macroscopic point of view, all forces are balanced; this usually also means that the system's parts are at rest, though a system in uniform motion could be considered in mechanical equilibrium, I guess. The point that the authors are trying ...

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UPDATED: I now think my previous answer was wrong, because the set up would be equivalent to the following question: Is a black body sphere inside a black body shell hotter than the shell? Just change the question to add a carefully crafted lens that focuses all the radiation into the sphere (you could make the shell as large as you want), which of course ...

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