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41

To sustain a fire, you need three factors: fuel, oxygen, and heat. Take away one of the three and the fire goes out. Water removes heat. Most of this "removing heat" is the evaporation - roughly 540 calories / gram, so 7x more heat than is needed to get water from 20°C to boiling (with a tip of the hat to @Jasper for pointing out erroneous value in earlier ...


14

To sustain fire, it is true that you need the tri-factor of oxygen,fuel, and heat. However extinguishing fire through the use of water, is different than one would think. Indeed, water "sucks" energy in order to change its phase, and thus reduces the heat factor, but the real crux lies in the water expansion properties. Water is heavier than hot air, and ...


11

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.


4

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

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

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

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.


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, ...


1

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 ...


1

The Wien displacement law gives the maximum of a function, so the way to compute it is to start with the Planck function in frequency domain, $$ B(\nu,T)=\frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT}-1} $$ Take the derivative with respect to $\nu$, set it equal to zero and solve for $\nu$. You'll likely have to use some numerical methods (e.g., iterative ...


1

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.


1

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 ...


1

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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