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47

Moonlight has a spectral peak around 650nm (the sun peaks at around 550nm). Ordinary solar cells will work just fine to convert it into electricity. The power of moonlight is about 500,000 times less than that of sunlight, which for a solar constant of 1000W/m^2 leaves us with about 2mW/m^2. After accounting for optical losses and a typical solar cell ...


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 ...


34

No, it is not possible to hide a person's heat signature indefinitely. Even with the best suit imaginable, you will eventually either begin leaking the heat, overheating the person, or both. One problem is that there are no perfect thermal insulators. This means that you must either use the best available and keep emissions below some threshold of ...


32

Because the liquid would boil away. Boiling is what happens when the partial pressure of a liquid exceeds the ambient pressure. Liquids have higher partial pressure as they get hotter, so we usually associate boiling with high temperature. For example, water needs to be heated to 100°C to boil at 1 atmosphere ambient pressure. However, pressure is ...


20

At least one point in your favour is that the light we receive from the Moon has barely anything to do with its temperature. Instead it is mostly a secondary light source "reflecting" light from the Sun towards us. The second point in your favour (I think) is that the thermodynamic argument seems pretty weak. We are not trying to make Earth as hot as the ...


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 ...


13

If you use complicated routes redefining "focusing" towards generating the temperature, yes. Physicists at CERN's Large Hadron Collider have broken a record by achieving the hottest man-made temperatures ever - 100,000 times hotter than the interior of the Sun. Scientists there collided lead ions to create a searingly hot sub-atomic soup known as ...


13

Such technology is in its infancy, but it definitely exists. The images below are produced by several companies promoting their thermal/IR camouflage clothes. Obviously the applications are well-suited for the military, so who knows what more the military has developed. This last image is made by a company called Blucher Systems. The link provides much ...


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.


7

You could try to develop a material that acts like a fluorescent one, that is, transform infrared radiation into lower energy photons, such as microwave. So you will not glow on infrared but on some other wavelength of your choice. Now, if this is technologically feasible (using nano-engineered materials perhaps), I have no idea!


5

So a while ago I did a little project where I grabbed a "standard solar model" from this paper, which gives me some information that's useful for actually making an estimate. (Unsurprisingly the link given to download the data has changed in the last ten years; I haven't sleuthed to see whether the data is still publicly available.) Only about 1.5% of the ...


5

The thing is, even if you concentrate all the black-body radiation at temperature $T_0$ on a portion of matter $A$, this matter itself will, at least, radiate the same way all the energy when its temperature matches that of the black-body. So even assuming $A$ is a black-body, i.e. absorbs all radiation received and reflects none, it will itself emit energy ...


5

Three things to consider: The temperatur $T$ of the glass, the thermal conductivity $\kappa$ of the glass, and the fact that you are actually comparing conduction from skin-to-glass with convection from skin-to-air, which might not be a fair comparison. To no. 1., remember that your hand is much warmer than the glass. The glass temperature is not ...


4

For this process, we found that the entropy change of the universe was negative. You found nothing of the sort. You instead found that a tiny, tiny fraction of the photons emitted by the Sun hit the Earth, and that a tiny fraction of those photons that do hit the Earth trigger an endothermic reaction. The second law of thermodynamics does not prohibit ...


4

It is not possible because of conservation of etendue. This is based purely on geometry, not really a law of physics in that sense. No guarantees regarding quantum effects etc., but in the realm of ray optics it can't be done. Basically, given any source of light radiating from finite surface to half space, you can never concentrate the entire emitted ...


4

As mentioned in the other answers the etendue theorem rules this out for a system of mirrors and lenses. However, I think it's important to note that simple thermodynamic arguments are insufficient for the reasons given below. I will answer the question using mirrors rather than lenses as it makes the physics clearer. Suppose we have a massive mirrored ...


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 ...


4

It forms a cone because it depends on a shock wave, and the region enclosed by the shock wave appears conical in shape. See, for example, the apparent cones here: They are also visible here: Wikipedia appears to be fairly clear on why vapor cones are related to shock waves. From the introduction to the article about vapor cones: Atmospheric water ...


4

If we want to examine gravitational collapse from a statistical mechanics point of view, we find that there's a tradeoff between the fact that a more spread-out collection of matter has more possible position states, whereas a more concentrated collection has more possible momentum states (because more of the system's potential energy has been converted to ...


3

The definition does not require an "atmosphere". Your "environment" can be vacuum. A liquid can boil in vacuum.


3

To add to Whelp's Answer. Even though the $$\bar{d}Q=T\,dS$$ does not hold in an irreversible process, it still gives us something general. Consider a thermodynamic system linked to a system of reservoirs and which can only exchange heat and nothing else with the reservoirs. Let's call the thermodynamic system an engine for our convenience. Then the engine ...


3

This relation is not true for general processes. For a closed system, the general relation is $\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem). Another way of writing it is $dS=\delta Q/T + dS_{irr}$ where $ dS_{irr}$ is the entropy change due to irreversibilitiy in the closed system ...


3

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.


3

Thermal radiation consists of electromagnetic radiation that was produced by the thermal motion of charged particles in matter. In particular, the thermal radiation surrounding an object in thermodynamic equilibrium with its environment is known as black-body radiation, which has a characteristic spectrum that depends only on the object's temperature. Most ...


3

Ingenious. A and B are small, but they cannot be points.The image of B is magnified at A. Therefore if A and B are the same size, some of the light from B will miss A.


3

Suppose the pressure at the Earth's surface is $P$. Consider an air column of cross-sectional area $A$. The upward force on the column is $F_{\text{up}}=PA$. Denote the weight of the column as $W$. By definition of "weight", the downward force on the column is $F_{\text{down}}=W$. Suppose the pressure is too low, such that ...


2

Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)? That's not exactly true. Deviations ...


2

Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly ...


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

Because what you are doing is a flow process, with mass inflow and no mass outflow, you need to use the thermodynamic equation: $dU_{cv}={m_{in}d}{H}_{in}-{m_{out}d}H_{out}+\delta Q-\delta W_{shaft}$ If you insulate your air cylinder well enough, $\delta Q = 0$. Assuming that your air cylinder does not deform, $\delta W = 0$. Since you are filling your ...



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