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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 ...

32

Landauer's principle (original paper pdf | doi) expresses a non-zero lower bound on the amount of heat that must be generated by computers. However, this entropy-necessitated heat is dwarfed by the heat generated through ordinary electrical resistance of the circuitry (the same reason light bulbs give off heat).

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 ...

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.

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

Computers manipulate internal stored values "0" and "1" represented as different voltages. Every change 0-to-1 and 1-to-0 involves an electric current I passing through a circuit resistance R, which gives rise to ohmic or "Joule" heating.

5

As the comments to the question have stated, in real gasses ( contrasted to ideal gasses which just bounce around elastically) there exist both elastic and inelastic scatterings controlled by quantum mechanical interactions. Photons are generated leading to what we call Black Body radiation and an isolated gas volume will lose energy according to the ...

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 ...

4

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

Rising bubbles of air in a liquid oftentimes are anything but spherical. These bubbles have haphazard shapes because they are rising and because they are interacting with other nearby bubbles. The combination of drag, turbulence, and mutual interactions prevents those bubbles from taking on a nice, simple spherical shape. Here's a rather non-spherical ...

4

There are a lot of misconceptions here so let's take it one step at a time. The entropy in classical mechanics is called the Gibbs entropy, $$S = - k_B \sum_i p_i \ln p_i,$$ where $p_i$ is the probability of some microstate $i$. This is essentially the same thing as Shannon entropy for physical systems. With this concept one can view knowing ...

4

You would be wise to somehow determine the exact fluid used by the original manufacturer. Consider that each of the floats has a fixed density, and has a temperature marked on its hanging tag. So you need a liquid which will have the correct, different density at each temperature marked on a tag. In short, the liquid you choose must match both the ...

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 ...

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

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

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

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

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 ...

3

It's pretty much as you say. But it sounds as though you're trying to guess. I'd suggest a couple of little sketches and some first principles reasoning; see my drawing below Both the circles are my reversible heat engine. On the left, it is running "forwards" and yielding work $W$ from the nett flow $Q_C$ into the cold reservoir at temperature $T_C$ ...

3

It is not thermodynamics that controls crystal formation at the atomic level, but quantum mechanics. Large crystals, from diamonds to clear ice crystals are a macroscopic manifestation of the underlying quantum dynamical level. The molecules that build up the crystal have such field properties, dipoles and quadrupole and even higher moments that have ...

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.

2

What other variables I should know before it can be calculated? : I only know the values before the collision, for example the initial velocities VA,VB, initial temperatures, surface areas, etc. You need to know the $C_R$ coefficient of restitution, CR is the coefficient of restitution if it is 1 we have an elastic collision, if it is 0 we have a ...

2

It's because normal ice, ice Ih, is less dense than liquid water. Ice Ih forms hexagonal crystals. The bonds in that crystalline structure make the water molecules slightly further apart than they are in the liquid form at the same pressure. That water expands on freezing makes water resist freezing as pressure increases. This in turn makes the fusion point ...

2

For the sake of the explanation I will assume you mean a gas bubble in a liquid*. David Hammen names a few conditions for a bubble to be spherical, in fact you could summarize these all as: for a bubble to be spherical the surface tension has to dominate over other forces (per unit length). If surface tension is indeed dominant than the pressure in the ...

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