# Tag Info

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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 ...

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

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 ...

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 ...

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 ...

2

Ever heard of the cosmic microwave background? The CMB is a relic from when the universe became "opaque" - when, as Wikipedia says, protons and electrons combined to form neutral atoms. These atoms could no longer absorb the thermal radiation, and so the universe became transparent instead of being an opaque fog. So photons decoupled and the CMB was ...

2

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!

2

Consider a satellite in orbit about the Earth and moving at some velocity $v$. The orbital velocity is related to the distance from the centre of the Earth, $r$, by: $$v = \sqrt{\frac{GM}{r}}$$ If we take energy away from the satellite then it descends into a lower orbit, so $r$ decreases and therefore it's orbital velocity $v$ increases. Likewise if we ...

2

What makes water boil/evaporate is the thermodynamic concept derived from the first and second law of thermodynamics. You can read this article to find out the derivation from entropy to the Clapyeron equation. http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation#Derivation_from_state_postulate \frac{\mathrm{d} P}{\mathrm{d} T} = \frac ...

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

1

Heat is simply released to the surroundings during freezing. Freezing occurs because at a temperature below the freezing point, solid has a lower Gibbs free energy than liquid. The plots of G vs T is mathematically derived. At the freezing point, $G_{solid} = G_{liquid}$ There is no change in Gibbs free energy in a reversible freezing process. Therefore, ...

1

based on casual observation, believes that coffee will cool faster than ordinary hot tap water Let's revisit the "casual observation" part. Are the initial conditions the same? Water from a kettle will usually be hotter than from a coffee machine. And we tend to do things to coffee that we don't do to water, like add 10% of another liquid (milk) at low ...

1

For stars (which have huge amount of mass and density), gravity is taken to be responsible for the heat increase. because heat and volume (thus density) thus gravitation of a (massive) star, are related. This is exactly one of the factors that make nuclear fusion (in stars) possible. The two effects thermodynamics (and kinetic energy) and gravity are ...

1

I guess this is possible: for example, thermal radiation can be emitted within a very small solid angle, e.g., upwards.

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Fun, So you are asking about the thermal energy content of the sun? If we assume that all the hydrogen is dissociated. (single atoms) Then each atom has three degrees of freedom and carriers 3/2 kT of energy. So count up the number of atoms at each temperature.... That will work until the atoms ionize. Then there will be equal energy in all the ...

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Wikipedia Intensive and extensive properties By contrast, an extensive property is one that is additive for independent, noninteracting subsystems.[1] The property is proportional to the amount of material in the system. For example, both the mass and the volume of a diamond are directly proportional to the amount that is left after cutting it ...

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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 ...

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