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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 ...


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 ...


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!


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.


2

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


2

No, if $S$ is really the only thing you know about your system then there is no way to know its energy. There is no relationship between the energy and the entropy that doesn't involve some other quantity such as temperature. ...but surely you know something about your system, other than its entropy? I mean, you must know something about what it's made of, ...


1

"a circuit that has different resistors at extremely different temperatures" -- Each resistor independently puts out its own noise related to its own temperature. "one long resistive element that has a temperature gradient across the whole thing" -- That's actually the same thing again. Treat it as a large number N of resistors in series, each with ...


1

Actually one can flood the mines on an asteroid. The water will become boiling, but before it all boils out the water will freeze. As such, the scientists could say so only if they anticipated negativity in ice bulbs forming in the mines. But the process of freezing is very slow, so the water most likely would reach the bottom of the mines.


1

The dimensionless Rayleigh number characterizes buoyancy driven convection. When the Rayleigh number is below a critical value, heat transfer is primarily by conduction (e.g. no Benard convection cells). When the Rayleigh number is above the critical value, heat transfer is primarily by convection (e.g. Benard convection cells spontaneously form and ...


1

You couldn't make the heat "disappear". You can however make it go somewhere else. To talk about thermodynamics very loosely Anything (especially a vacuum) between the source and the detector will hide the source, as long as the medium remains at the surrounding temperature. This means that if you have a heat source behind a window, take a thermal imaging ...


1

There are actually two questions here. At phase equilibrium, yes independent water molecules do enter and leave the liquid vapour interface. The rate of crossing of individual molecules from one phase to another is characterized by an Arrhenius type rate equation of the form: $$\alpha \exp \left[ - \frac{\Delta E}{kT} \right]$$ where $\Delta E$ is the ...


1

It depends on your power supply. The power $P$ through a circuit element is always given by $$ P_\text{thing} = I_\text{thing}V_\text{thing} $$ where $I$ is the current through and $V$ the voltage drop across the thing that you're interested in. If your circuit element obeys Ohm's Law $$ V_\text{thing} = I_\text{thing} R_\text{thing} $$ then you can write ...


1

There are some nice ideas in other answers, but they are overseeing some basics. Let's do some thermodynamics. The efficiency of a thermal engine is bounded by the Carnot efficiency: $$\eta \le 1 - \frac{T_c}{T_h} $$ Where $T_c$ is the temperature of the cold end and $T_h$ the heat source. Assuming we are in a cool environment, $T_c=0 C$, $T_h=37 C$, so: ...



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