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35

Typically, satellites use radiative cooling to maintain thermal equilibrium at a desired temperature. How they do this depends greatly on the specifics of the satellite's orbit around Earth. For instance, sun-synchronous satellites typically always have one side in sunlight and one side in darkness. These are particularly easy to keep cool because you can ...


21

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...


15

The reason is the same as why a metal pipe feels colder than wooden plank at the same temperature: thermal conduction. The heat from your tongue (including the moisture) is absorbed faster than your body can replenish it. This has the effect of freezing your saliva in the tongue's pores to the metal surface (which itself isn't too smooth at small scales). ...


13

The low-entropy initial state of the universe is an open problem without a satisfactory answer. Your question is the first time I've heard the suggestion that the initial state should have been a crystal; you remind me that the quark-gluon plasma, which was the state of the universe while it was too hot for nucleons to be stable, has been shown to be a ...


12

First, you have system with some energy, named $U$ by physicists. You think you have all the information you need to characterize the system but then some guy comes near and says: "Whoa, that's bad, the volume of your system can change." You say: "No problem, we just add here $pV$. Our new energy is $H=U+pV$." "But hey," they say, "your temperature can ...


9

Short answer: Gibbs free energy $G = U + PV - TS$ combines internal energy $U$, pressure $P$, volume $V$, temperature $T$, and entropy $S$ into a single quantity that measures spontaneity. With that, I mean that processes that lower the Gibbs free energy of your system will spontaneously occur, and equilibrium is reached when the Gibbs free energy reaches ...


8

As an example, the International Space Station (ISS) has external thermal radiators. They looks similar to solar panels, but instead of pointing the flat side towards the sun, they point towards empty space. An ammonia loop carries heat from various parts of the space station to the radiators. This is a picture of a radiator: (source) External Active ...


8

You might get an order of magnitude estimate as follows. We make the rough assumption that everything ends up in its vessel as a monoatomic ideal gas - actually it will be a plasma, with a thermal energy per mole of $\frac{3}{2}\,R\,T_{final}$, where $T_{final}$ is the thermodynamic temperature of the plasma. Neglecting heats of vaporisation (we assume ...


7

the two paradigmatic cases that illustrate these two possibilities is a gas, for the first, and a crystal for the second. Paradigms and examples are well and good, but be careful not to assume they are the only possibilities. In particular, black holes have entropy -- a lot of entropy. In fact they saturate the Beckenstein Bound. The entropy of a black ...


7

Now I am left wondering why does the heat become lost as if travels slightly. It is not lost. It is spread more out. If you stand so close to the heat source that you are hit by, say 1/10 of it's radiation (1/10 of all photons sent out hit you), then when standing further away you are maybe only hit by 1/100. The heat radiation sent from the source ...


5

The satellite itself can do with radiative cooling but some instruments on board, e.g., IR sensors, require temperatures as low as than 4 K for which Helium dewars are used. Bolometers require even lower temperatures (in the mK range). A good summary is available here.


5

Sanaris's answer is a great, succinct list of what each term in the free energy expression stands for: I'm going to concentrate on the $T\,S$ term (which you likely find the most mysterious) and hopefully give a little more physical intuition. Let's also think of a chemical or other reaction, so that we can concretely talk about a system changing and thus ...


5

It all boils down to energy and heat capacity. Water has a specific heat of 4.186 J/g degreesC, versus air, which has a specific heat of 1.005 J/g degreesC. To keep a radiator at a temperature designed to heat a room, 70C or more, it would take a multiple amount of air blown through, as not only the specific heat per gram but also the density of air ...


4

As always the answer is a simple thing. You calculated the change in entropy using the definition of entropy \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T} \end{equation} Note that this applies to heat transferred reversibly. More generally we must use Clausius theorem \begin{equation} \mathrm{d}S \ge \frac{\mathrm{d}Q}{T}\end{equation} ...


4

1 You could find a relation between C and T from the assumptions(quarter of an ellipse) and just substitute that in the integral $$ m\int C dT $$ 2 We usually say specific heat capacity of metal is some value at particular temperature and pressure say ( Standard Temperature and Pressure ). When we know that the specific heat changes with temperature ...


4

Here is a link to a study comparing heating water in a microwave to heating water in a conventional oven. Depending on the power of the microwave, the volume of the water, and time it's placed inside, the temperature will vary approximately linearly with time until either the system reaches equilibrium (for low power microwaves and large volumes of water) or ...


4

This is such a complicated question! The worst part is that as heat leaves the chimney, it draws air from the room with it - air that needs to be replaced from outside. This actually makes fires quite good as ventilation systems. Whether a fire heats a building depends in very large part on the degree to which cool air can flow past parts heated by the ...


3

This is an excellent question, but not readily answered, as discussed in Floris's Answer. Here is a stab at how to get an estimate of efficiency. It is the method which I believe is reasonably accurate: the actual values will need to be refined by experimental measurement: I am not too confident of the actual numbers that fall out owing to the ...


3

What I will state is speculative and based on the statistical mechanics derivation of entropy, and just the way I view it and do not consider that there exists a problem. After all thermodynamic theory emerges from the underlying statistical level of atomic and molecular interactions. where p_i is the probabability of microstate i. Setting aside quantum ...


3

The heat simply disperses and the further away from the heat source you get the more space the heat has to disperse into


3

First, I want to say that different people use different notation and I welcome any comments. I also feel as if I am about to enter a minefield. Here the answer is made up with examples of use of $d$, $\partial$ and $\delta$. I would say for $d$ that $dV \over dx$ would be the total derivative in one dimension for $V(x)$ where the potential $V$ is a ...


3

Joule's law, and thermodynamics in general, is a model of the classical world. Here, classical should be interpreted as non-quantum-mechanical. Thermodynamics is the study of large collections of particles and their collective behavior. No microscopic model is assumed, and one tries to extract as many (non-trivial) features as possible based on purely ...


3

I was told a long time ago that the sound is from "twinning" - this is where a metal under large stress experiences a reorientation of grains to relieve stress. However I am not convinced this is the case - typically when the engine parts (catalytic converted being probably the hottest) cools down, it will shrink - and there is some "give" in the mountings ...


3

There are two parts to this question: if the fans consume more power than the pump this depends mostly on the fan and pump, I don't have much of an answer here, but in any case the difference will be rather small relative to the total power consumed by the computer. if electrical components are more efficient at lower temperatures This is true, and ...


3

Yes it does depend on temperature. $PV = nRT$ so $V \propto T$ or volume is proportional to temperature in Kelvin. To go from 0 degrees Centrigrade to 1 degree Centrigrade the temperature in Kelvin changes from 273 K to 274 K (approximately) so the volume changes from $V$ to ${274 \over 273} V$ which is equal to $(1 + {1 \over 273}) V$ - so there is ...


3

The best way to numerically work with continuous phase transitions is to study observables that have a vanishing length dimension (or mass dimension in the language of QFT). Take for example the Binder's cumulant ($\langle m^4\rangle/\langle m^2\rangle^2$ modulo factors of 3 and constants, where $m$ is the order parameter) or the correlation length scaled by ...


3

However, my hands seem to feel more comfortable being exposed to winter air than to snow Air (still air) is a better insulator than snow. In fact the insulating properties of snow are due to the fact is has a lot of trapped air within it. The other component of snow, water, is not noted for being a good insulator. The trouble with using air as an ...


3

The answer depends on many factors, but here are the basic bits of physics that play: The power and wavelength of the laser The reflectivity of the surface (function of wavelength of the laser) The size of the focal spot The thickness of the sheet The thermal conductivity of the sheet The reflectivity of the copper is a particularly important one. If you ...


3

It's given that volumetric heat capacity = c is 3.45 joules per kelvin per cc that is equal that the heat necessary to heat up 1 cc of cooper for 1 K is c, so $\Delta Q = CV \ \Delta T$ where $V = lA$ is a volume.


3

This has everything to do with entropy: when the temperature is higher, the benefit of having more water molecules in the air (giving rise to greater entropy) become energetically more favored. This is why water "dissolves better in air" at higher temperatures. Another way of looking at this (pure statistical thermodynamics): when water is cold, few ...



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