New answers tagged temperature
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An increase in pressure is not what causes condensation and rain.
Besides, the formula $pV = \text{const}$ applies to an isolated sample of a fixed amount of gas at a fixed temperature. Those conditions don't hold true for air in the atmosphere.
The real reason it rains is quite complicated, but the gist of it is that upward air currents can carry air with ...
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Generally speaking low and high pressure areas are associated with vertical movement of the air. Air rises in a low pressure area and falls in a high pressure area. In a low pressure area the rising air cools and this is likely to condense water vapour and form clouds, and consequently rain. The opposite is true in a high pressure area, which is why high ...
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Mad props for a cool question. I'm going to justify essentially the converse of the statement because it doesn't make much sense to talk of the temperature of a system that is in a pure state.
Let's assume that we're talking about a quantum system with disrete energy spectrum (with no accumuation points) in thermal equilibrium. Let $\beta = 1/kT$ be the ...
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This kind of exponential decay toward "equilibrium" can be derived when one looks at a Markov process.
In this case, if we call $S_t$ the state of the system at time $t$ and $S_{t+1}$ the state at time $t+1$, one has for the evolution:
$S_{t+1} = T S_t $
where $T$ is called the transition matrix. This implies that $S_t = T^t S_0$. The idea is then to ...
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This form of $dE/d\tau$ is valid only when the system is not too far from equilibrium and linear response assumption is valid. The fact that $dE/d\tau$ depends on the difference $E - E(0)$ alone is a consequence of assuming a linear response.
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Yes, colouring the water could make the pool heat faster, though whether the colour you noticed has an significant effect is debatable.
Swimming pools exchange heat with their environment by conduction through their walls, evaporation at the surface and absorption of sunlight. I have little direct experience of swimming pool thermodynamics, but some ...
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The valve will, if not to big, not let cold air in, but will let warm air out, which will be replenished by cold air from the bottom (I am assuming that the balloon is open at the bottom). If the valve would be very big, then cold air would also flow in through the valve, but I don't know how you could calculate this.
For the flow rate you could use ...
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Assuming the surface of the metal remains smooth, the reflection from it will be specular and the metal will look shiny regardless of the temperature. However the amount of light metals absorb, instead of reflecting, generally increases with increasing temperature because you get more scattering of the conduction electrons by lattice vibrations. So the metal ...
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Actually, this is an assumption of the Landau theory: the simplest field model exhibiting a phase transition is analogous to a $\phi^4$ theory, which has the lagrangian density
$$
{\mathcal L} = \partial_\mu \phi \partial^\mu \phi - \frac{m^2}{2} \phi^2 - \frac{\lambda}{4} \phi^4\,.
$$
For $m^2 > 0$ the potential in the above lagrangian has a single ...
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Nothing - that is the correct definition.
One little caveat is that small systems are usually in contact with a larger system with a temperature that's more easily controlled or measured (the "heat bath"), and $T$ usually stands for the temperature of the heat bath rather than the small system itself. However, this doesn't make a lot of difference in ...
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"Absolute Hot" is a cute phrase, but meaningful only in a limited context, where the concept of negative temperature applies. That is, there's a finite number of energy levels in a finite energy range, and whatever particles/quanta/excitations we're studying stay in that context. For example we may be interested in spins of nuclei in a crystal, and how ...
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I think the best physical references would be at the high end, such as the maximum volume of undistorted sound around 194 dB. There are several other examples of sound pressure level including one on Wikipedia. I don't know enough to know if a thermoacoustic device would meet your requirements, but that's another possibility, anyway.
As an added note, I ...
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You say:
when a body is subjected to 0 k temperature , it becomes rigid
but this isn't true because vibrating systems (usually) posess zero point energy. A good example of this is that Helium remains a liquid even at absolute zero. I would guess you're wondering if the uncertainty principle can be violated at absolute zero, but in fact zero point ...
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As for the question of whether anything can be hotter than the sun. The Sun is composed of plasma, an energetic phase of matter in which electrons get ripped off of atoms, and electrons and ions coexist in something that might best be described as an ionized gas.
According to this wiki page, the so-called Z machine has achieved temperatures on the order of ...
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It depends on many factors such as the reentry velocity of the object, its shape (cone-spherical, etc.), what the planet's atmosphere is made of, whether it enters at some shallow angle and also the altitude where there's density variations in atmosphere, etc.
Googling on this, could return you a lot of results. And, all results matched a certain value. ...
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The wiki article has a lucid calculation of what you are asking.
Notice that this is the incident power, and somewhwere the book must say per meter square on the target.
Incident and absorbed are two different concepts.
The incident/incoming radiation is computed at the location of the earth, so it depends only on the parameters of the sun and the ...
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I apologize "basics foundations of thermodynamics" still does not make a lot of sense to me.
Steve B already provided some answer associated to one way to interprete the word "foundation" that is from statistical mechanics. I will kinda play here devil's advocate and assume that you are refering to axiomatic thermodynamics.
As far as I am concerned, the ...
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Yes, it can be proved. But the mathematical rigour only applies to ideal gas, so you might feel disillusioned after all the calculations, considering how much this can be valid in reality.
If you insist you want to know, then you may read on. From the first law of thermodynamics,
$$\delta Q = \frac{3}2 nRdT + \frac{nRT}V dV,$$
where internal energy ...
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There's a group I call "thermodynamic purists" who think that thermodynamics is a self-contained system based on semi-mathematical "axioms". I disagree! I think that thermodynamics is fundamentally a consequence of statistical mechanics, and that this is the best way to think about it and understand it. I acknowledge that reasonable people can differ on ...
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Temperature is the measurement of kinetic energy per unit particle mass. Since you've added the same amount of heat energy to each object, the finite object will have a higher temperature because its heat energy is distributed across a smaller collection of mass. Taking something's temperature is indeed a meaningful measurement ;)
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You're thinking of the magnetocalorific effect. When you put a paramagnetic material into a magnetic field the field aligns the magnetic dipoles within your material. This releases energy and heats up the material slightly.
If you let the material coll back down to ambient temperature then remove it from the magnetic field the dipoles return to a random ...
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A maybe more mathematical awnser: You can define temperature as a scalar field (e.g. on earth). So given a certain position on the surface of the earth (or in three dimensions if you wish, it does not change anything) you have a scalar, the temperature on this position. Now you can take the gradient of this field, and now you have a vector.
More directly on ...
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Temperature gradient is actually an object called a one-form. A temperature gradient does not have a direction. Instead you combine it with a vector to get a scalar (the temperature change). It's the vector that gives the direction.
To take a simple 1-D example, suppose we have a temperature that varies along the $x$ axis as:
$$ T = 298 + x $$
so at $x = ...
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