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The relationship between thermodynamic and information entropy is still a work in progress. However, it is clear that he thermodynamic entropy is a rough measure of disorder, in the sense of not knowing in which specific microstate, all compatible with a given macrostate, the system is. Thus there is at least a loose relationship between thermodynamic ...


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To calculate the amount of heat energy needed to heat a pot with its content of water we can use a simple formula: $$\Delta H=mc_p(T_2-T_1)$$ Where $\Delta H$ is the heat energy needed to heat an object of mass $m$ and specific heat capacity $c_p$ from $T_1$ to $T_2$. For a pot with water these energies would need to be calculated for both separately and ...


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I will give contra example for some of the comments: You can blow with mouth wide open very hard and then the speed of air is high but it is still worm. Also you can blow with lips pursed very gently and the air speed is slow, but still you feel it is colder … so the right answer is : air comes out cold, because it expands coming out of small opening.


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This is an interesting question and one that probably needs detailed simulation to settle. But one can make the following broad prediction: the shape of the meteorite would have minimal effect on the outcome, for the following reasons: At the kinds energies let slip in the moments of impact and the kinds of pressures and temperatures that prevail, all ...


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Real world assumptions: No temperature gradients over the cross-section of the pipe. Plug flow (turbulent flow). Water heat capacity $c_p$ and density $\rho$ are temperature invariant. Consider an infinitesimal mass element $dm$ at temperature $T(x)$ travelling down the pipe at speed $v$. We apply Newton's law of cooling to it: $$\frac{dQ}{dt}=hdA\big(...


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You can wait three or four minutes and let the pressure drop or..... If you feel lucky you can try tapping the can a dozen times, as is shown on this video https://youtu.be/NQYO3Dp8lCA


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From a biochemical point of view, heat detection is achieved by proteins at the surface of nerve cells. They basically just trigger a nerve signal above a given temperature. So they DO detect temperature and not a "heat flux". It may seem surprising that nerve cells react so quickly but the increase/decrease in temperature does not need to go all through the ...


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There are two definitions of entropy, which physicists believe to be the same (modulo the dimensional Boltzman scaling constant) and a postulate of their sameness has so far yielded agreement between what is theoretically foretold and what is experimentally observed. There are theoretical grounds, namely most of the subject of statistical mechanics, for our ...


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What is entropy really? I want to answer(!) this question from a different point of view. First off, I focus on your title and the phrase “really”. We don’t know what entropy is really. We don’t know what energy is really also, and any thing or concept else too. Entropy, like all other concepts created by humans, is a convention between some people to ...


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The definition of an empirical temperature is basically what the Zeroth Law of Thermodynamics does. Let us suppose we do not have any prior knowledge about temperature. What we do know is that if we put two bodies in contact with each other they may change some thermodynamic properties - a volume, for instance - of one another. When such a thing happens we ...


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The hypothetical cubic box which is created effectively by collisions cannot be the solution at least Because sometimes the ideal gas is defined as the limit when there isn't any intermolecular interaction, even elastic collisions. (see for example Sethna, p39) But you can generalize the proof (in kenetic energy language) for the arbitrary shaped container ...


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The temperature appearing the the Clausius inequality is definitely the temperature of the "boundary interface (with the surroundings)", or simply the temperature of the sources. One of the best places I have seen this discussion is in Fermi's book, chapter 5, section 11. He is explicit about it. To see this you have to recapitulate the steps in obtaining ...


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No, it is not possible. Any material has black body emission which is a distribution of wavelengths across the electromagnetic spectrum. All materials have specific absorption frequencies at which they will absorb a photon. This will result in a transfer of energy from the hotter material to the colder one.


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No, any material with a finite temperature will release thermal radiation, including mirrors, and it's not possible to make a mirror that's perfectly reflective. Generally speaking, the best you can do is limit an object to only radiative heat exchange (i.e. no conductive or convective heating) which is why nice thermoses are vacuum-insulated.


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According to the three laws of thermodynamics, insulanium cannot make a drink hot or cold forever. I guarantee you will feel the heat and that it will look almost like a mirror at almost all frequencies. James Clerk Maxwell, the eminent Scottish physicist, came up with something called Maxwell's Demon (still being studied) that may debunk the second ...


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I dont think it is possible to make anything 100% reflective although we can get arbitrarily close. The reason may be explained by tunneling by photons. Also,the entropy of an isolated piece of insulanium must increase with time. I think you can see now why such a thing is very very unlikely.


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The 'go to' partial differential equation here is surely the Heat equation (Fourier), here in one dimension: $$\frac{\partial T}{\partial t}=\kappa \frac{\partial^2T}{\partial x^2}+\frac{\dot{Q}(x,t)}{c_p\rho }$$ It can be easily expanded into three dimensions or expressed in polar, spherical or cylindrical coordinates. It's not clear from your question ...


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In thermostatics a phase is a region of space occupied by a homogeneous material characterized by a pair of thermic and caloric equations, i.e., $T=f(V, x_1, x_2, ...) $ and $U=g(V, x_1, x_2, ...) $, resp., where $V, x_1, x_2, ...$ are the various extensive variables describing the mechanical, chemical, electrical, magnetic, etc., configuration. Different ...


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Gold will compress to about half of its volume at atmospheric pressure if you compress it to 2 million atmospheres at room temperature, which is something that I'm sure has been done with diamond anvil cells. For many metals, the atomic lattice will also undergo structural phase transitions from one lattice type to another at certain pressures, but I don't ...


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I would say that gold will remain gold and do not go into nuclear fusion. However I am not saying that it is impossible. You may have heard of neutron stars and black holes, when the matter is under immense pressure the electrons around the nucleus could not withstand it and they fall into the nucleus and convert everything in a giant neutronic mass. ...


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If you consider a very diluted atomic gas with $N$ atoms (say hydrogen atoms), it won't be equivalent to an ideal gas with $3N$ degrees of freedom for all temperature ranges. Instead, at very low temperature but very low density, you will expect these atoms to form molecules e.g. $H_2$ molecules and the internal energy will be $5N k_B T/4$ as there are $5N/2$...


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To know whether a column of fluid is in equilibrium, you take a fluid element anywhere in the column, displace it by a small amount (compared to column height) in any direction, and see what forces come to act on it; if the forces are such that they push (or pull) the fluid element back to its initial position, then the fluid column may be said to be in ...


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Why does a critical point even exist? I think this question is equal to this one: "Why the width of the two phase region is bigger at lower temperatures and pressures?" Specific volume of liquids mostly depends on the temperature of them in comparing with their pressure. This means, for a well-defined increment of the pressure, we can neglect its effect ...


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Attempting to answer the "why" question intuitively: In a liquid, the molecules experience significant intermolecular force - so much so, that the average energy of the molecules is insufficient to escape the attractive force of the surrounding materials. The result is that it energetically favorable for them to remain close together, even if that means ...


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The way you calculate (classically) the heat capacities for gases is by comparing the expressions for the internal energy given by thermodynamics and kinetic theory. The Equipartition Theorem says that at thermal equilibrium at temperature $T$ each quadratic term in the (mechanical) energy of the molecule contributes with $kT/2$. For a gas with $N$ ...


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In the temperature range you are talking about, and assuming we are talking about pressure that are not close to vacuum, then a monotomic gas (and I'd prefer talking about Ar, or He, and not a monotomic O, or H) Cp/Cv=5/3 (billiard ball atoms - no vibration/rotation) Cp-Cv =R (R is gas constant, and there is a universal gas constant) and from these, I ...


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Leaving this here for now... Will update with more information and references later. Einstein and Debye showed that specific heat is a function of temperature, but is asymptotic at high* temperatures. Here is a simple explanation why: Heat, with regard to everyday applications, is simply a measure of the motion of atoms and molecules. Let's start with ...


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Clearly the first law, as stated in Ján Lalinský's answer implies the conservation of energy. Let us consider a system $S$ and a surrounding $\Omega$. Since energy is additive one can writhe the first law to the universe $T=S+\Omega$ as $$\Delta U_T=\Delta U_S+\Delta U_\Omega=Q_S-W_S+Q_\Omega-W_\Omega.$$ But The system can only exchange heat with the ...


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I have no idea what you are asking in (1). (2) During adiabatic compression, the temperature of the system does change. Starting from the first law of thermodynamics: $$\mathrm{d}U = \mathrm{d}Q + \mathrm{d}W$$ For an adiabatic process $\mathrm{d}Q=0$. If we assume an ideal gas, then the internal energy is $\mathrm{d}U = C_V\, \mathrm{d}T$ and the work ...


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I will try to answer these questions from different views. Macroscopic view The "quantitative" rather than qualitative difference in a liquid-gas phase transition is due to the fact that the molecules arrangement does not change so much (there is no qualitative difference) but the value of the compressibility changes a lot (quantitative difference). This ...


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For a pure substance that can exist in the solid, liquid, and vapor states (i.e., wood is not in this category), let's assume that a closed container is half full of liquid and half full of vapor. As the temperature rises, the liquid expands and the liquid density falls. Also, as the temperature rises, the pressure in the container rises due to the vapor ...


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CAVEAT - I am giving a possible calculation, but I believe the answer may be off by a factor 2x (compared to the lapse rate observed in the atmosphere). I am leaving it here for you to ponder. Perhaps it can inspire you to find the correct solution yourself. Or perhaps the difference is due to the fact that this calculation doesn't assume convection - so ...


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Good question. I don't have my Widom around, but I'll try to answer from memory. I think the consensus is to say a substance is at its gas state if it could be a liquid at the same temperature. This, as opposed to same pressure, same volume, etc. If the temperature is supercritical, there is no transition between liquid and gas, and the generic term "fluid"...


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If the speed of ball is large most of the time, then speed of air flow across the ball (as seen from reference frame attached to the ball) is large most of the time, which means that heat transfer from ball to air is predominantly due to forced convection rather than free convection. With this assumption, you may use forced convection relations for heat ...


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The many worlds interpretation has a lot of problems, but this isn't one of them. You're imagining "creating alternate universes" as some energetic event, like a mini Big Bang, but what really happens is a smooth splitting of the wavefunction. For example, suppose we have a spin in a superposition of up and down states, $$\frac{1}{\sqrt{2}} (| \uparrow \...


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To answer your first and second questions, yes. I think where it may be a point of debate to some is the entire concept of "universes" being spawned. Universes may not necessarily be "spawned" so much as already being in existence, essentially a universe for each discrete outcome of any event, past, present, future, each existing per its own space-time. ...


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The equation is a cubic polynomial curve fit. There are many mathematical curve fitting programs that can give this formula, and they merely require that you list four or more points of volume vs. temperature throughout the range of interest. The exact constants that result from the data depend on the units on volume and temperature, so the equation that ...


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As the steam gets created from the evaporated liquid at the bottom of the liquid drop, it expands and provides additional pressure acting on the bottom of the liquid drop. The steam does move upwards, but new steam is continually being created so the pressure is able to hold the drop slightly above the surface.


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I figured it out. If you pull out the summation out front, everything except the $s=n$ term vanishes. The terms with a higher power than $n$ vanish when taking the limit while the terms with a lower power than $n$ vanish when taking the $n$th derivative. However, it would be great if someone can come up with a better, more constructive way of deriving that ...


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There is a misnomer in the original question. There is NO loss of heat when the liquid pressure drops and the liquid boils at a lower temperature and pressure. Energy is conserved, so the high pressure, high temperature liquid, boils into a low pressure low temperature mixture such that the heat content of the low pressure liquid plus the heat content of ...


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To expand on the above answer by @PhysLab (which is very well written), the answer depends on how "in depth" you want to look at it. Before discussing energy transfer, let's cover a few fundamentals, because these are slippery concepts and often misunderstood. (Including by me, so fingers crossed!) Let's look at what "energy" and its "transmission" involves....


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Energy transfer can be thought to occur via the exchange of a 'virtual particle'. In nature, there are 4 fundamental forces, namely: 1. Electromagnetic force 2. Gravitational force 3. Strong force 4. Weak force Each of these forces have a different exchange particle: For instance, the exchange particle for EM is a photon whereas that for the strong force ...


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You are mistaken. You seem to be assuming that there is some kind of inertia in the process of heat transfer, as in water sloshing about in a tank. There is no such inertia here, so there is no oscillation. You write: Since heat has been transferred from A to B, unless I'm mistaken this will place B momentarily at a higher temperature than A. Yes, you ...


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In "Adventures in Friedmann cosmology: A detailed expansion of the cosmological Friedmann equations" by Robert J. Nemiroff and Bijunath Patla in the American Journal of Physics volume 76, on page 265 (2008); http://dx.doi.org/10.1119/1.2830536 the authors call them "cosmic strings" But this is in the context of cosmology, so its for a universe that on very ...


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I'm not sure what you mean by the "Lower Point". There is no such label in your diagram. The entropy axis is, I believe, linear, but is not necessarily bounded by 0 entropy at the lower horizontal edge. In other words, you should consider that the curve is one depicting that the system spends most of its time at maximum entropy (the maximum entropy value) ...


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For reference, the diagram is below. First, I'm just going to give a quick explanation of entropy. Before Boltzmann, people knew about entropy, they just didn't explain it correctly; namely, they thought of it as a measure of the uselessness of arrangements of gas. As an example, they thought that if you had a box, and all of the gas molecules in the box ...


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The probable reason why poeple say ice cream heats up your body is when you compare the amount of heat that it takes away your body compared to the food caloric value of the ice-cream. 100 grams vanilla ice cream is about 207 food Calories ($207 kcal$) according to this site. On the other hand, the amount of heat that the ice cream takes from the body is ...


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Step 1) Find a model for the gas in the room, might as well take ideal gas for demonstration, and I think this is generally an alright model for the air. Step 2) Apply the first law of thermodynamics, ignore the work done in opening the window. The first law now states Energy change is = Entropy change because the room rigid walls. Step 3) The average ...


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Part of the answer is, at low Reynolds number, in streamline flow, there's little if any drag. At higher speeds, though, the wake of a moving object creates vortices (and does so in random fashion, a spontaneous symmetry breaking occurs). Those random vortices cause a velocity-squared retarding force, which is energy-losing whether your motion is ...


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Your title is different from the question you ask in the text. You are correct that the amount of energy needed to raise the temperature of a fixed mass of ideal gas by $1^\circ C$ is independent of the absolute temperature. But this is not the same as calculating the coefficient of performance of a heat engine operating between those temperatures. You ...



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