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According to a NASA page, the density in the middle of the Sun is about 150 g/cm3. That's about 9 × 1025 protons in a 1cm3 box, or 450 million to a side, and using that spacing for a voltage calculation reveals a typical interaction energy of 65 eV or so. (If you've never seen this unit before, that is the energy used by a 1V battery to move an electron's ...

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As mentioned in the comments, this is an instance of supercooling. When you cool a liquid below its freezing point, the molecules are still moving around quite a lot and any two that stick together are likely to be broken up by a subsequent impact. Liquids freeze better when the molecules have something to latch onto -- either a block of the same ice they ...

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In a given orbital, electron motion has nothing to do with temperature. Atoms do have a variety of electronic states and, at higher temperatures, the higher energy states are more likely to be populated. Temperature, however, is most commonly determined by the translational motion of the nucleus of the atoms. Let $v$ be the speed of a nucleus of an atom ...

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Since state $\sigma$ is not in thermal equilibrium I don't think one can use your definition of "thermodynamical" entropy. In fact, one should instead use Von Neumann entropy, which is a correct measure of statistical (so not quantum!) uncertainty. There is no other "classical" or "thermodynamical" entropy in quantum systems. As you mentioned, for a thermal ...

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You calculation seems valid as a rough approximation, giving the highest reasonable number. The actual number will be somewhat smaller, because you assumed that all the energy used in converted into heat and all the heat goes into the juice. Some points to see why the actual number should be smaller (but actual approxiations for these are hard to do!): ...

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since P is a constant and can be taken outside of the integral There is no reason whatsoever why $p$ should be a constant, unless specified so; in particular, in your exercise the task is to find a solution for isothermal transformations. For gases and fluids $p$ is a function of the volume and other variables as well, therefore the equation becomes $$... 2 The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal. The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average ... 2 In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ... 1 You may have seen the reasoning to follow in most textbooks already but apparently it is not emphasized enough so I will say it again here. The crucial starting point is the second law of thermodynamics that claims that the entropy change of the universe \Delta S_{univ} is either zero or strictly positive for any physical change that occurs in it. I ... 1 Summary: Based on provided data you will be able to comfortably bring the water to boiling point in 1 hour. The energy input required to heat the water alone is significantly less than the available energy and the difference is greater than typical thermal losses. ie Water_mass x delta_temperature x Water_specific_heat < energy input Details below. ... 1 That is the heat equation in polar coordinates with axial symmetry. The (isotropic) heat equation without sources or sinks is$$ \frac{\partial U}{\partial t} - K\nabla^2U =0. If you look up the Laplacian operator in cylindrical coordinates, you will find that your expression matches this exactly. 1 Checking for electron degeneracy is a matter of comparing the Fermi kinetic energy with kT. If E_F/kT \gg 1, then you may assume the electrons are degenerate. The central density of the Sun is around \rho=1.6\times 10^5 kg/m^3 and the number of atomic mass units per electron is around \mu_e =1.5. The number density of electrons is therefore ... 1 Rubber consists of many long-chain polymers. In an unstressed sample, these are randomly arranged. As a mental model think of them as anchor chains, where the angle between each link is entirely random - the overall polymer is in essence a random walk through the medium. Now, you pull on it. The net result is to better align the backbone of the polymer ... 1 As stated in the comments, any loss in the system will contribute to the Johnson noise, so you are right about the skin effect and the Eddy current. I want to add that, interestingly enough, this apply not only to electric circuits, but to other linear dissipative systems. A very interesting paper from 1951, Irreversibility and Generalized Noise, proves it ... 1 The Title is not matching the description. Answer as per title will be : Yes one can do this but it is usually done the other way. In most of the books first the equation PV=mRT (R is gas constant whose value depend on nature of gas) is established then they use avogadros principle to state it in molar form as PV=(nM)RT ; V/n=MRT/P ... 1 Chemical vapor deposition is used to produce diamonds at low pressure https://en.wikipedia.org/wiki/Synthetic_diamond#Chemical_vapor_deposition 1 According to the phase diagram of diamond (see for example http://files.umwblogs.org/blogs.dir/6093/files/2011/10/carbon_phase_diagram2.jpg) there is a region where diamond is stable and graphite is metastable, at pressures and temperatures in the range you are asking about: Given the slope on the diagram, you would actually expect that this reaction is ... 1 In mechanics, a mass m experiences a force \textbf{F} along some path C. The work done on the mass is given by W = \int_C \textbf{F} \cdot d\textbf{r}, such that the energy of the mass increases by $W$. Positive work corresponds to energy being added to the system in question (which is inevitably taken from the surroundings). Edit: To answer ...

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Statistical Equilibrium: That state of a closed statistical system in which the average values of all the physical quantities characterizing the state are independent of time. In other words when systems do not evolve in time i.e. when they are in steady state. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either ...

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This phenomenon is due to the presence of air bubbles in the water. First note that the solubility of air in water decreases as temperature increases. Therefore when water is heated, less air can be dissolved in the water and when the water leaves the highly pressurized pipes, the air within the water is able to form bubbles and escape into the surrounding ...

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