# Tag Info

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From a thermodynamical point of view, living beings are able to reduce their entropy by exporting entropy to the external world. This does not contradict the 2nd principle, since living beings are open systems. For this reason, in a thermodynamically homogeneous universe (heat death), no change in the entropy can occur, and consequently no living beings (nor ...

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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 ...

7

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 ...

4

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 ...

4

I would say it looks like nothing. The heat death requires that the whole universe is thermodynamically homogeneous, and that the universe has reached its maximum entropy. This means that every thing becomes a disordered lump of very sparse matter, without anything to see whatsoever. It's as if the universe is in a state akin to the "chaotic nothingness" ...

4

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 ...

3

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 ...

3

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 $$... 3 The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal. The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average ... 2 In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ... 2 I think you have a fundamental misunderstanding of what the heat death really is. Any observer, whether they are a time traveler, observer from another universe, or whatever, would just see a lot of empty space. The first thing to know is that the heat death is not a single event. The universe, after heat death, is dead in the sense that nothing is ... 2 That is the heat equation in polar coordinates with axial symmetry. The (isotropic) heat equation without sources or sinks is$$ \frac{\partial U}{\partial t} - K\nabla^2U =0. $$If you look up the Laplacian operator in cylindrical coordinates, you will find that your expression matches this exactly. 1 The internal energy of a system is directly proportional to its temperature. Formally,$$E_{sys}=\frac{3}{2}RT. $$You could then note that$$PV=nRT=H_{sys}-E_{sys},$$or$$H_{sys}=RT\bigg(\frac{3}{2}+n\bigg)$$or, identically,$$H_{sys}=\frac{3}{2}RT +PV.  Your method should work, however, this is in my opinion a more "elegant" solution.

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Without knowing a lot more about the details of the entire system, we can't say exactly what the difference will be between the AC1+AC2 case and the AC2-only case. However, there are a few things we can work out from simple energy balance. In the following, I'm considering the air as an ideal gas with a fixed specific heat capacity at constant pressure ...

1

If you place water (or other material) in a pressure-tight container, the water will change as heat and pressure cause its molecules to become more or less energetic and the bonds among its molecules to become more or less stable, or begin breaking apart. These changes are summarized in a chart called a phase diagram. Here is a simple phase diagram for ...

1

Your model is correct. The speed of the transfer depends on the difference in temperature and the thermal resistance of the glass and the water, which is constant. Interestingly the speed is not constant. The closer you get to the final temperature the more it slows down. Technically you never really get there exactly. There is no finite time where you can ...

1

Here's a hands-wavy explanation, which might help you're intuition. The alternative to step 4 and 5 would be to vent the steam to the atmosphere. But the steam is still much hotter than the atmosphere, so a lot of energy will be wasted. So you want to recapture that energy, and the logical place to put it is back into the boiler. However, taking it out ...

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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 ...

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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. ...

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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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