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13

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 ...


7

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 ...


6

In some sense yes. The temperature is defined as an imaginary time in Matsubara Green's functions or some path integrals. Thus, a negative inverse imaginary temperature can be considered as a time. Here is a quotation from Alexander Altland, Ben Simons "Condensed Matter Field Theory": "Thus, real time dynamics and quantum statistical mechanics can be ...


5

Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac ...


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 ...


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

Assuming this can be treated as a classical ideal gas, by the equipartition theorem $$U=\frac{3}{2}N k_B T$$ and the ideal gas law $$PV=Nk_B T$$ we find that the internal energy is $$U=\frac{3}{2}PV$$ Therefore, the internal energy is multiplied by a factor of 5 in this process. Energy is a state function; this is why we can determine the change in ...


4

It's true. Special equipment and a long time is required to mix helium and nitrogen. According to one study, a mixture of 2.7% He, 93.3% N at 800 p.s.i.g. required a special cradle to repeatedly upend the cylinder, and 20.5 hours to reach equilibrated gas, which then remained mixed: http://pubs.acs.org/doi/abs/10.1021/je60005a002. The helium repeatedly ...


4

If the initial momentum is $mv$, then when the particle bounces off the wall, it is going in the opposite direction, so its new momentum is $-mv$. The difference is $2mv$ because $mv - (-mv) = 2mv$. By the way, this is not really how it works; molecules will speed up or slow down when they hit the wall. If the molecule is moving slowly, it's likely to speed ...


3

If I understand the question correctly, then:- Object 1 is a thermal mass at temperature T1. Object 2 is a piston (partially) filled with fluid, at temperature T2, where T2 < T1. The piston is pressing against object 3. There's an implicit assumption that the meeting point between the piston rod and object 3 is a poor thermal conductor. Object 3 is ...


3

As I said in the comments, you need to define what you mean by "easier". The total heat flux out (or in) to the room is $$P + k(T_{in}-T_{out})$$ Where $T_{in/out}$ are the temperatures of the incoming and outgoing air respectively, $P$ is the heat flux of the heat source, and $k$ is a coefficient that has to do with the heat capacity of air, the flow ...


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 ...


3

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!): ...


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

The answer depends on whether the wheels skid. When you brake with just the rear wheel, it's quite possible to skid; if you apply the front brake, the increase in normal force on that wheel tends to prevent skidding (although in extreme cases it could make you fly over the handlebars). Applying the rear brakes hard enough to block the wheel would generate ...


3

Physical things (solid, liquid, gas, plasma) both absorb and emit energy in the form of electromagnetic radiation of a wide range of frequencies. How fast they radiate and the strongest frequencies of radiation depend on the absolute temperature. How fast they absorb depends on the temperatures of objects around them. Therefore, the net intensity (energy per ...


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

This is simply about words. A process can cause a change. For example: A (reversible) adiabatic process can cause a (reversible) temperature change.


2

Define "well insulated", do you mean no heat at all escapes, or comes in from outside? Which is physically impossible, so the smaller the room volume, the less heat will escape. I am assuming you are pulling in exactly as much heat as you are pushing out per unit time. Smaller is better for keeping warm, larger is better for keeping cool.


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.


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

To illustrate the difference take two blocks in contact with each other, with block A at a higher temperature than block B initially. As soon as we bring A in contact with B, energy will flow from A to B, the temperature of A will decrease, and the temperature of B will increase. Given enough time, the two blocks will reach the same temperature, and the ...


2

What you describe can be achieved and breaks no laws of physics (which is the same thing put differently :-) ). What cannot be achieved is to end up with more energy at the end of the process than you had at the start. Consider this as a "heat engine" problem. Assume that all temperatures are absolute temperatures (eg say degrees K). Requirement: A ...


2

Let me quote this line which says that: I touch it that it's temperature did not drop down Its better to use thermometer to check the readings as it gives you accurate reading. Please check this link as it shows what you did wrong: Why does cold metal seem colder than cold air? The process of touching and determining its temperature is wrong. There ...


2

Since you only want to know what the final temperature will be, it's easy to do this by simply considering how much energy will flow between the water and the glass. Knowing that the final temperatures of the two will be equal, and that energy is conserved (so any energy leaving the water must enter the glass), we can set up a couple of linear equations ...


2

Your mistake is that by "state" you mean following: particle either in section A or section B. In reality what you use for computing entropy is number of ways for particle to be in some point of phase space, that is having particle coordinate and impulse (all 3D). But you can omit these details. Consider box with single particle and 2 sections divided by ...


2

I don't know how hot campfires get, but let's take 1000K as a nice round number. The Planck distribution for 1000K looks like: (Calculated, as the logo suggests, using this web site.) In my answer to Can a glass window protect from heat radiation? I post this graph showing the transmission of glass in the IR region: And this shows the transmission ...



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