Tag Info

Hot answers tagged

28

Law of Thermodynamics says that two bodies eventually will have equal temperatures. That is not an absolute Law. There are conditions, and one of those conditions involves the energy input to the bodies. If this Law was absolute, then the Sun would be at the same temperature as the universe, about 2.7 K, because the universe is much larger than the ...


20

As far as the theory goes, you are absolutely correct, the (negative) binding energy between atoms in a molecule contributes to the total mass of that molecule, so a stable molecule is less massive than the sum of the masses of its constituent atoms. However (as you yourself calculated), the mass difference is absolutely tiny, and as far as I know, it has ...


15

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 ...


8

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 ...


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

While respecting the other good and thoughrough answers, I feel I can give you a simple explanation to your exact question. The 0th law of thermodynamics is the law of the temperature balancing that you refer to. As you mention in your question, this law says that two bodies eventually will have equal temperatures when in thermal contact. "Eventually" is ...


6

Yes the core will warm gradually. Heat transfer in a solid is conduction. Ice has a known thermal conductivity and will have a linear temperature profile from all paths from surface to center. There will be concentric rings of constant temperature at all times. It would be impossible to warm just the surface and not warm up the molecules next to the ...


6

The potential energy for a diatomic molecule is not $$ U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} |\vec{q}_1 - \vec{q}_2|^2 $$ but is instead $$ U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} (|\vec{q}_1 - \vec{q}_2| - r_0)^2, $$ where $r_0$ is the equilibrium bond distance. The important difference here is that in your version, any displacement of the vector ...


5

Entropy is subjective in the sense that you get to pick which macroscopic observables you care about keeping track of (usually, for instance, you care about things like temperature, pressure, etc.). Once you've defined the macroscopic observables, entropy is defined as the logarithm of the number of possible microstates that give rise to those macroscopic ...


5

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 ...


5

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" ...


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 ...


5

The air has very low thermal conductivity and capacity, in most cases outside, the main contributor to thermal exchange (and thus perception of temperature) is radiation (Stefan's law, every object is radiating light all across the spectrum, with colder bodies giving most of it in infrared, hotter is more visible red (coal embers, hot iron), then yellow, ...


4

Perhaps because of the same reasons as the warming of greenhouses? If the windows are uncovered the sunlight increase the energy inside, by isolating the warm air inside the structure so that heat is not lost by convection. [From Wikipedia]


4

In macroscopic units it should be $$S=-R\alpha \log(\alpha e^{-S_1/R})-R(1-\alpha)\log\Big(1-\alpha)e^{-S_2/R}\Big) \\=\alpha \Big(S_1-R\log\alpha\Big)+(1-\alpha)\Big(S_2-R\log(1-\alpha)\Big),$$ where $R$ is the universal gas constant. In the pure case, this reduces to the textbook formula. But such a formula cannot be true in general. The general formula ...


4

The equipartition theorem is a mathematical consequence of very specific kind of Hamiltonians. It states that any 'squared' term of deegree of freedom in the Hamiltonian gets $\frac{1}{2}k_bT$ of energy (it is a statement about the energy distribution for this kind of Hamiltonians). For example - classical ideal gas Hamiltonian - ...


3

Lunar soil or lunar regolith, is mostly created by meteorite and micrometeorite impacts which directly pulverize the rock, or from the ejecta from the impact. Some amount (I can't seem to find any figures) is also created from high-energy particles in solar wind causing bits of rock to spall. In theory, the bootprints would last until the soil turns over ...


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 ...


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

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

In principle, no, you cannot make a Dyson sphere which is indistinguishable from the CMB. The reason is fairly simple. Let's start with a blackbody DS which encloses nothing at all, and is so far from any nearby stars that no noticeable radiation reaches it. Since it is surrounded by CMB with an effective temperature of 2.75 K, it will reach an equilibrium ...


3

There is no particles in space so how can it have a temperature since it does not really pass the heat around and if so how can a object cool down in space? When space is said to be cold, it does not always mean its temperature is low; it may not have temperature if the radiation present is not equilibrium radiation (which never is exactly; the ...


2

So, unfortunately, if a box contains a certain amount of kinetic/potential energy stored within it, that energy appears as a mass to the outside world. This is actually the basis for using the so-called "mass defect" (change in mass) in radioactive reactions for detecting the energy released: One over-large Uranium nucleus becomes some smaller parts with ...


2

Yes, bonds have mass, like every other kind of energy. This can be significant; if you had a glueball (a hypothetical particle made of massless gluons), it would have mass, and all of the mass would be from the bond energy! Same would go if you somehow managed to bind photons together.


2

Intergalactic space is filled with a photon gas at temperature 2.7K. For heat transfer you don't necessarily need atoms, other particles, such as photons suffice.


2

Suppose you start with a system in some state $P_1, V_1, T_1$ and you add some quantity of heat $\Delta Q$ to it so the system changes to a different state $P_2, V_2, T_2$. The final state will depend on how you added the heat $\Delta Q$. Adding the heat $\Delta Q$ in a reversible process will result in different values for $P_2, V_2, T_2$ compared with ...


2

The variables involved here are classical and you resolve them classically. They enter into the operator because they are parameters of the wavefunction. So let's do this a little more broadly. For continuous systems, we want a family of solutions based on some parameters which I'll collectively identify as $\alpha \in A$; the solutions are then labeled ...


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

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible