# Tag Info

41

Candle wax expands considerably when hot and molten. So while burning the candle the level in the glass rises. But when the candle is extinguished the outer region (nearest the glass) cools down quicker (candle wax doesn't conduct heat very well) and solidifies first, becoming immobile. The molten remainder then shrinks before solidifying. So it's the ...

38

You feel cold when heat is flowing from you to the surroundings, your body tries to burn more energy to keep up your temperature, so you shiver. Water conducts heat much more effectively than air (more than 100x as well) so even with water at the same temperature as air you will lose a lot more heat and feel cold. When your body is too hot it losses energy ...

34

The next approximation beyond the ideal gas is given by the Van der Waals fluid equation. It is a phenomenological law which takes into account the finite size of the molecules and their interactions with themselves. When you plot several Van der Vaals isotherms for a given substance, you observe that some of them show a phase transition from gas to liquid ...

16

Getting from gas to liquid is a matter of interparticle interaction winning over thermal agitation. There are several reasons why interparticle interactions are very weak in the case of helium atoms. On one hand, it is a noble gas and thus cannot form covalent bonds. On the other hand, it is very light hence highly non-polarizable: its Van der Waals ...

12

Yeah, I would guess that the alternate heating/cooling of the wax in the sun pushes it up the side of the glass. Presumably the surface tension between the wax and the glass is quite strong and holds the wax up once it's been pushed up. Subsequent cycles cause wax to "backfill" the wax that's been pushed up. It would be interesting to design an experiment ...

10

You seem to make the implicit assumption that your vessel is placed in an environment that does not emit any thermal radiation, i.e. is already at 0 K temperature. The temperature of your container will asymptotically decrease to 0 K but will never actually reach it. Assuming black-body radiation, fixed heat capacity $c$, and sufficient thermal ...

9

The only mention of this subject I can recall seeing is an aside in Xiao-Gang Wen's book, Quantum Field Theory of Many-Body Systems. Footnote on page 86: A sound wave in air does not correspond to any discrete quasiparticle. This is because the sound wave is not a fluctuation of any quantum ground state. Thus, it does not correspond to any excitation ...

7

The energy comes from the sun, that much is certain. One possible mechanism is capillary action resulting in a meniscus, assuming the sun heats the wax to a (near) liquid state. The other possible mechanism is a vaporization/redeposit cycle. During the day, heat from the sun creates wax vapor, which is heavier than air [citation needed] and therefore ...

6

It is entirely possible that the entire candle has melted in the sun. If this happens the molten wax has a significantly greater volume than the solid wax candle and so the whole level will rise up the glass ie liquid wax takes up more volume than solid wax. As it cools again the level will fall as it solidifies. Typically it will cool from the top and ...

5

Bubbles are formed when the pressure in the fluid is less than the saturated vapor pressure of the liquid at that temperature, modified by surface tension effects. Surface tension actually increases the pressure inside a gas bubble in water; the approximate increase in pressure is $\Delta P = \frac{2\sigma}{r}$ (see for example this earlier answer). The ...

4

It is all a matter of engineering balance, between the water circulating in the radiator circuit of the car, which enfolds the engine and with water-metal contact which takes heat away at a certain rate. In automobiles and motorcycles with a liquid-cooled internal combustion engine, a radiator is connected to channels running through the engine and ...

4

Three processes are involved: Conduction: Heat flows from the object to its environment. Removal rate of heat from the interface further away from the object is proportional to the coefficient of conductivity (0.024 for air, 205 for aluminum -see http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html). Convection: The interface between the object ...

4

Are all heaters (same wattage, electric to thermal, no geothermal or other extra energy source) exactly as efficient as each other? No. Let's focus just on electrically powered heaters. If you have a heater that basically consists of a resistor with a current passing through it, you have 100% efficiency of electrical energy to heat energy conversion. ...

4

I will be blunt. As fas I know, nobody knows a priori for which systems equilibrium statistical mechanics will work or not. Part of the current effort to determine which systems are fine being described by equilibrium statistical mechanics focuses on various proofs of ergodicity for such systems. For now, they are somewhat limited to either a restrictive ...

4

If the universe is open, there's obviously more universe that you haven't included in your system. The universe, by definition, contains all energy and matter. An open system, by definition, has an outside system to exchange energy and matter with. If that outside system isn't part of the universe, then where is it?

4

This is what I thought at first: You need a restoring force in order to have a phonon: after all, phonons result from the quantization of the lattice energy written as a sum of harmonic oscillators. The hamiltonian of a gas cannot be written in this way: there are no restoring forces (you cannot "pull" a gas), so there can be no phonons. As Rococo wrote, ...

4

Ok, lets look at how we determine $\mu$ in a cosmological setting. In order to determine $\mu_i$, we can use the fact that, in equilibrium, $\mu$ is conserved in all reactions. This means that if we have a scattering process $i + j \rightarrow a+b$, then we know that $\mu_i + \mu_j = \mu_a + \mu_b$. Fermions in equilibrium, like electrons and neutrinos ...

4

In summertime, the ceiling fan blows air downwards and cools down your body using the wind chill effect. In wintertime, if you have an active heating system at home, it will heat up the air in the room. Hotter air moves up and accumulate near the ceiling, colder air being down. A ceiling fan in reverse direction moves cold air up pushes hot air downwards to ...

4

I think such a function may only exist in the Maxwell-Boltzmann limit. Here's why: For simplicity let us parametrize everything in terms of $\beta = 1/T$ and denote $Z(\beta) = \int{d^3p\; f_{eq}(p, \beta)}$. Rewrite the latter as $$Z(\beta) = 4\pi \int_0^\infty{dp\;\frac{p^2}{e^{\beta E_p}\pm 1}} = 4\pi \int_m^\infty{dE\;\frac{E\sqrt{E^2-m^2}}{e^{\beta ... 4 The definition of work:$$W=\int \vec F \cdot\mathrm{d}\vec x$$So, work requires a force and a displacement. That is all. Think of it like this: If you push hard on a wall, you might use much effort to apply large force - but nothing moves and no work of use is done. If you push against a balloon, you can make it move very far. But you didn't really do ... 4 I will try to answer as many questions as I can. I won't presume to give you complete exhaustive answers, but maybe they will be nonetheless useful to you. What variables determine the range of temperatures over which matter is liquid? My understanding of thermodynamics is that matter changes from a solid to a gas when some thermal vibrations create an ... 4 Looking around, the root mean square speed of air at 20 C is about 500 m/s, and given that you have \langle v^2 \rangle \propto \, T so that v_{rms}(T) = \sqrt{\langle v^2\rangle} varies with \sqrt{T} then have$$v_{rms}(15) = v_{rms}(20)\times \frac{\sqrt{15+273}}{\sqrt{20+273}} \approx 496 m/s$$and$$v_{rms}(25) = v_{rms}(20)\times ...

4

As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical ...

4

It won't work because your perfect vacuum is permeated by the cosmic background radiation, which itself is only asymptotically reducing to zero with the expansion of the universe. Trying to exclude the cosmic background radiation backs you into the infinite steps that forms the basis of the third law again. Also, using a container results in quantum ...

4

There are some issues with the experimental setup you proposed (apart from the fact that when its temperature is lowered the gas would become a liquid and then a solid - if it's not $^4$He: in that case it will stay a liquid). Let's see why. In the picture above, I've sketched your experimental setup. The black box must be impermeable to matter in order ...

4

When you feel hot, you perspire so as to benefit by evaporative cooling. As the relative humidity gets closer to 100%, the sweat cannot evaporate and evaporative cooling becomes less effective. Liquid water is a much better conductor of heat than air (even humid air) is, so if the water is even a few degrees cooler than your body, you feel cold because the ...

4

Throttling the gas (Joule-Kelvin expansion) only lowers the temperature of the gas when the Joule–Thomson coefficient is positive. For Helium, that point (the "J–T inversion temperature") is reached at 43°K (source: Cryogenic Society of America; the wikipedia article gives an incorrect value of 51°K). Above that temperature, Joule-Kelvin expansion will ...

3

Energy stored as heat, by itself, is neither low- nor high-quality. What matters is the temperature at which the heat is stored, and the relationship of that temperature compared to the heat sink that will absorb the excess energy in the process. To be more specific, say you have a heat sink at $T_S=20°\:\mathrm C$, such as the atmosphere for a car engine. ...

3

The (long-term) temperature of an object depends on the heat transfer between it and all of the environment. Air isn't a great conductor of heat. So if there is little air movement, the radiation environment may dominate the heat transfer. A cold calm day may feel quite balmy under full sunlight. On a cold evening, the sky may have a radiation ...

3

One benefit of scaling the heat capacity with another extensive variable is that you end up with an intensive property -- heat capacity per # of particles. Similarly specific heat refers to the heat capacity per unit mass so that the value of the intensive property can be compared between samples of the same material but with different sizes or geometries ...

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