# Tag Info

18

First, strictly speaking a neutron star is not a nucleus since it is bound together by gravity rather than the strong force. Measuring a surface temperature for any star is deceptively simple. All that is needed is a spectrum, which gives the luminous flux (or similar quantity) as a function of photon wavelength. There will be a broad thermal peak somewhere ...

9

Earth can lose heat to space through radiation. The earth behaves roughly as a blackbody and so radiates electromagnetic radiation into space at a rate of roughly 120 PW.

5

It depends on the mass of the molecule in question. Here's a quick, back-of-the-envelope answer. In a body at thermal equilibrium, every energy mode has the same average amount of energy, $\frac12kT$, where $T$ is temperature and $k$ is Boltzmann's constant. One of the energy modes is the translational kinetic energy of a molecule in some direction $x$, ...

4

You are missing the fact that when I have a lot of air in a tight space, and I then heat it up, I get really high pressure. You have to draw yourself a diagram of pressure vs volume - compressing the cold gas requires a certain $\int P \cdot dV$ of work, but then I heat the gas and the subsequent expansion takes me along a different curve where the work ...

3

Taking an intuitive guess here: The pressure above the water column is indeed very low, and water molecules at the surface may escape - but they are also held back by the surface tension of the water (is your meniscus concave or convex?). There is an equilibrium here, and the temperature is low enough that the water won't boil off all at once. So at room ...

3

I believe that in the top you'll find it's the saturated vapor pressure of water. http://en.wikipedia.org/wiki/Vapour_pressure_of_water (80 F ~27 C vapor pressure of about 27 mmHg or 27mm/760 mm * 14.7 = 0.52 psi.. not bad.

2

According to the first law of thermodynamics \begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align} Where $Y$ is the generalized force, $dX$ is the generalized displacement. Helmholtz Free Energy \begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align} Gibbs Free Energy \begin{align}G=U-TS-YX=\sum_j\mu_jN_j.\end{align} Therefore that ...

1

You are correct. In a real process, one would modify this formula to include the so-called polytropic exponent $n$ such that $PV^n=const$. This reflect that the process is not perfectly isentropic. For a fixed final volume this means the final temperature and pressure will be higher than in the ideal case, and more work need to be expended.

1

Incense burning is much more of a smoldering reaction. You've got a porous medium that retains heat well and admits oxygen poorly. The propagation is mostly limited by the ability of heat to conduct along the stick and the ability of gases to pass in and out of the solid structure. Conduction doesn't care about gravity and The viscosity in those small ...

1

In your question you already have the answer, photovoltaic cells. They are not heat engines but change the black body electromagnetic radiation of the sun to electricity. Of course this depends on the fact that the black body temperatures of the cells are much smaller than the black body temperatures coming from the sun. As @ThePhoton says , note the ...

1

Incense is a much better object for this puzzle than a candle. When a candle burns, the wax is melted. Normally the melting wax has to travel up the wick before it burns, and that gives a relatively constant rate of burning. When you place the candle horizontally (or upside down), some of the melting wax can fall away without burning. Consequently the flame ...

1

This is true because we have to use the grand canonical potential $\Omega$ which is directly related to the pressure: $\Omega /V=-P$ (This can be derived by Legendre transformations starting from $E=TS-pV+\mu N$) So if you maximize the pressure you actually minimize the free energy (density) (because of the minus sign) so you are looking for the state with ...

1

Yes, even though some answers to nearly equivalent cousins of this question claim that the answer is No. We really understand the right "theory of nearly everything" which means that using the microscopic theory – it's the "statistical mechanics" part that is most important here – we may calculate everything that happens in the world of everyday phenomena. ...

1

Thermal velocity is the velocity that a particle in a system would have if its kinetic energy were equal to the average energy of all the particles of the system. Take an ideal gas in three dimensions as an example. In equilibrium there is a distribution of velocities among all of the particles. Some move "fast" others move "slowly". That means that ...

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