# Tag Info

15

It can't be from the moisture in the air. If there was enough moisture in the air to produce condensation then it would be condensing on everything. There would actually be less of it condensing on the tailpipe, because the tailpipe is quite warm. In fact the water is generated by the combustion of the fuel in the car. It comes from the hydrogen in the ...

6

I basically agree with Chase's answer - with some additions. But let's see thoroughly what is behind that sentence in Wikipedia ("Most steel engines have a thermodynamic limit of 37%"). Internal combustion engines are modeled by the Otto cycle rather than by the Carnot cycle. Looking at this link for the Otto cycle (or this great link from MIT), you may ...

5

The expression for the speed of sound is really: $$c = \sqrt{\left(\frac{\partial P}{\partial \rho}\right)_T}$$ Which is always true. The simplification you have listed is assuming an ideal gas. So the limits of that expression are the same as that for an ideal gas -- molecules interact only through collisions (no long-range forces) and the molecules ...

5

The distribution you use depends on the ''ensemble'' you are working in. There are three kinds of ensembles: Mircocanonical ensemble (or practical: an isolated system): the number of particles $N_s$ and the energy $E_s$ is fixed. This is ONE SPECIFIC REALISATION of the system. You have only one possible value for the energy and number of particles, so the ...

5

Given your description, you clearly have non-exponential behaviour. However, there are two possible reasons for this behaviour: Some materials in the system are nonlinear and do not follow Newton's law of cooling, which is that the heat flux at a given point is proportional to the temperature gradient vector. I should think this is the least likely of the ...

5

The noise is either from the AC electricity, which would be a 60Hz buzzing, or from small bubbles forming on the heating element itself. When the electricity stops, both the buzzing and the bubble formation will stop as well. Bubbles create sound due to quickly expanding from a small nucleus. Here's a book I found with a section on noise from bubble ...

4

To absorb infrared light, a stretching or bending vibration of the molecule must change the molecule's dipole moment. In $N_2$ and $O_2$ there is no dipole moment regardless of how you stretch the bond. On the other hand, O=C=O can change dipole moment by the C moving toward one O and away from the other O, or by bending with the C becoming a vertex of an ...

3

There are many problems : $1.$ As pointed out by Olin, gas cannot exist as a gas at $0 K$. $2.$ In ideal gases, interaction between molecules are absent. Hence, there is no potential energy. Remember that Potential energy always has an additive arbitrary constant. $3.$ As pointed by Wojciech, you would need (to take}energy to cool that ...

3

Kyothe was on the right track, but in fact we do radiate in the visible, just in such small amounts that it's not detectable for all practical purposes. If you look at the referenced Planck (black body) curves for objects around human body temperature, the short-wave tail is nonzero in the visible range, but it's there.

3

In the limit of very high temperature all gases become ideal (assuming they don't ionise, dissociate, etc) but this regime is far above the Boyle temperature. Around the Boyle temperature the long range attractive forces are still significant and cause non-ideal behaviour. It's just that there is a sweet spot where the attractive forces are balanced by the ...

3

Please tell me what I did wrong It takes General Relativity (GR) to describe black holes and, in GR, energy conservation is, well, subtle. From John Baez's Relativity FAQ "Is Energy Conserved in General Relativity?": In special cases, yes. In general — it depends on what you mean by "energy", and what you mean by "conserved". So, in general, ...

3

There are a couple of issues you might want to consider. Firstly there is the slightly boring one that we physicists measuring the mass of the black hole are outside it, and from this position the photon never reaches the event horizon let alone crosses it. I don't want to go into this here since the subject has been flogged to death in numerous questions ...

3

Like all good physicists we'll start by assuming the animal is spherical. Actually the calculation I'm going to describe is basically the same whatever shape the animal is, but choosing a sphere means I can write down some simple formulae. If the radius of the spherical animal is $r$, then the total area of its skin is the surface area of the sphere: $$A ... 3 Consider a continuous body C and a closed subset V with boundary \partial V completely included in C (notice that \partial V \subset V). If p \in \partial V, the external part of C, namely C \setminus V acts on p (actually on a neighbourhood of p in \partial V) by means of a superficial force$$d\vec{f} = \vec{s}(p, \vec{n}) ...

2

This article has some relevant results based on a study of bird plumage (it also happens to be cited in the abstract of the Nature paper mentioned in one of the other answers), and is summarized in simpler terms here. I'll attempt to summarize the summary. Black and fluffy/loose fitting clothing is best if it is hot out and there is any ($>3 ... 2 In the process you describe the system won't necessarily return to its original state. Suppose you instantly remove the weight and lift up the piston so the gas expands irreversibly to it's new equilibrium volume. The gas does no work in expanding so its temperature doesn't change - all that happens is that the pressure falls. Now compress the gas back to ... 2 The combustion temperature of gasoline is about 550K. So the Carnot limit for a gasoline-burning engine outdoors is roughly $$\epsilon_C = 1-\frac{T_f}{T_i} = 1 - \frac{275 \mathrm{K}}{550 \mathrm{K}} =50\%$$ Here I've taken the output temperature to be near freezing, which is a rudimentary estimate at best (as dmckee points out). This provides us a rough ... 2 You should use$U=q\epsilon_1$. With the total number of particles$Nbeing constant, we have: $$\frac{\partial S}{\partial q}=\epsilon_1 \frac{\partial S}{\partial E}=\frac{1}{T}\epsilon_1\tag{1}$$ As you said: $$\frac{\partial S}{\partial q}=k_B\ln(N/q - 1)=k_B\ln(N\epsilon_1/q\epsilon_1 - 1)=k_B\ln(N\epsilon_1/U - 1)\tag{2}$$ $$\to ... 2 The expression$$ k_B \frac{\Omega}{\bar{\Omega}} $$equals$$ k_B\frac{1}{\bar{\Omega}}\frac{d\bar{\Omega}}{dE} $$which equals$$ \frac{dS}{dE}. In thermodynamics, where S is the Clausius entropy, this is equal to 1/T where T is the Kelvin temperature. In statistical physics, this expression can be taken as a definition of 1/T of a system from ... 2 Although this isn't obvious, the system doesn't return to its initial state. If you were to very slowly remove the weight from the piston, then the gas would do work on the piston as you removed it, which means that its internal energy would be reduced. If you remove the weight very quickly then the gas still does work on it, but it will do less work than it ... 2 The thermal radiation from B does indeed heat object A. The trouble is that A loses energy by thermal radiation faster than the thermal radiation from B can heat it, so the end result is that A cools down. You can show this very easily. The Stefan-Boltzmann law tells us that the energy flux per unit area is proportional to T^4 so the rate of heat loss ... 2 Let me expand on the previous answer. We expand each term separately. We have, \begin{align} \log \left( 2 \cosh x \right) & = \log \left( e ^x + e ^{ - x } \right) \\ & = \log \left[ e ^{ x } ( 1 + e ^{ - 2 x } ) \right] \\ & = x + \log \left( 1 + e ^{ - 2 x } \right) \\ & \approx x + e^{-2x} \end{align} For the second term we have, ... 2 The answer is no, or at least it is in the classical vacuum sense. I also don't see a rationale for why creating a vacuum would require infinite energy. An explicit construction is to use a solid-phase reactive chemical "getter" to eliminate (nearly) all gas molecules present; in experimental practice, virtually all man-made materials still outgas ... 2 All materials emit thermal radiation (such as light). The hotter the material, the more the radiation is shifted to high frequencies (shorter wavelengths). The radiation comes from oscillating electrons (regardless of whether there is an electric current). Welding reaches temperatures high enough to cause significant emission of UV light. Oxyacetylene and ... 2 No, in fact you could even view the spontaneous evaporation as being driven by the fact that it increases entropy. Basically what's happening is the liquid particles have random speeds (with distribution characterized by temperature), and they bump into each other. Every once in a while, two particles near the interface will collide in just such a way that ... 2 x=x(y,z), y=y(x,z), z=(x,y)dx= (\frac{\partial x}{\partial y})_z dy + (\frac{\partial x}{\partial z})_y dzdy= (\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz\therefore dx= (\frac{\partial x}{\partial y})_z [(\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz] + (\frac{\partial ... 2 The key point I'm getting at is that when the pressurized liquid moves through the throttling valve, the auto-refrigeration effect is really a way of splitting the hot vapor "part" away from the cold liquid "part". I think this is the main misconception you have. Typically when a material boils, the gas that is released is at roughly the same ... 1 If I put the weight back on the piston, the system will again achieve its initial state. No, it won't. In the end, the pressure will be the same, but the temperature and therefore the volume will be higher. Firstly, in a real system there will be friction due to gas viscosity and piston/cylinder interaction. But even in an ideal system, after the ... 1 You have to realize first that Charles' law is the change in volume with respect to temperature for constant pressure while Boyle's law is the change in volume with respect to pressure for constant temperature. So when you combine them, you need to account for these If I take a gas of volumeV_1$, pressure$P_1$and temperature$T_1$and let it change have ... 1 In your 5th "Let" statement "$\frac {d\rho_c}{ dt} = 0.1kg/m^3/sec\$ the rate pressure changes in the large container", "pressure" should be "density" and there is a sign error. Your equations 1-4 are correct, but it is important to determine what is known and what is unknown and analyze how many independent equations and unknowns you have. Also, mass of ...

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