# Tag Info

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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 ...

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So my question is, "Why should the change in entropy be zero, even if the particles are distinguishable?" In statistical physics, entropy can be defined in many different ways. One possibility is to define it as log of the accessible phase space, given the macroscopic constraints (volume). Such entropy is not a homogeneous function of energy, volume ...

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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 ...

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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 ...

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$x=x(y,z)$, $y=y(x,z)$, $z=(x,y)$ $$dx= (\frac{\partial x}{\partial y})_z dy + (\frac{\partial x}{\partial z})_y dz$$ $$dy= (\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz$$ $$\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 ... 2 If You know density rho_r at some temperature T_r, there is a following formula for density: rho=rho_r[1+b(T-T_r)], where rho is the density at temperature T and b is called coefficient of cubical expansion, evaluated at reference temperature and density (rho_r and T_r). It is valid for liquids. 1 I am not an aerodynamics specialist, so the following is almost certainly a huge oversimplification (or maybe downright wrong), but I think it might help with intuition. Suppose you have an amount of energy E available to spend, and you are trying to accelerate an object of mass M. Suppose you can impart the energy in the form of kinetic energy to an ... 1 I think it is important to not loose yourself in calculations. The method in your book probably starts from saying that considering an object \delta Q that has the following general form: \delta Q = C_vdT + hdV say, then it is an exact differential iif \left(\frac{\partial C_v}{\partial V}\right)_T = \left(\frac{\partial h}{\partial T}\right)_V ... 1 From http://chemicals.etacude.com/z/zinc_chloride.php it would seem that the enthalpy of formation of zinc chloride is -415 kJ/mol. It has a melting point of 290C and a boiling point of 732C. Since it seems happy to boil (and make a smoke screen), you would need to go above (given the formation energy well above) 732C to get it to dissociate by itself on a ... 1 The way I always thought about this was to pick one of the variables to be thought of as the dependent variable. Here I will pick z. Then we think of z(x,y) to be a function which has partial derivatives \partial_x z = \frac{\partial z}{\partial x} and \partial_y z = \frac{\partial z}{\partial y}. Now we must compute$$\left( \frac{\partial ...

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It depends on how you define your system or your control volume. If only the container is considered then indeed the entropy has decreased due to cooling. On the other hand if you account for the container plus the escaped vapour the entropy has increased, as the randomness of the molecules in the vapour state is larger than compared to in liquid state at ...

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Temperature is a macroscopic concept, so you're bound to run into some problems when you apply it on a molecular scale (what does temperature and equilibrium (or for that matter, friction) even mean on such a small level?). A thermal equilibrium does not mean all the molecules have the same energy. The distribution of their energies looks like this (normal ...

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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 ...

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Addressing parts of the question out of order: ...the heat capacity would smoothly approach zero around the transition. I have never seen anyone refer to these types of transition... The heat capacity of all substances smoothly approaches zero at absolute zero. $S(E)$ has a segment of zero first derivative no, the first derivative is a constant, ...

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As of last week, scientists have created a thermoelectric material that increases its efficiency to 15-20%. This is still not good enough to heat or cool your house, but it gives me hope that someday in the future, we have materials so good and cheap that we can build our houses and cars with it. So we may not be able to use the heat inside the house to cool ...

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I am going to address the question as to why energy and information have time symmetric conservation properties whereas entropy does not. According to the Wikipedia entry on entropy - "The entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium, which is the state of maximum entropy." ...

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