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Hot answers tagged thermodynamics

6

For all practical purposes the weight will remain 1000 pounds when you freeze it. In theory the mass of the water will reduce somewhat through cooling because it will contain less energy when frozen, but the effect would be utterly negligible. Although the weight will be the same, water expands in volume by about a tenth as it freezes (which is why, for ...

4

$\frac{hc}{k_BT}$ is a length, where $h$ is Planck’s constant and $c$ is the speed of light. Apart from a numerical factor of order 1, this is the thermal de Broglie wavelength for a photon gas at temperature $T$, the critical wavelength at which quantum effects begin to dominate. Also, apart from another numerical factor of order 1, this is the ...

4

It happens because the kinetics of boiling have a finite time scale. This is because to begin boiling, a nucleus must be furnished to trigger the phase change. That nucleus usually takes the form of an air bubble in a crack or crevice in the water container's walls. If the bubble exists when the boiling point is reached, boiling begins without delay. If no ...

2

I use units where $k_\mathrm{B} = 1$. You are correct up to the point where you obtain $S = \beta E + \log{Z}.$ Probably the most common source of confusion in thermodynamics is forgetting which variables are held fixed in partial derivatives. In particular, $$\beta = \frac{\partial S{\left(E,V,N\right)}}{\partial E}.$$ $T$ is not one of the variables ...

2

Friction does no work on the block as displacement is 0. That is incorrect. Friction is not a conservative force, like gravity. The work done by gravity depends only on the end points and not on the path traveled. The work done by friction is along the entire path of travel and therefore depends on the length $L$ of the path. So the work done by ...

2

You don't give the context, but this idea tends to be used in scattering experiments where the particles are either massless (e.g. photons) or highly relativistic so their rest mass is negligible compared to their energy. In that case the de Broglie wavelength is related to the energy by: $$\lambda = \frac{hc}{E} \tag{1}$$ And as you say, $kT$ has the ...

1

At the heart of a thermos jug lies a Scottish invention, the vacuum flask: A vacuum flask (also known as a Dewar flask, Dewar bottle or thermos) is an insulating storage vessel that greatly lengthens the time over which its contents remain hotter or cooler than the flask's surroundings. Invented by Sir James Dewar in 1892, the vacuum flask ...

1

I will only comment on your assumption of the efficiency of the cycle. The efficiency equation you are using assumes a reversible cycle where all the heat added $Q_H$ occurs at a single temperature $T_H$ and all the heat rejected $Q_C$ occurs at a single temperature $T_C$. In your cycle the heat added during the isochoric process and rejected during the ...

1

You incorrectly applied the Clausius inequality to this problem. The temperatures that must be used in the denominators are the values at the interfaces between 1 and 2 and between 2 and 3 (see Moran et al, Fundamentals of Engineering Thermodynamics). So the correct application of the Clausius inequality should look like this: $$\Delta S_1\geq\int{\frac{... 1 It's not really clear what you're asking, since you can no more "convert" energy into length as you could convert mass to length -- they're fundamentally distinct. However, I'll assume that what you want is something like this: As you note, k_BT has dimensions of energy, equivalent to M L^2 / T^2, where M, L, and T are mass, length, and time, ... 1 Here's a formula you will need. I'm sure you know it already:$$P = \frac{\Delta E}{\Delta t}$$Here's the major problem you have. In your post, I see lots of effort to calculate \Delta E, with temperature changes and such. However, calculating how much time the air spends inside your heater is rather difficult because the air goes in and churns ... 1 In isotherm process and with no change of phase we can't use those formulas and there's no formula for defintion of Q (correct me if I'm wrong). You are correct that you can't use those two formulas for an isothermal process with no phase change, but you can determine Q from the first law. Usually we use \Delta U=Q-L but so we are defining Q by ... 1 Work done by a force is always given by the line integral in your question. The work energy theorem, as you have written it, is basically a statement of a conservation of energy. Work is a mechanism of moving energy from one place or form to another. In a system where the only places for the energy to go is into kinetic energy or heat, then the total ... 1 I believe the issue is that you are including heat in your work calculation when you shouldn't. The formula W_F = \Delta KE + H is not correct. The change in internal energy (\Delta U) according to the first law of thermodynamics is given as$$ \Delta U = Q - W where $Q$ is the same as your $H$. Work should just be $W = \Delta KE$, the heating does ...

1

Frankly, I don't see any significance to the blue vertical lines either, other than perhaps to label specific materials. But then, why vary the height? I agree with you the horizontal lines probably indicated ranges due to material variations. That would make sense for wood being a natural material with lots of variations. Interestingly, there is a note in ...

1

If black holes are to satisfy the second law of thermodynamics, then since black holes do not remember what went into them (other than the total mass, angular momentum, and charge), their entropy has to be larger than the maximum entropy of all possible matter distributions that could have formed the black hole. Some fiddling around on the back of an ...

1

Kuma spotted my mistake by pointing out that \begin{equation} \frac{\partial (\beta E)}{\partial E} \neq \beta. \end{equation} I wanted to provide the answer in a slightly shorter way. Also setting $k_B = 1$, \begin{equation} \begin{aligned} \frac{\partial S}{\partial E} &= \beta + E\frac{\partial \beta}{\partial E} + \frac{\partial \ln Z}{\partial E} \\ ...

1

My best guess would be that it's due to convection flow passing through the pasta's cylindrical gap in a stable manner when upright. Convection when the bottom is heated causes hot water (and possibly an amount of steam formation if very hot) at the bottom to flow upwards. Pasta's that are lying down will be pushed around all the time and never really find ...

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