Tag Info

New answers tagged

30

I really, really like this question. Thus, the air molecules contribute a small portion of their kinetic energy to the paddle, which is then expended as heat on the other side of the border, making the air molecules on the left colder, while air molecules on the right heat up. Doesn't this mean a decrease in entropy? Yes it does. However, we need to ...


2

Photons come with chirality, so you should consider angular momentum conservation as well. For $1\gamma \to 2\gamma$ scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated)


0

The key to the answer is this part of your quoted text: ...within the limits of the dimensions of the cell in which its representative points lie... Thus in the book you quote, the number $\Omega$ is to be obtained by counting the number of cells (whose dimensions are arbitrary), not the number of all possible states. Alternatively, one can replace ...


1

Entropy is from statistical physics, a single particle or a single degree of freedom does not have an entropy. Edit: There are a lot of different things that are called entropy. So I'm not sure I feel comfortable with the above as a blanket statement. And there is a way to see time from the dynamics of quantum mechanical systems, so if the above statement ...


1

I am not sure this is the answer you are looking for but you may have heard, in QM especially, that an excited system eventually goes back to its ground state. This is true of subatomic particles like neutrons decaying into proton + extra stuff and of atoms too which, once in an excited electronic state, eventually release this extra energy as a photon. The ...


9

After the hypothetical split, 2 photons with the same energy would be propagating at an angle ok with momentum conservation. Then there would be a rest frame where the angle is 180 degrees. Now if you stay in this restframe and go back in time before the split, your single photon would be at rest. However, that is not possible: According to relativity, speed ...


8

A photon is an elementary particle. As much elementary and as much particle as the electron . A single elementary particle has a fixed mass and cannot emit another particle without violating energy conservation, because its mass is fixed. In the center of mass of a massive elementary particle, electron, there is no energy for an emission , for a ...


1

To start with the law of increasing entropy applies to isolated systems. The system you describe is isolated if one considers the total entropy of both the paramagnetic material and the permanent magnet, including any radiation. The order introduced in the paramagnetic material is balanced by a disorder in the permanent magnet plus any radiation from ...


2

The connection of heat to entropy in thermodynamics is through: where S is entropy Q is heat T is temperature, and it is through differential changes. This in no way means that heat is entropy . The easiest way to acquire an intuition of entropy is to read up on the statistical definition which can be proven to be the same as the thermodynamic ...


0

There are 3 forms of heat transfer: radiative, convective, and conductive heat transfer. All 3 forms of heat transfer are normally operating at the same time.


0

Entropy is fundamentally a subjective quantity, it is the amount of information needed to specify the exact physical state of a system given it's macroscopic specification. The fact that you don't know the exact physical state of a system, yet the laws of physics are such that information is never lost, implies the second law of thermodynamics (one also ...


1

Introduction: Entropy Defined The popular literature is littered with articles, papers, books, and various & sundry other sources, filled to overflowing with prosaic explanations of entropy. But it should be remembered that entropy, an idea born from classical thermodynamics, is a quantitative entity, and not a qualitative one. That means that entropy ...


0

Let's frame this question in terms of a heat engine (Carnot engine). Here is a diagram I made for a class when teaching this stuff. Heat flow $\dot{Q}$ has an associated entropy flow $\dot{Q}/T$. The job of a thermodynamic engine is extract/filter as much useful work as possible from a flow of energy and entropy. To answer your question in layman terms. ...


1

When energy flows from a hotter subsystem to a colder subsystem the entropy of the combined system increases. $$\frac{dS_1}{dE_1}=\frac{1}{k_BT_1}<\frac{1}{k_BT_2}=\frac{dS_2}{dE_2}.$$ So if they exchange energy in a way that conserves energy then we get $-\Delta S_1<\Delta S_2$ and the total entropy increases. This does require that the systems ...


1

They look like they could be related. What is the relationship? From your two equations, we have $$k\ln \Omega = k \ln Q + kT\frac{\partial}{\partial T} \ln Q = k \ln Q + \frac{kT}{Q}\frac{\partial}{\partial T}Q$$ but $$Q = \sum_ie^{-\frac{E_i}{kT}}$$ and so $$k\ln \Omega = k \ln Q + \frac{kT}{Q}\frac{1}{kT^2}\sum_i E_ie^{-\frac{E_i}{kT}} = ...


2

In the limit that $T\rightarrow\infty$, the partition function and the multiplicity of states are equal. Why? Well, we have that $Q=\sum_{i} e^{-E_i/kT}$, where $i$ indexes all possible microstates. If $T\rightarrow\infty$, these Boltzman factors all approach one, and we have $Q=\sum_i 1=\Omega$. You might think that in the limit $T\rightarrow\infty$ the ...


2

I am not familiar with the textbook you mentioned but there is a conventional "proof" of the equality $\oint _{rev}\frac{\delta Q}{T}=0$ for states that can be described by a thermal parameter, usually temperature $T$ and by a mechanical parameter, such as $V$. The "proof" goes by approximating the area enclosed by the $T,V$ cycle with Carnot cycles, that is ...


1

Issues with that derivation: You're missing the extra term $\frac 52 k N,$ which may matter if you have to do any work with chemical potentials. Your students will not necessarily know why to parcel the space into volumes of size $\lambda^3$. Starting from the definition of entropy and deriving that the thermal volume $\lambda^3$ is important seems ...


-1

Your argument is wrong in that the second body receives Qin=σT41A1. You can direct radiant energy at it - does not mean it will receive it. Heat does not flow from cold to hot. Temperature wins every time. Double the heat at the same temperature is still the same temperature.


3

Mirrors and lenses cannot do what you ask. Light has a specific intensity, meaning a power per unit area per unit solid angle per unit wavelength. Mirrors and lenses can never increase the specific intensity. As an example, consider using a lens to focus sunlight onto a target. The specific intensity of the sunlight is the same with or without the lens. ...


0

I posted the following solution on this board wanting to get opinions of the validity of a solution using only the microcanonical ensemble: Simpler derivation of sarkur-tetrode equation I still haven't received any comments, but if you have any, feel free to chime in. The Sarkur-Tetrode equation is the following without the 5/2 constant term: $$ kn \ln ...


0

It can be seen from the Gibbs free energy: $$dG=dH-TdS$$ Lets assume that it is possible and the situation reverses from Figure c to a. The change in entropy of this system is negative ($dS<0$) because it goes from a disorderly to orderly state. We can see in that case $dG>0$ assuming $dH>0$, this indicates this is non-spontaneous and will ...


0

Spontaneity of a thermodynamic process can be analysed from the perspective of free energy. Lets look at the Gibbs' free energy: $$dG=dH-TdS$$ For a process to be spontaneous, $dG<0$. Likewise, a non-spontaneous process is equivalent to $dG>0$. At equilibrium, we have $dG=0$. Consider a couple of scenarios: $dH>0$ & $TdS>0$: Then ...



Top 50 recent answers are included