# Tag Info

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When you pour the hot water in, the air inside the thermos is still quite cold (ambient temperature, approx.) But then when you shake it up the cold air is heated by the hot liquid. Gases expand considerably when heated, approximately acc. the Ideal Gas Law: $$pV=nRT$$ This causes a modest (and harmless) pressure increase in the flask, which is what you ...

47

There is another effect here which is significant, as follows. Warm water wants to evaporate, but in a flask-shaped container, the evaporation can take place only at the free surface of the water in the flask. Furthermore, as soon as the boundary layer of air right next to the warm water becomes saturated with vapor, the diffusion of water vapor into the air ...

9

I have no opinion about whether the wavefunction is "real" or "just a trick." I have never even understood what the debate is about, since you could have the same argument about any mathematical object (eg an electric field vector). There is a genuine physics question, however, in asking whether quantum mechanics is a complete formalism, ...

6

You can see it like this: if you consider the universe as a whole, "of course" each process is "spontaneous" in the sense that the total entropy of the universe entropy always increases. After all, if something happens somewhere in your universe, then it must be allowed to happen: otherwise, it just would not happen! However, what you ...

4

Let's assume that the total amount of water is so large that submerging the sphere raises the overall water height by only a negligible amount. Completely submerging a weightless sphere of volume $V$ is then equivalent to elevating a mass of water $m=\rho V$, where $\rho$ is the water density, by a distance $r$, where $r$ is the sphere radius. The reason is ...

4

Without making use of Taylor series what the textbook is computing is \begin{align} \require{cancel} E(x) &= \lim_{dx\to0} \frac{(f(x+dx)-f(x))^2-(dx)^2}{2(dx)^2} \\ &= \frac12\lim_{dx\to0} \left(\left(\frac{f(x+dx)-f(x)}{dx}\right)^2-\frac{(dx)^2}{(dx)^2}\right) \\ &= \frac12\left(\left(\lim_{dx\to0} \frac{f(x+dx)-f(x)}{dx}\right)^2-\lim_{dx\to0}...

3

I've read that a quasi-static process in which entropy change only because of heat exchange: $\Delta S=\int \frac {\delta q} T$ is not called irreversible. Only a reversible heat transfer is not irreversible. A change in entropy is defined for a reversible heat transfer, or $$\Delta S=\int \frac{\delta q_{rev}}{T}$$ Where $T$ is the temperature at the ...

3

The two definitions you cite do not mean that non-spontaneous processes decrease entropy, if that is what you are implying. In other words, the fact that all spontaneous (natural) processes increase entropy does not mean all non-spontaneous processes should decrease entropy. The refrigeration/heat pump cycle is a non-spontaneous transfer of heat from cold to ...

3

Isnt it true that if $p,T$ are held fixed then $$dG= \mu_1 dn_1 +\mu_2 dn_2+\ldots?$$ Here the $\mu_i$ are the chemical potentials and $n_i$ the numbers of material of species $i$. In chemical process the $n_i$ change and the equilibrium $dG=0$ is given by a set of relations between the $\mu_i$ that are determined by conservation relations, such as $$... 3 No heat is generated unless work is being done on/by the object. As an object in space in freefall is undergoing no force on it, no work is done. Thus the kinetic energy remains kinetic energy, and just... sits there. Of course, unless the object is at absolute zero temperature, it will be releasing heat. In the form of normal thermal radiation. But I'm ... 2 Not sure if this is what you are looking for, but I'll give it a go. In a classical system, you are working in phase space. You have to subdivide this space into volume elements in order to do any type of counting. We assume that the system under consideration depends upon some f generalized coordinates and momenta:$$E(q_1...q_f, p_1...p_f)$$and thus ... 2 Here is one way of looking at it. You are right that whenever h-bar pops up, quantum effects need to be taken into account. If h-bar were a larger number, then we could see strange quantum effects with our own eyes- and in the limit of h-bar going to zero, quantum effects vanish. (As an aside, in the limit of the speed of light becoming infinite, relativity ... 2 The equations that should be used are$$m_1C_1\frac{dT_1}{dt}=\frac{kA}{L}(T_2-T_1)$$and$$m_2C_2\frac{dT_2}{dt}=\frac{kA}{L}(T_1-T_2)$$So,$$\frac{d(T_1-T_2)}{dt}=-\frac{kA}{L}\left[\frac{1}{m_1C_1}+\frac{1}{m_2C_2}\right](T_1-T_2)$$The solution to this equation is$$(T_1-T_2)=(T_1-T_2)_{t=0}\exp{\left(-\frac{kA}{L}\left[\frac{1}{m_1C_1}+\frac{1}{m_2C_2}\...

2

This is a widespread misunderstanding, and it is based on a poor notation. The relationship between heat exchange and entropy growth is $$\Delta S \ge \int \frac{dq}{T}$$ for a general process. An equality sign is only correct here in the restricted case of reversible processes: $$\Delta S = \int \frac{dq_{\rm rev}}{T}.$$ If a heat exchange is taking ...

2

Two types of processes: There are thermodynamics processes that change the state of some variables like temperature and entropy. In statistical mechanics there are microscopic processes which move particles between microstates, this change may move the system closer to the equilibrium macrostate or not. There is no such thing as a spontaneous ...

2

This means that the heat transferred to get from state a to state b must be less in the irreversible case than in the reversible case. But the work done is the same. Though it depends on the details of the process, you are correct that less heat is transferred to the system in the case of the irreversible expansion process connecting the same two ...

2

But the work done is the same As you realized, this is not true. Work done is a path dependent process. You are taking a different path and you expect the amount to work done to change. It is hard to say exactly what has happened with the amount of work done, but that is because you have not specified exactly how you made the process irreversible. For ...

2

Can an object in space have a high kinetic energy, but at the same time not release heat? Yes, because the kinetic energy of the object as a whole, its macroscopic kinetic energy with respect to an external frame of reference, has nothing to do with it radiating heat, which is a function of its temperature relative to the temperature of it surroundings. In ...

1

In the original posting the author remarked: I can't see how the radiation being emitted by the Sun can be seen as doing work on anything.’ In a subsequent posting the author writes: Of the three options I presented in the question, the answer is 1: the emission of radiation into space is actually an irreversible process.’ My comments: The radiation ...

1

On your first question - whether wavefunctions and superposition are 'real' or 'a model making accurate predictions' - there is no way to tell. This is metaphysics. Since by definition there is no observable difference between true and false models so long as they both make accurate predictions, there is no scientific way to distinguish them. On your second ...

1

Note that your linked question is about the moments of inertia for diatomic molecules, but the ground state for argon at ordinary pressures and temperatures is a monatomic. The angular momentum stored in an isolated atom is the sum of the intrinsic spins and orbital angular momenta of its constituent protons, neutrons, and electrons. If you were to reach ...

1

There exist systems with other degrees of freedom other than pressure temperature and volume, for example in magnetic systems you can also talk about the degree of magnetization and the applied field. So the completely general answer is no, because it is not always true. Once you have limited yourself to systems that only have these degrees of freedom, we ...

1

The force that was needed to keep the piston at a certain height depends upon the no. of molecules colliding with the piston(this is nothing but the pressure). Actually, this is nothing but the pressure times the area. The pressure depends on the number of molecules colliding with the piston per unit area (and also how hard they hit). The force on the ...

1

It is not the maximum work that can be extracted from free expansion. It is the maximum work that can be extracted at constant temperature from the initial state to the final state by any process.

1

This is not the Taylor series expansion for E(x). The derivation uses the Taylor series expansion for f(x) to determine the strain in the material at material location x. f(x) is the location at time t of the material point that was at location x at time zero. The focus of this analysis is the small element of material that was situated between locations x ...

1

Low temperature phase diagrams are reasonably abundant in the literature. For simplicity I considered an oxygen-nitrogen mixture which, at the right proportions, is basically air. C.S. Barrett et al. in the Journal of Chemical Physics published just such a paper in 1967. Figure 1 shows the phase diagram they determined. Adding oxygen to nitrogen lowers the ...

1

Indeed, binary phase diagrams (not to mention ternary) are a bit difficult to 'get' when you first encounter them. While I dearly love Porter and Easterling's book, this figure is not the best presentation. First, the final diagram, (f), should really be considered as a map, not curves. If you look at the map of a country, the cities and roads are not ...

1

You are correct and you can also verify everything by considering an ideal gas. $$Q=W$$ and $$Q = T \Delta S - TS_g$$ where you can see that the entropy generated always decreases the heat transfer in the algebraic sense. Consider an ideal gas for the process you've shown. $$Q = -mRTln\frac{P_b}{P_a} - TS_g$$ where $P_b < P_a$ therefore the reversible ...

1

To Malawby: ALL radiation hitting the earth is either absorbed or reflected. Regardless of temperature of the source or the earth. What happen in this scenario is that the earth and the green gasses are exchanging energy by radiation with the net effect of 0 W/m2. If the earth was warmer we would have a net effect of >0 W/m2 etc to the green gasses. Even ...

1

If the temperature of the body is at absolute zero (or at least below the vacuum temperature) then yes. On a macroscopic level there will be no transfer of kinetic energy into internal heat energy. To dissipate kinetic energy into undirected, random, internal kinetic energy (=heat) you need forces to act on the body. As there are no in space the kinetic ...

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