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The complete formulation would be that for any closed system: $$dS = \delta S_{r} + \delta S_{p} = \frac{\delta Q_{r}}{T} + \delta S_{p}$$ where $\delta S_{r}$ and $\delta Q_{r}$ are respectively the entropy and heat received by the system from the environment, $T$ is the temperature of the element of the system that receives $\delta Q_r$, $\delta S_{p}$ is ...

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I find this question very interesting, as it deals with crucial concepts, common misunderstandings and often-encountered unclear reasoning. Part of the reply by Andrew Steane points to an answer (in the legend of his Fig. 17.3). Yet, on the other side, I don’t feel that the demonstration that follows is fully appropriate or correctly addresses the issue (for ...

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Measuring is only a snapshot of the position, not which path of all possible paths it is taking. Italics mine. A particle measured is not taking all possible paths, one specific path is measured and the possibilities are in the probability distribution calculated for the interaction. To make concrete, here is a measurement that happened once and a picture ...

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conclude that the work done by friction or any opposing force in both the scenarios will be the same, then how can we say that we can get maximum work from a reversible process? Shouldn't both reversible and non-reversible process give the same result? To address this, we introduce a concept known as entropy production (see here), so if have viscous forces ...

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If you use energy from a system to compress a spring then all of this energy can be recovered by making the spring to do work as it expands back to its original length. If you use energy from a system to compress an ideal gas then all of this energy can be recovered by making the gas do work as it expands back to its original volume. If you use energy from a ...

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This is for a heat engine. The maximum work corresponds to the generation of the minimum change in entropy, and the minimum change in entropy is zero. A reversible process has zero change in entropy. Therefore a reversible process does maximum work. The work done on a heat engine is $$W=Q_c-(-Q_h)$$ where $Q_c$ is the heat entering the cold reservoir and $... 0 This will elaborate a little bit on @Chet Miller already comprehensive answer. When Chet stated "The change in entropy of the system for this alternate reversible process is the same as that for the irreversible process between the same two end states", that's because entropy, like internal energy, temperature, pressure, etc., is a system property ... 1 Deriving each of these two equations required devising an alternate reversible process for both the system and the surroundings separately, and evaluating the integral of dq/T for each of those alternate processes (which differ significantly from the actual irreversible process). For the case of the system, for example, rather than using a single ideal ... 0 This is a good question and there is in fact no contradiction. Noether's theorem in Lagrangian mechanics states that the conservation of energy can be inferred mathematically as a consequence of the invariance of the behaviour of nature under time translations. That is, it doesn't matter whether you run your experiment today or tomorrow, as long as the ... 1 Time translation symmetry does not preclude an arrow of time. The former, which assumes that the laws of physics do not change over time (not that time must move forward), is assumed in both Lagrangian mechanics and thermodynamics and underlies the law of conservation of energy. 0 Here is another way to think about it, If you have the first law of thermodynamics, $$dU = dq+dw$$ You can rearrange this to become, $$dq = dU - dw$$ Now, the above equation the heat transfer is an 'inexact differential' that means there is no function$F$such that when you take the differential of$F$i.e:$dF$that you get$dq$. With this in mind, we ... 3 Your original equation is incorrect. It should read $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$Only for a reversible process at constant temperature is the equation you wrote correct. 1 Some background: In the thermodynamics framework, every way of adding energy to a system consists of a generalized driving force and a generalized displacement. When stretching a solid, for example, the system gains strain energy$\Delta U$through a mechanical force$F$and an elongation$\Delta L$. When compressing a gas, mechanical work is done through ... 0 In every equation involving physical quantities, it is always possible to add a proportionality constant to allow for different units of measurement between the left and the right side of the equation. It may look weird, but it is not wrong. Although, for clarity, I would avoid introducing it, if not necessary. 0 Your$V_{GS}$isn't normalized (the sums of the squared elements don't add up to 1). Also, I'm not sure if the$|0011 \rangle$coefficient should be 0.1 instead of 0.3. The sums of squares then add up to 1.0052, the value you got for the trace. If you start with the matrix for the normalized state$V_{GS}$, your trace should sum to 1. 2 For an isolated system, for both reversible and irreversible process dQ = 0. So according to the equation of entropy isn't dS always zero for an isolated system? No. Although a differential change in entropy is defined for a reversible transfer of heat, or $$dS=\frac{\delta Q_{rev}}{T}$$ entropy can be generated without heat transfer. Any irreversible ... 3 The equation $$dS = \frac{dQ_\mathrm{rev}}{T}$$ describes the change in entropy only for a reversible process. If the process is not reversible, then$dQ = 0$does not imply$dS = 0$. 1 I can't really understand what you did, but I would have done this differently. For maximum work, there must be no entropy generation in the system, so the sum of the entropy change of the body plus the entropy change of the reservoir must be equal to zero. So, $$(S_2-S_1)-\frac{Q_1}{T_1}=0$$where$Q_1$is the heat extracted by the reservoir. This gives $$... 4 The second law leads to (or can be stated)$$ \Delta S_{\rm tot} \ge 0 $$not \Delta S_{\rm tot} > 0 (where 'tot' refers to everything that undergoes some change during the given process). The result is zero for reversible processes, and greater than zero for irreversible processes. So the adiabatic process is allowed by the second law, and so is any ... 2 The first term of your first equation only applies to the system for a reversible heat transfer process and the inequality only applies to an irreversible process. For the second equation the entropy of the surroundings can only change if there is heat transfer to or from the surroundings. Since there is no heat transfer in an adiabatic process \Delta S_{... 3 The only way the entropy of the surroundings can change is if there is heat transfer to or from the surroundings. An adiabatic process is, by definition, a process in which there is no heat transfer between the system and its surroundings. So it is impossible for the entropy of the surroundings to change for an adiabatic process. For a reversible adiabatic ... 4 In an adiabatic irreversible process, entropy is generated within the system, and there is no transfer of entropy to the surroundings because heat cannot flow out. Thermally, the surroundings do not even know that anything has happened within the system. So all the generated entropy remains within the system and causes the system entropy to rise, while the ... 1 The original equation you wrote assumes that, rather than putting body in contact with a reservoir at T2 for the entire process, it is put in contact with a sequence of reservoirs at gradually varying temperatures running from T1 to T2. This would be equivalent to the "reversible" process you described of gradually changing the temperature of the ... 1 Your reasoning appears to be sound. The only issue I see is when you talk about raising the temperature of the reservoir. If it is a reservoir its temperature can't technically change. Instead, you would theoretically need to place the body in contact with an infinite series of reservoirs, each differing from the prior by dT. The same goes for the reverse ... 4 Congratulations! You stumbled upon entropic uncertainty relations (a good review here). They are a reformulation of the usual uncertainty principles using entropy instead of variance. The simplest and most famous one is probably the Maassen Uffink relation: let \rho be a quantum state, let A and B be two observables and$$ \mathcal M_O:\rho\mapsto\... 1 The words "ordered" and "disordered", in relation to entropy, are the source of a lot of confusion and are not even always an accurate description. In your coin example, a more typical use of the word "ordered" would be to say that all three coins are the same (either heads or tails). If you start in an ordered state, and each ... 1 The question is not fully clear to me but I think it may be due to a misunderstanding of certain topics. I read in my book that any system tends to become disordered or tends to become more probable and this decides the spontaneity of a process. I personally think that we should first introduce the second law of thermodynamics and with that define entropy ... 1 Which "reversibility" are you referring to? If you mean "thermodynamically reversible" (a flow which does not generate entropy) then viscous dissipation ($\mu\nabla^2\mathbf{u}$) always ensures irreversibility, whatever the Reynolds number. But perhaps you are referring to "kinematic reversibility", which implies reversal of ... 2 The absolute entropy of a system in a given equilibrium state at temperature$T$and other parameters specified is given by $$S = S(0) + \int_0^T \frac{1}{T} dQ_{\rm rev}$$ where it is understood that the other parameters are being held constant and$dQ_{\rm rev}$refers to heat entering the system by a reversible process, and the limits on the integral ... 0 If the space is unbounded a single particle will have an infinite number of positions. It’s velocity can be any value between zero and c if you believe relativity in that situation, which is problematic. If you know the position relative to some frame of reference, a dubious notion physically because a single particle’s position must be defined against some ... 0 In an isothermal process, in which T is a constant, you can integrate and obtain$Q=T\Delta S$, in which Q is the heat exchanged to change the entropy by$\Delta S$. For non-isothermal processes you will obtain a different relationship though. 0 It's fine to wish to integrate$T\,dS$and to plan for a lower limit of$S=0$under the Third Law. It's not fine to bring$T$outside the integral to obtain$T\Delta S=TS$as if$T$is constant, because it's not (unless you're integrating from 0 to 0, in which case no net energy is transferred). 2 Heat is not a thermodynamic state variable, while entropy is. In other words, heat is a word that describes the spontaneous transfer of energy between a system and its environment (which we denote$dQ$), which depends on the path, or the way that you apply changes to the system's macroscopic variables. It is meaningless to talk about heat as a property of ... 1 Reversibility is characteristic of the regime of Stokes flow—also known as “creeping flow.” In this case, the velocity is always, in an appropriate sense, small. It is small enough that without external forcing, the viscosity terms damp the fluid to momentum to zero essentially instantaneously. (How small this is in practice obvious depends on how viscous ... 1 Omitting (for brevity) the electrochemical potentials, we have the exact differential form $$dS(U,V) = {dU + p(U,V)dV \over T}$$ The relation expressing the entropy$ S = S (U, V) $as a function of the extensive parameters$(U,V)$is sometimes called the fundamental thermodynamic relationship, because its partial derivatives provide the intensive ... 2 If a system has a multi-dimensional state space and we consider a change in which all but one of a complete set of parameters is constant, then this suffices to single out a unique direction in the state space. For a small movement along that direction, various properties such as$U$and$S$have a unique change, by$dU$and$dS$. One must then find that for ... 2 How can we one support Schröder's affirmation? The first law, for a closed system ($N$= constant): $$dU=\delta q -pdV$$ For a reversible process $$dS=\frac{\delta q_{rev}}{T}$$ and $$dU=TdS -pdV$$ Then, for a constant volume process,$dV=0$, and $$\biggr (\frac{\partial U}{\partial S}\biggl )_{N,V}=T$$ Hope this helps 0 The reasoning goes that because entropy is a state function, this relation ($dS=dU/T+P/TdV\$) holds even for irreversible processes because ... Well, this premise is wrong because an irreversible process goes through out-of-equilibrium states for which state variables such as T or P might not even be defined at the scale of the system, and such out-of-...

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I think part of the confusion comes from the fact that "the surroundings" is not the same as "the universe". The surroundings are just what the system constituted by the piston interacts directly with. The surroundings can themselves be encompassed into other surroundings. Now, looking just at what is happening between the piston and its ...

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This is an answer to the title of the question Where does entropy-increasing waste heat go during non-reversible processes? The "waste heat " goes to black body radiation. This radiation is only modeled correctly with quantum mechanics. A lot of the energy will go into the kinetic energy of the particles in the medium studied, raising the ...

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Entropy and life Entropy is an observer's uncertainty about the state of a system. A measurement (macro state) gives you a belief distribution over possible configurations (micro states) the system could be in. The Shannon entropy of this belief measures the observer's uncertainty. A uniform distribution over consistent states simplifies the entropy to the ...

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Except that it relates an electromagnetic arrow of time to the usual thermodynamic one, this answer's consistent with Nogueira's, and, as no answer's been accepted by the OP yet, I'm wanting to provide a verbal equivalent to it, per the PSE policy of permitting any participant's approval of a number of answers to the same question, and in view of my belief ...

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