How can there be an entropy change in this system?

Suppose if I have a system consisting of liquid water, $1\, \mathrm{kg}$ at $290\,\mathrm{K}$, I stir it, and do say, $10\, \mathrm{J}$ of work on it, I can calculate the temperature change of the system given that:

$$U = cT \quad\mbox{ and }\quad S = c \ln \Omega$$ for $c$ constant.

From the fundamental equation of thermodynamics: $$dU = dQ + dW = 0 + dW = 0 + 10 = 10\,\mathrm{J}$$ Hence: $$dT = \frac{dU}{cM} = \frac{1}{410}\,\mathrm{K}$$

But how can there be a change of entropy in the universe when $dQ = 0$. I understand that we can calculate it using the formula for $S$ given, but I don't understanding how the fundamental equation allows this?

$$dQ = 0$$ and $$dS = T^{-1}\,dQ$$

Hence, it may be concluded that: $$dS = 0$$

Can someone tell me where my understanding is lacking, because obviously the entropy change is not zero in this case?

  • $\begingroup$ Where did the energy to stir the water come from? Oh, it was external to the system. $\endgroup$ – LDC3 Jun 10 '15 at 2:31

The formula

$$ dS = dQ/T $$

only applies to thermodynamic processes that can be described by a path in thermodynamic state space (representing a quasi-static process, where the system is in thermodynamic equilibrium at all stages). Only for such processes it is meaningful to talk about continuous change of thermodynamic entropy.

Stirring a fluid is not such a simple process. If any work is to be done on the fluid, the fluid needs to get into state of flow. In such a state of flow, how soever slow, supplying work via stirring is an irreversible process. It is similar thing as when we're pushing a book from one side of a table to another; work is being done, but irreversibly. Both dry friction and fluid viscosity are forces that change their direction once the direction of displacement is changed, so the process is not reversible and consequently, the relation $dS=dQ/T$ cannot be used.

The whole process of stirring is a process whose intermediate stages are outside of the domain of classical thermodynamics. The only thing one can say about it based on thermodynamics is that if the process begins in an equilibrium state 1 and ends in equilibrium state 2, the change of entropy after the state 2 is attained obeys the Clausius inequality

$$ \Delta S(1\rightarrow 2) \geq \int_1^2 \frac{dQ}{T_\textrm{res}} $$

where $Q$ is total energy increase of the system due to heat transfer since state 1 and $T_\textrm{res}$ is the temperature of the reservoir that enables the heat transfer (provided it can be ascribed temperature). If no energy transfer except due to stirring occurs, $dQ=0$ and whatever $T_\textrm{res}$, the right-hand side vanishes:

$$ \Delta S(1\rightarrow 2) \geq 0. $$

  • 2
    $\begingroup$ This is not necessarily true. You should be able to quasistatically stir the liquid by stirring it very slowly. $\endgroup$ – Ian Jun 10 '15 at 4:05
  • $\begingroup$ Yes, but you use the word quasistatic in the sense "very slow". "Quasistatic" in thermodynamics does not mean only that the process is slow, the system must also go through equilibrium states. Stirring the fluid, how slowly soever, pushes the system out of thermodynamic equilibrium. $\endgroup$ – Ján Lalinský Jun 10 '15 at 6:20
  • 1
    $\begingroup$ Why does stirring it put it out of equilibrium? $\endgroup$ – Ian Jun 10 '15 at 12:42
  • $\begingroup$ Because according to the original assumption, work is being done on the fluid. This means flow is present and viscous forces are present. Work being done against viscous forces is irreversible process. $\endgroup$ – Ján Lalinský Jan 20 '16 at 7:10

There are two points here:

  1. Those equations above from $dU=\delta Q+dW$ are for gas (ideal gas), not liquid. You cannot use them to calculate the entropy of liquid.

  2. Work done in the first law of thermodynamics is defined as: $dW=pdV$ (quasistatic process), where p is generalised force (pressure) and V is generalised displacement (volume). When you stir the system, you do not do a work on the system in term of the first law.

  • $\begingroup$ I think 2. is incorrect. Although there is no PV work done, there is certainly work done on the liquid. This needs to be included in dW. See my answer. $\endgroup$ – Ian Jun 10 '15 at 4:02

First of all, there is a problem with your formula for the entropy. $S\neq c\ln T$. One way to see this is that temperature has units, and you cannot take the log of a quantity with units. The entropy is instead defined as

$$S=k_B \ln \Omega$$

where $k_B$ is Boltzmann's constant, and $\Omega$ is the multiplicity of the system, also known as the number of configurations in phase space available to the liquid at a given energy.

Now let's assume that the liquid is isolated from its surroundings, such that no heat leaves or enters the system. This is called "working in the microcanonical ensemble." Then dQ=0.

Let's also assume that we are doing the work sufficiently slowly (quasistatically) that the system virtually never leaves equilibrium.

If we do the work sufficiently slowly, stirring the liquid will merely cause the entire body of liquid to rotate together at equilibrium. This does not change the multiplicity of the system! How do we know this? Consider the system in the rotating reference frame, that is the reference frame with the angular velocity of the moment of inertia of the liquid. In this frame, the liquid is still, and it has the same multiplicity that it did in the rest frame before the liquid was stirred since the volume didn't change. Multiplicity is independent of the reference frame (correct me if I'm wrong) so the multiplicity did not change when we stirred the liquid quasistatically. Therefore, based on our definition of entropy $dS=0$. This is the result we obtained from the first law of thermodynamics.

  • $\begingroup$ The OP is talking about a different kind of stirring. Basically heating up the liquid through randomizing the motion. $\endgroup$ – lalala Feb 17 '17 at 9:53

The integral of dQ/T is equal to the entropy change of a system between two thermodynamic equilibrium states only if the path between the two states is reversible. If you want to determine the entropy change between two thermodynamic equilibrium states for a system that has undergone an irreversible process, you need to dream up a reversible path for the system between the same two thermodynamic equilibrium states, and calculate the integral of dQ/T for that path.

Answer #2 is pretty much completely incorrect.

  • $\begingroup$ It's quite possible to have a reversible path between these two equilibria. We just have to stir the liquid quasistatically. Then the stirring can be reversed by quasistatically stirring in the opposite direction. $\endgroup$ – Ian Jun 10 '15 at 4:04
  • $\begingroup$ This is not correct. No matter what direction you stir the liquid, you are dissipating mechanical energy (viscously) to thermal energy. When you stir it in one direction, the temperature rises. When you stir it in the opposite direction, the temperature rises again. So you have not returned it to its original state. $\endgroup$ – chet miller Jun 10 '15 at 12:38
  • $\begingroup$ How is energy dissipated? If I set the whole body of liquid in motion, I believe it has to keep spinning by conservation of angular momentum $\endgroup$ – Ian Jun 10 '15 at 12:44

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