Why does taking work from a system not reduce its entropy?

In Daniel Schroeder's book Thermal Physics, he makes the point that as energy is put into a system it's entropy goes up (there are some exceptions to this generally, but for this question we can ignore this). However, as energy is put into the system, the entropy keeps increasing, but by lesser and lesser amounts. We also define the reciprocal of the change in Entropy over the change in Energy as the Temperature of the system:

$$\cfrac{1}{T} = \cfrac{\partial S}{\partial U}$$

This all makes sense: A lower temperature object is one where much entropy is gained if energy is added to it, and a higher temperature object is one where very little entropy is lost if energy is removed from it. From this, it is easy to see why the higher temperature object will spontaneously give energy to the lower temperature object if they are put into thermal contact with each other.

This also makes sense from the interpretation of entropy as the number of "microstates" of the system. Sure, removing energy will make the microstate of higher temperature object more certain (because there are now fewer of them), but having the lower temperature object receive that energy will add much more uncertainty to the microstate, more than making up for the microstates lost in the higher temperature object's system.

But if all this were true, then it shouldn't matter how energy enters/exits my system. That is, it shouldn't matter if the energy comes in/out via heat flow or whether work was done on it. Certainly, we can see in the above formula for temperature that the temperature depends on how the entropy $S$ changes when there is a change in the system's energy $U$. There is no reference to $Q$, the heat put into the system either in the formula or in the interpretation of entropy above.

Which is why I'm confused when he speaks about heat engines. Schroeder tries to determine the maximum efficiency of a heat engine that takes heat $Q_h$ from a hot reservoir, puts out work $W$ and dumps out heat $Q_c$ as waste. Starting with the definition of efficiency $e = W/Q_h$ he puts in two constraints. The first constraint comes from the First Law of Thermodynamics and makes sense:

$$Q_h = Q_c + W$$

The second constraint comes from the Second Law and has me confused:

$$\cfrac{Q_c}{T_c} \ge \cfrac{Q_h}{T_h}$$

Why is there no term for any entropy carried out when work was done on the system? In other words, why isn't the Second Law constraint like so:

$$\cfrac{Q_c}{T_c} + \cfrac{W}{T_?} \ge \cfrac{Q_h}{T_h}$$

(I suppose $T_?$ would be the temperature of the engine, between $T_c$ and $T_h$, when it is doing the work.)

Yet the constraint is not written this way, so why does taking work from a system not reduce its entropy??

• $T_? = \infty$ – hyportnex Aug 17 '18 at 21:36
• Yet, we've said that that $T_?$ must be between $T_c$ and $T_h$... @hyportnex – Israel Aug 20 '18 at 20:49
• indeed you have said that but that is wrong – hyportnex Aug 20 '18 at 21:42
• I know something's wrong (either that or I've just caught something that everyone who ever took the subject missed) ... I'm just trying to get the reason why it's wrong. At any rate, because it sits between the two heat reservoirs, the temperature of the machine should always be between $T_c$ and $T_h$. Do you agree with this, or is this the part that is wrong? @hyportnex – Israel Aug 20 '18 at 22:00
• In your example there are three energy sources (sinks): two heat reservoirs and one work reservoir. Their temperatures, whatever that term "temperature" may mean for an energy source, are independent of each other; there is no natural ordering of those temperatures unless there is heat (entropy & energy) flow from one to another without work being done. Since a work reservoir transfers energy and zero entropy (reversible process) its temperature can be take as $\infty$. – hyportnex Aug 20 '18 at 22:17

1 Answer

Not EVERY change in the system's internal energy corresponds to a change in its entropy. This is important to this issue as work is precisely the change in energy that doesn't correspond to a change in entropy.

The above formula for the temperature is somewhat misleading as you do not note what is kept constant during the partial derivative. It's very important that all external variables - such as volume - are kept constant. That limits the energy change to be only the heat, which is what makes $dS=dQ/T$ rather than any-old $dU/T$. A change in volume will result in energy change that is not related to entropy change, i.e. work.

The second law accounts for only the entropy change, not the overall energy change. So adding a $+W/T$ term is inappropriate - that would introduce a term that's tracking something unrelated to entropy-change, work.

• Thank you for your response... You state that "a change of volume will result in energy change that is not related to entropy change", yet compressing a gas should raise its temperature, and shouldn't all that energy mean that there is more uncertainty of the gas's state (hence, entropy increase)? Also, doesn't the thermodynamic identity $dS=P/T dV$ (everything but Volume is kept constant) show that changing something's volume can increase its entropy? @PhysicsTeacher – Israel Aug 20 '18 at 21:09
• Temperature is not the same as entropy. Consider an Otto cycle (en.wikipedia.org/wiki/Otto_cycle). It has two strokes of compression/expansion which are isentropic: entropy does not change, even though volume and temperature does. Note that change in volume CAN result in entropy increase, especially if it's fast. The thermodynamic identity you state holds only under constant internal energy. Generally $dS=\frac{1}{T}(PdV+dU)$. Slow compression will result in $dU=-PdV$ so that $dS=0$. – PhysicsTeacher Aug 21 '18 at 3:10
• Wow, so the answer is that when work is done on a system through compression, the extra entropy gained by all that extra energy (more velocity uncertainty) in the system is exactly cancelled out by the smaller space (less position uncertainty). It really is amazing that the two competing processes would exactly cancel each other out since the entropy increase/decrease is happening through two seemingly unrelated processes -- one being an increase in velocity uncertainty and the other being a decrease in position uncertainty. Nor does the shape of the container matter. @PhysicsTeacher – Israel Aug 22 '18 at 21:39
• Note that this is an idealization - it's true for a quasistatic process, but in practice you'll have a finite rate, and irreversibility and entropy build-up to some extent. Also, it's not correct for some cases, see the 'quantum adiabatic theorem' and 'level crossing'. – PhysicsTeacher Aug 23 '18 at 3:14