# “Violation” of the Second Law

I can't reconcile some facts about entropy and irreversibility. This depresses me, because I feel I can't quite grasp the importance of entropy.

I will illustrate my problems with an example given by my book:

1 kg of liquid water at 100 °C is condensed at constant pressure via heat transfer to the surrounding air which is at 25 °C. What is the net increase of entropy? ($s_{lv}$ = 6.048 kJ/kg·K, $h_{lv}$ = 2,257 kJ/kg)

Solution

$\Delta S_\text{system} = m \cdot \left(-s_{lv}\right) = -6.048 \frac{\text{kJ}}{\text{K}}$

$Q_\text{to surr} = m \cdot h_{lv} = 2257 \text{ kJ}$

$\Delta S_\text{surr} = \frac{Q}{T_0} = \frac{2257}{273.15+25} = 7.570 \frac{\text{kJ}}{\text{K}}$

$$\boxed{\Delta S_\text{net} = \Delta S_\text{system} + \Delta S_\text{surr} = 1.522 \frac{\text{kJ}}{\text{K}}}$$

I can easily follow that. However...

# Problem 1: entropy variation is just like that of a reversible process

My book says that, for a system undergoing an irreversible process: $$dS \ge \frac{\delta Q}{T}$$ Specifically, $$dS = \frac{\delta Q}{T} + \delta S_\text{gen}$$

However, the example above, which is an irreversible process (finite temperature difference), has exactly $\Delta S = \frac{Q}{T}$, completely disregarding the generated entropy (which is $\Delta S_\text{gen} = \Delta S_\text{net}$).

In other words, we have $Q = T\Delta S$, which everyone knows is just for reversible processes.

# Problem 2: dissipated heat is just like that of a reversible process

The book is quick to do $Q = \Delta H$, but it is known that that equation is only for reversible processes -- this is written in the chapter about the First Law.

Also, there's this equation, which says that, in an irreversible process, the exchanged heat must be less than that of a reversible one between the same states: $$\delta Q_\text{irr} = T\ dS - T\ \delta S_\text{gen}$$

Once again, the generated entropy is ignored and we have, like above, $Q = T\Delta S$.

Because of the above, I really can't see what difference would it make to the process if it were reversible, that is, if it exchanged heat with a reservoir at the same temperature as the water. Please, can anyone make sense of it?

• A reversible process is defined by $\Delta S = 0$. In the formula $dS = \delta Q/T + \delta S_\text{gen}$, the $\delta S_\text{gen}$ is normally the entropy increase within a particular subsystem. – David Vercauteren Sep 17 '14 at 7:25
• Could you please elaborate on the "within a particular subsystem" part? Seems to me it's key for my confusion. (Also I think your first sentence is not correct, because a system could lose heat AND generate entropy such that $\Delta S = 0$. My book has such an example if you'd like.) – André Chalella Sep 17 '14 at 12:33

The $Q=H$ refers to the system itself, the $2.3MJ$ latent heat rejected by the $1kg$ steam as it condenses to liquid at $100C$ is the same heat that would be measured if the surroundings were at $99.9999999999C$ that process being effectively reversible, and therefore then we would have $\Delta S=0$. But since $Q$ is rejected to the environment that is at $T_0=25C$ the same heat drops in quality by increasing the entropy by $\frac{Q}{T_0}$. The entropy is not generated in the steam/liquid, rather it is generated as $Q$ is dissipated in the surrounding while it passes the liquid to environment interface from $100C$ to $25C$.

• Thanks, that was a good answer. I apologize but I was in the middle of writing an answer myself, for which I made a nice drawing and which I will elaborate a bit further. – André Chalella Sep 18 '14 at 5:48

As @DavidVercauteren hinted in his comment and @hyportnex pointed out in his answer (albeit with other words), the problem here is that the process undergone by the system is in fact not irreversible: the irreversibilities occur elsewhere.

There is a missing piece to the puzzle, which, after some thought, I present below: The key here is that the irreversibilities occur where heat flows through temperature differences, which is NOT in the system, since it is isothermal. Nor is it in the surroundings, which are also isothermal. So there must be a region where temperature transitions from $T$ to $T_0$.

(Damn book could have pointed that out.)

Sure enough, we can model this region as a control volume without any matter flowing through its boundaries — only heat, which doesn't accumulate inside it either. It is apparent that the matter inside the control volume doesn't change its state, so steady state it is. The Second Law for this control volume, then, is:

$$\Delta S = \frac{Q}{T} - \frac{Q}{T_0} + S_\text{gen} = 0$$

Now it's clear that $S_\text{gen} > 0$ (that is, entropy must be generated) since $T_0 < T$.

So we found the answer to the question. Now, it is said that whenever entropy is generated, an opportunity of doing work has been lost. Let's quickly analyze that.

If some substance were to flow inside our control volume, receive heat $Q$, do work, reject heat $Q_0$, receive work until it goes back to its initial state (much smaller than the other one) and then get out, the First Law would tell us the net work is:

$$W = Q - Q_0$$

But we know that the rejected heat is a function of generated entropy. In fact, the more $S_\text{gen}$, the more $Q_0$.

$$\Delta S = \frac{Q}{T} - \frac{Q_0}{T_0} + S_\text{gen} = 0 \\ \Rightarrow Q_0 = Q \frac{T_0}{T} + T_0 S_\text{gen}$$

So we have the following expression for work, which clearly shows that, the more irreversible the process is, the less work we get:

$$W = Q \left(1 - \frac{T_0}{T}\right) - T_0 S_\text{gen}$$

In our original process, $S_\text{gen}$ was maximum, such that $Q_0 = Q$ and then no work was done. In an ideal process, $S_\text{gen}$ would be zero, such that $Q_0$ would be minimum (not zero — it's a function of $T_0$) and then work done would be maximum.