# Contradiction in definition of entropy?

I'm studying for my thermodynamics exam and I came across something which really confuses me.

An infinitesimal change in entropy $$dS_{sys}$$ of a system at temperature $$T_{sys}$$ during a reversible transformation, where $$\delta Q_{rev}$$ is defined as the heat going in/out the system is given by: $$dS_{sys} = \frac{\delta Q_{rev}}{T_{sys}}$$

However, there is a statement in my book claiming that: $$dS_{sys} > \frac{\delta Q_{rev}}{T_{surr}}$$

My confusion is the following: If an infinitesimal change in entropy of a system at temperature $$T_{sys}$$ is defined as above, how can the statement $$dS_{sys} > \frac{\delta Q_{rev}}{T_{surr}}$$ be true? In order to calculate the change in entropy the path must be reversible, meaning that the temperature of the system is equal to the temperature of the surroundings, i.e. $$T_{sys} = T_{surr}$$ otherwise the path isn't reversible. The statement clearly doesn't hold if my reasoning is correct.

Can someone clarify this to me because I'm really struggling with this.

• Are you sure that in the inequality, the subscript of Q isn’t irrev? – Chet Miller Feb 29 at 16:41
• Like @ChetMiller pointed out, that inequality holds for irreversible processes due to the nature of not being able to extract 100% work from heat. – Weezy Feb 29 at 16:54
• Thermodynamics has lots of confusing notation - in this case, I think the $\delta Q_{\text{rev}}$ is probably meant to signify the heat if the process were reversible. I.e. the heat due to a fictitious reversible process connecting the start and end states. It is not the heat for the actual process. – jacob1729 Feb 29 at 18:08
• I have the same comment as Chet. It would save us all a lot of time if you recheck the book statement. If it is as you stated, then give us the context in which the statement was made. – Bob D Feb 29 at 20:32

For a general transformation between $$A$$ and $$B$$, the entropy change can be written:
$$dS_{A \to B} = \frac{dQ_{A \to B}}{T_{\mathrm{surr}}} + dS_{\mathrm{created}},$$
where $$dS_{\mathrm{created}} \geq 0$$ is the entropy created by irreversible processes.
For a transformation to be reversible, you need $$dS_{\mathrm{created}} = 0$$ and also $$T_{\mathrm{surr}} = T_{\mathrm{sys}}$$. In that case, $$dS = \frac{dQ}{T_{\mathrm{surr}}} = \frac{dQ}{T_{\mathrm{sys}}}$$ so the first inequality does not hold strictly.
However, for real macroscopic processes, $$dS_{\mathrm{created}}$$ is always $$>0$$, even by an infinitesimal amount (no transformation is fully reversible). In that case, the inequality becomes strict and $$dS > \frac{dQ}{T_{\mathrm{surr}}}$$. Of course it is still useful to consider adiabatic processes as they are sometimes a really good approximation to some almost reversible real transformations, and they can also be used for non-reversible process to calculate the change of entropy between two states $$A$$ and $$B$$ by considering the corresponding adiabatic path between the initial and the final states (as $$dS_{\mathrm{A\to B}}$$ does not dépend on the path followed).