# Increase of entropy when two reservoirs are in thermal contact

While reading my textbook , I came across a proof which intended to verify that entropy always increases when a hot reservoir is kept in thermal contact with a colder reservoir.

The proof goes as follows :

Let the temperature of reservoir A be $$T_A$$ and the temperature of reservoir B be $$T_B$$

Change in entropy of Reservoir A = $$\frac{-Q}{T_A}$$

Change in entropy of reservoir B= $$\frac{Q}{T_B}$$

Total change in entropy = $$\frac{-Q}{T_A}+\frac{Q}{T_B}$$ =$$Q(\frac{1}{T_B}-\frac{1}{T_A})$$

Since, $$T_A>T_B$$

Therefore, total change in entropy is positive. Hence, we can say that entropy of the universe increases in this particular system.

My confusion : The total entropy change is positive which clearly implies that both the reservoirs are undergoing irreversible processes. And we know that for an irreversible process,$$dS>\frac{dQ}{T}$$ Then how can they write :

change in entropy of reservoir A = $$\frac{-Q}{T_A}$$.

Shouldn't they replace '=' with '>' here?

Similarly, for reservoir B they should have written

Change in entropy of reservoir B> $$\frac{Q}{T_B}$$

Where I am going wrong with my reasoning?

Denote the transported entropy by $$S_A$$ from body $$A$$ at temperature $$T_A$$. The same entropy falls from temperature $$T_A$$ to temperature $$T_B$$ and the thermal work being dissipated is exactly $$\Delta W = (T_A-T_B)S_A$$ at the lower temperature $$T_B$$ and thereby generating in the heat conducting interface entropy in the amount of $$\Delta S = S_B-S_A>0$$ where $$S_B$$ is the amount of entropy entering body $$B$$ at temperature $$T_B$$. But the dissipated work is also $$\Delta W = T_B \Delta S$$ because $$T_BS_B=T_AS_A$$ by the formal constitutive definition of a heat conductor, an equality that holds at every cross section of a conductor $$T(x)S(x) = \text{constant}=Q$$ the incoming thermal energy, "heat".