Entropy change in irreversible heat flow

For an irreversible heat flow, from an object $A$ at temperature $T_A$ to another object $B$ at temperature $T_B < T_A$, I'd like to know how to evaluate the change in entropy using the following expression:

$$\Delta S =\sum_{i} \frac {\Delta Q_i}{T_i}$$

I know it is an irreversible adiabatic process, thus I have to select a reversible transformation which proceeds from the initial to the final state in order to find the entropy change with respect to $A$, and then sum it with the change in entropy of $B$ ($\Delta S = \Delta {S_a} + \Delta {S_b}$), but I don't know how to calculate this expression.

If someone could explain generically how it can be done, I'd appreciate it.

• This question is dealt with in great detail by RICHARD C. TOLMAN, PAUL C. FINE: "On the Irreversible Production of Entropy," REV MOD PHYS VOLUME 20, NUMBER 1 JANUARY, 1948 – hyportnex Nov 26 '14 at 19:27

If it is adiabatic then $\Delta Q_i$ will be always zero. The fact that it is irreversible doesnt matter. Any path that thakes you from A to B will result in the same change of entropy, as both initial and final states are in equilibrium. If you choose what is called a quasistatic path, which is idealized as a tranformation that occurrs slow enough so that each intermediate state is in equilibrium, then you can integrate your equation and the result will be as expected: $\Delta S =0$ (you expected that because it is an adiabatic process)
The problem you are trying to solve is different from the standard textbook situation where $A$ and $B$ are two heat reservoirs. When considering equilibration of temperatures, you have to account for the finite heat capacity of the two objects. This allows you to consider a series of quasi-static reversible heat exchanges with a series of heat reservoirs, and calculate the total entropy change from there.