Change in Entropy for an isolated system

[Pouyan]

Homework Statement

Two systems, A and B, have the total constant heating capacities CA and CB as well as the initial temperatures TA and TB. They are put in contact with each other so that heat can be exchanged in a slow process without heat exchange with the surroundings. In the final state, both systems have assumed the same temperature. Determine the total entropy change in this process and show that it is positive.

Relevant Equations

dQ=TdS
dU=CdT

ΔU_A + ΔU_B = 0 (Is this because of isolated system am I right?)

ΔU_A = CA * (T_final - T_A )
ΔU_B=CB * (T_final-T_B)

And because of a very slow process : S=ln(T)
T_final= (CA T_A + CB T_B)/(CA + CB)

ΔS_final = CA*ln(T_f/TA) + ln(T_f/TB) * CB

My QUESTION is :

When we say No heat exchange with the surroundings or isolated system, do we mean ΔU=0 ?

 

[Chestermiller]

Yes. Isolated means no work and no external heat exchange.