Change in Entropy for an isolated system

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In an isolated system, the internal energy changes of two subsystems, ΔU_A and ΔU_B, sum to zero, indicating no heat exchange with the surroundings. The final temperature, T_final, is calculated as a weighted average based on the specific heat capacities and initial temperatures of the subsystems. The change in entropy, ΔS_final, is derived from the logarithmic relationship of final and initial temperatures. The concept of an isolated system implies that both heat and work exchanges with the environment are absent. Thus, it is confirmed that ΔU=0 in such scenarios.
Pouyan
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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 ?
 
Physics news on Phys.org
Yes. Isolated means no work and no external heat exchange.
 

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