Change in Entropy for an isolated system

  • Thread starter Pouyan
  • Start date
  • #1
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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 ?
 

Answers and Replies

  • #2
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4,296
Yes. Isolated means no work and no external heat exchange.
 

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