Chapter 3

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Unformatted text preview: ‐B > 0 if TB > TA) and   qB‐A is the heat lost by B as it flows into A (qB‐A < 0 if TB > TA). If we denote by  MA and MB the molar mass of A and B, and by CPm(A) and CPm(B) the molar  heat capacities at constant pressure for A and B, respectively, and if we  note that the heat is exchanged at constant pressure, then:  qA‐B = ΔH(A) = (mA/MA) CPm(A) (TF ‐ TA) and   qB‐A = ΔH(B) = (mB/MB) CPm(B) (TF ‐ TB)  This allows us to define TF by:  mA m CPm (A) TA + B CPm (B) TB M MB TF = A   mA mB CPm (A) + CPm (B) MA MB Note that, if A and B are made of the same material and have the same  € mass, then the final temperature, TF, is half way between the initial  temperatures, TA and TB.  Marand’s Notes: Chapter 3 ‐ The Second Law of Thermodynamics  103  Now, we know the characteristics of the final state (its temperature) and  we can imagine a path leading from the initial state to the final state, which  is reversible. A reversible path could be a constant pressure slow heating of  A from TA to TF and a constant pressure slow cooling of B from TB to TF. For  such processes, the changes in entropy are equal to:  T mA CPm (A) ln F MA TA   TF m ΔS(B) = B CPm (B) ln MB TB ΔS(A) = The change in entropy for the surroundings is zero because the process was  € adiabatic (total heat is zero). Therefore the total change in entropy for  system + surroundings is equal to the sum of the entropy change for A and  the entropy change for B.  Let us take, as an example A and B made of the same material of molar  mass 100 g/mol. Let us consider the block...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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