Chapter 3

Ifweassumethatattemperaturetthe

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Unformatted text preview: s to have the same mass (1000 g)  and let us consider the molar heat capacity to be 24.44 J.mol‐1.K‐1.  If the  initial temperature of the blocks are respectively, 0°C and 100°C, then the  final temperature, TF = 50°C (see above). The entropy change for the two  blocks will be:  Marand’s Notes: Chapter 3 ‐ The Second Law of Thermodynamics  104  T m T mA CPm (A) ln F + B CPm (B) ln F MA TA MB TB   323 1000 323 −1 ΔS = (24.44)ln + ln = 5.93 J.K 100 373 273 ΔS = ΔS(A) + ΔS(B) = Note that since ΔSSurroundings = 0, then ΔSUniverse = ΔS. Therefore ΔSUniverse is  € positive, in accord with the requirement that ΔSUniverse  > 0 for a  spontaneous process. Indeed, heat transfer and temperature equilibration  processes are spontaneous.    Example #2:  Consider what happens when you mix under a pressure of 1 bar, 500 g of  ice at ‐10°C with 100 g of H2O vapor at +110°C.  Assume the mixing takes  place in a thermally insulated container, so that the heat transfer occurs  exclusively between the ice and the vapor. What is the final state of the  system and what is the change in entropy for the process (system +  surroundings)?   Here, we will assume that the final state of the system is liquid water at  some temperature TF (between 0 and 100oC). The latent heat of  vaporization (standard enthalpy change for vaporization) is equal to   44 kJ/mol and that for the fusion is 6 kJ/mol. These processes occur  Marand’s Notes: Chapter 3 ‐ The Second Law of Thermodynamics  105  reversibly at 1 bar at 373 K and 273 K, respectively.  The molar heat  capacity at constant pressure of wat...
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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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