ps02_sol

The calculation is readily carried out by simply

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Unformatted text preview: several phase transitions present. The calculation is readily carried out by simply determining how much heat is needed for heating to each phase transition: Starting out at 298 K, we first heat the sample (α phase at room temperat ure) to the first phase transition α -&gt; β, at 993K: 993 q298 K &quot;993 K = # nC p dT = 298 3.012 PS 2 ! 993 # (2 moles)(21.6 + 15.9 \$ 10 298 1 of 6 %3 T J ) dT = 44, 290 J mole &amp; K 10/4/05 …clearly, we have more heat to use, so we continue. The phas e trans ition c onsumes a s mall amount of heat: q298 K &quot;993 K = n#H\$ &quot;% = (2)(2, 010) = 4, 020 J We continue summing up the heat for the next few transitions in a similar manner: 1373 ! q993 K &quot;1373 K = 1373 \$ nC # p dT = 993 J \$ (2 moles)(34.9 + 0.0028T mole % K )dT = 29,041J 993 q&quot; #\$ = n%H &quot; #\$ = (2)(2, 300) = 4, 600 J Our running total for heat transferred to reach the gamma phase ast 1373K is now 81,951 J. This ! leav es 18,747 J of heat yet to be c onsumed. Cont inuing on to the next trans ition: ! 1409 1409 \$ nC...
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