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Unformatted text preview: p# dT = q1373 K "2300 K = 1373 J $ (2 moles)(44.8 mole % K )dT = 3,226J 1373 q" #$ = n%H" #$ = (2)(1, 800) = 3, 600 J
We hav e 11,701 J of heat yet to use up. We determine the final temperat ure by integrating to us e up
the ‘left over’ heat in the delt a phase: ! Tf q1409 K "T f = $ nC 1409 Tf #
p dT = J $ (2 moles)(47.3 mole % K )dT = 11, 701J 1409 94.6Tf & 133, 291J = 11, 701J
' Tf = 1535K
b. Because this proc ess occurs at constant pressure, we immediately know: ! " dq % " (H %
C p = $ rev ' = $ '
# dT & p # (T & p
dqrev = dH
qrev = )H
* )H = 100, 698 J c. We det ermine the total entropy change in a manner similar to the integrating procedure us ed
in part (a): ! First, the appropriate relation to get us started is: 3.012 PS 2 2 of 6 10/4/05 " dq %
" (S %
C p = $ rev ' = T$ '
# dT & p
# (T & p
)S = * p dT
dS = Note that the last equality and integral hold only for regions where no phase transition occurs.
Using this ex pression, we integrat e over the temperatur...
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This document was uploaded on 12/21/2013.
- Fall '13