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60. (a) Starting from
0
Q
=
¦
(for calorimetry problems) we can derive (when no phase
changes are involved)
111
2 22
11
2 2
+
=
= 44.2 C,
+
f
cmT cmT
T
cm
−°
which is equivalent to 229 K.
(b) From Eq. 201, we have
()
(
)
229
tungsten
303
229
=
= 134 0.045 ln
= 1.69 J/K.
303
cmdT
S
T
§·
Δ−
¨¸
©¹
³
(c) Also,
(
)
229
silver
153
229
=
= 236 0.0250 ln
= 2.38 J/K.
153
S
T
Δ
³
(d) The net result for the system is (2.38 – 1.69) J/K = 0.69 J/K. (Note: these calculations
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 Spring '08
 Any
 Physics

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