P21_054 - . 37 − 1 . 69 = 0 . 68 J/K. (Note: these...

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54. (a) Starting from Q = 0 (for calorimetry problems) we can derive (when no phase changes are involved) T f = c 1 m 1 T 1 + c 2 m 2 T 2 c 1 m 1 + c 2 m 2 = 44 . 2 C , which is equivalent to 229 K. (b) From Eq. 21-1, we have S tungsten = Z 229 303 cmdT T = (134)(0 . 045) ln µ 229 303 = 1 . 69 J / K . (c) Also, S silver = Z 229 153 cmdT T = (236)(0 . 025) ln µ 229 153 =2 . 37 J / K . (d) The net result for the system is 2
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Unformatted text preview: . 37 − 1 . 69 = 0 . 68 J/K. (Note: these calculations are fairly sensitive to round-o± errors. To arrive at this ²nal answers, the value 273 . 15 was used to convert to Kelvins, and all intermediate steps were retained to full calculator accuracy.)...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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