ch20-p060 - 60. (a) Starting from Q = 0 (for calorimetry...

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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. 20-1, 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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