4020-F2011-HW11Solutions - MSE 4020 HW11-SOLUTION...

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1 MSE 4020 HW11-SOLUTION 11/04/2011 Problem 1 Assuming dislocation line length l 0 , thus, Core energy from given equation: w core = G ! b 2 4 " ! l 0 (1) Volume of dislocation core is, V core = ! " r 2 " l 0 and total number of moles is N = V core ! Total enthalpy change from melting temperature to RT is, ! H = ! H fusion + ! H T m " RT = ! H fusion + C p #! T (2) Given information: c p = 0.38 J g ! K ; b = 0.255 nm = 0.255 ! 10 " 7 cm # T = 1084 " 25 = 1059 K G = 48 GPa = 48 ! 10 9 Pa = 48 ! 10 9 J m 3 = 48 ! 10 3 J cm 3 note unit transformation ( ) $ = 7.1 cm 3 mole M = 63.546 g mole h fusion = 13.05 KJ mole = 13050 J mole % # H fusion = N ! h fusion = V core $ ! h fusion = r 2 ! l 0 7.1 ! 13050 = 5774.3 r 2 ! l 0 % C p !# T = N ! M ! c p !# T = r 2 ! l 0 7.1 ! 63.546 ! 0.38 ! 1059 = 11315.1r 2 ! l 0 Thus,Eq.(2) is # H = 5774.3 r 2 ! l 0 + 11315.1r 2 ! l 0 = 17089.5 r 2 ! l 0
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2 Eq.(1) becomes, w core = G ! b 2 4 " ! l 0 = 48 ! 10 3 ! 0.255 ! 10 # 7 ( ) 2 4 ! 3.142 ! l 0 = 2.4837 ! 10 # 12 ! l 0 Equating two equations above, 17089.5 r 2 ! l 0 = 2.4837 ! 10 " 12 ! l 0 # r = 3.8123 ! 10 " 9 cm
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4020-F2011-HW11Solutions - MSE 4020 HW11-SOLUTION...

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