Homework Solutions Chapter 10

Homework Solutions Chapter 10 - Homework Solutions 10.7 The...

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Homework Solutions November 17, 2005 10.7 The copper-gold phase diagram is constructed below. 10.8 This problem asks that we cite the phase or phases present for several alloys at specified temperatures. (a) For an alloy composed of 15 wt% Sn-85 wt% Pb and at 100 ° C, from Figure 10.7, α and β phases are present, and C α = 5 wt% Sn-95 wt% Pb C β = 98 wt% Sn-2 wt% Pb (b) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425 ° C, from Figure 10.18, only the α phase is present; its composition is 25 wt% Pb-75 wt% Mg. (c) For an alloy composed of 85 wt% Ag-15 wt% Cu and at 800 ° C, from Figure 10.6, β and liquid phases are present, and C β = 92 wt% Ag-8 wt% Cu C L = 77 wt% Ag-23 wt% Cu (d) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600 ° C, from Figure 10.17, β and γ phases are present, and C β = 51 wt% Zn-49 wt% Cu 34

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C γ = 58 wt% Zn-42 wt% Cu (e) For an alloy composed of 1.25 kg Sn and 14 kg Pb and at 200 ° C, we must first determine the Sn and Pb concentrations (using Equation 5.3), as C Sn = 1.25 kg 1.25 kg + 14 kg × 100 = 8.2 wt% C Pb = 14 kg 1.25 kg + 14 kg × 100 = 91.8 wt% From Figure 10.7, only the α phase is present; its composition is 8.2 wt% Sn-91.8 wt% Pb. (f) For an alloy composed of 7.6 lb m Cu and 114.4 lb m Zn and at 600 ° C, we must first determine the Cu and Zn concentrations (using Equation 5.3), as C Cu = 7.6 lb m 7.6 lb m + 144.4 lb m × 100 = 5.0 wt% C Zn = 144.4 lb m 7.6 lb m + 144.4 lb m × 100 = 95.0 wt% From Figure 10.17, only the L phase is present; its composition is 95.0 wt% Zn-5.0 wt% Cu (g) For an alloy composed of 21.7 mol Mg and 35.4 mol Pb and at 350 ° C, it is necessary to determine the Mg and Pb concentrations in weight percent. However, we must first compute the masses of Mg and Pb (in grams) using a rearranged form of Equation 5.4: m Pb ' = n m Pb A Pb = (35.4 mol)(207.2 g/mol) = 7335 g m Mg ' = n m Mg A Mg = (21.7 mol)(24.3 g/mol) = 527 g Now, using Equation 5.3, concentrations of Pb and Mg are determined as follows: C Pb = 7335 g 7335 g + 527 g × 100 = 93 wt% 35
C Mg = 527 g 7335 g + 527 g × 100 = 7 wt% From Figure 10.18, L and Mg 2 Pb phases are present, and C L = 94 wt% Pb - 6 wt% Mg C Mg 2 Pb = 81 wt% Pb - 19 wt% Mg (h) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag and at 900 ° C, it is necessary to determine the Cu and Ag concentrations in weight percent. However, we must first compute the masses of Cu and Ag (in grams) using a rearranged form of Equation 5.4: m Cu ' = n m Cu A Cu = (4.2 mol)(63.55 g/mol) = 266.9 g m Ag ' = n m Ag A Ag = (1.1 mol)(107.87 g/mol) = 118.7 g Now, using Equation 5.3, concentrations of Cu and Ag are determined as follows: C Cu = 266.9 g 266.9 g + 118.7 g × 100 = 69.2 wt% C Ag = 118.7 g 266.9 g + 118.7 g × 100 = 30.8 wt% From Figure 10.6, α and liquid phases are present; and C α = 8 wt% Ag-92 w% Cu C L = 45 wt% Ag-55 wt% Cu 10.11 Upon heating a lead-tin alloy of composition 30 wt% Sn-70 wt% Pb from 150 ° C and utilizing Figure 10.7: 36

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(a) The first liquid forms at the temperature at which a vertical line at this composition intersects the eutectic isotherm--i.e., at 183 ° C. (b) The composition of this liquid phase corresponds to the intersection with the ( α + L )– L phase boundary, of a tie line constructed across the α + L phase region just above this eutectic isotherm--i.e., C L = 61.9 wt% Sn.
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