9.5 point out the phases that are present and also the step compositions for the complying with alloys: (a) 90 wt% Zn–10 wt% Cu at 400_C(750_F). (b) 75 wt%Sn–25wt%Pb in ~ 175_C (345_F). (c) 55 wt% Ag–45 wt% Cu in ~ 900_C(1650_F). (d) 30 wt% Pb–70 wt% Mg at 425_C(795_F). (e) 2.12 kg Zn and also 1.88 kg Cu in ~ 500_C(930_F). (f ) 37 lbm Pb and also 6.5 lbm Mg at 400_C(750_F). (g) 8.2 mol Ni and also 4.3 mol Cu in ~ 1250_C (2280_F). (h) 4.5 mol Sn and also 0.45 mol Pb at 200_C(390_F). 9.5 This trouble asks the we mention the phase or phases current for several alloys at mentioned temperatures. (a) because that an alloy composed of 15 wt% Sn-85 wt% Pb and at 100C, from figure 9.7,  and  phases are present, and C = 5 wt% Sn-95 wt% Pb C = 98 wt% Sn-2 wt% Pb (b) because that an alloy composed of 25 wt% Pb-75 wt% Mg and at 425C, from number 9.18, just the  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 also at 800C, from figure 9.6,  and liquid phases are present, and also C = 92 wt% Ag-8 wt% Cu CL = 77 wt% Ag-23 wt% Cu (d) for an alloy written of 55 wt% Zn-45 wt% Cu and at 600C, from number 9.17,  and also  phases are present, and C = 51 wt% Zn-49 wt% Cu C = 58 wt% Zn-42 wt% Cu (e) because that an alloy composed of 1.25 kg Sn and 14 kg Pb and at 200C, us must first determine the Sn and Pb concentrations, as CSn =1.25 kg  100 = 8.2 wt% 1.25 kg  14 kgCPb =14 kg  100 = 91.8 wt% 1.25 kg  14 kgFrom figure 9.7, only the  phase is present; its composition is 8.2 wt% Sn-91.8 wt% Pb. (f) for an alloy written of 7.6 lbm Cu and also 114.4 lbm Zn and at 600C, us must very first determine the Cu and Zn concentrations, as CCu =C Zn =7.6 lb m  100 = 5.0 wt% 7.6 lb m  144.4 lbm 144.4 lb m  100 = 95.0 wt% 7.6 lb m  144.4 lb mFrom number 9.17, just the L phase is present; its composition is 95.0 wt% Zn-5.0 wt% Cu (g) for an alloy written of 21.7 mol Mg and also 35.4 mol Pb and at 350C, that is an initial necessary to determine the Mg and Pb concentrations, which we will do in weight percent together follows: \" mPb = nm APb = (35.4 mol)(207.2 g/mol) = 7335 g Pb\" mMg = nm AMg = (21.7 mol)(24.3 g/mol) = 527 g MgCPb =7335 g  100 = 93 wt% 7335 g  527 gCMg = 100 wt%  93 wt% = 7 wt%From number 9.18, L and also Mg2Pb phases space present, and also CL  94 wt % Pb  6 wt% Mg CMg Pb  81 wt% Pb  19 wt% Mg 2(h) because that an alloy written of 4.2 mol Cu and also 1.1 mol Ag and at 900C, that is first necessary to recognize the Cu and also Ag concentrations, which we will do in load percent as follows: \" mCu = nmCuACu = (4.2 mol)(63.55 g/mol) = 266.9 g\" mAg = nm AAg = (1.1 mol)(107.87 g/mol) = 118.7 g AgCCu =266.9 g  100 = 69.2 wt% 266.9 g  118.7 gC Ag =118.7 g  100 = 30.8 wt% 266.9 g  118.7 gFrom figure 9.6,  and liquid phases space present; and also C = 8 wt% Ag-92 w% Cu CL = 45 wt% Ag-55 wt% Cu9.10 below is a portion of the H2O–NaCl step diagram: (a) utilizing this diagram, briefly define how dispersing salt on ice the is in ~ a temperature listed below 0oC (32oF) can cause the ice cream to melt. (b) What concentration that salt is crucial to have actually a 50% ice–50% fluid brine in ~ 10oC (14oF)?(a)Spreading salt on ice will reduced the melting temperature, due to the fact that the liquidus heat decreases native 0C come the eutectic temperature at about 21C. Thus, ice cream at a temperature below 0C (and over -21C) can be made to kind a fluid phase by the addition of salt. (b) We space asked come compute the concentration that salt vital to have a 50% ice-50% brine solution at -10C (14F). In ~ -10C, Cice = 0 wt% NaCl-100 wt% H2O C = 13 wt% NaCl-87 wt% H O brine 2 Thus, Wice = 0.5 =Cbrine  Co Cbrine  Cice=13  Co 13  0Solving for Co (the concentration of salt) yields a value of 6.5 wt% NaCl93.5 wt% H2O.9.13 take into consideration a specimen of ice ns which is in ~ -10oC and also 1 atm pressure. Using figure 9.33, the pressure–temperature phase diagram for H2O, recognize the pressure to i m sorry the specimen have to be increased or lower to reason it (a) to melt, and (b) to sublime.9.13 This difficulty asks us to consider a specimen that ice i which is at -10C and 1 atm pressure. (a) In bespeak to recognize the push at which melt occurs in ~ this temperature, we relocate vertically in ~ this temperature until we cross the ice cream I-Liquid phase border of number 9.33. This occurs at around 570 atm; therefore the push of the specimen should be elevated from 1 come 570 atm. (b) In stimulate to determine the pressure at i m sorry sublimation occurs at this temperature, we move vertically downward from 1 atm till we overcome the ice I-Vapor phase border of figure 9.33. This intersection occurs at about 0.0023 atm.9.30 A 45 wt% Pb–55 wt% Mg alloy is rapidly quenched to room temperature native an elevated temperature in together a method that the high-temperature microstructure is preserved. This microstructure is uncovered to consists the  phase and also Mg2Pb, having respective mass fountain of 0.65 and also 0.35. Identify the almost right temperature indigenous which the alloy was quenched.9.30 We room asked to determine the almost right temperature from which a Pb-Mg alloy to be quenched, provided the mass fractions of  and also Mg2Pb phases. We can write a lever-rule expression for the mass fraction of the  step asW = 0.65 =CMg Pb  Co 2 CMg Pb  C  2The value of Co is declared as 45 wt% Pb-55 wt% Mg, and also CMg Pb is 81 2 wt% Pb-19 wt% Mg, i beg your pardon is live independence of temperature (Figure 9.18); thus, 0.65 =81  45 81  Cwhich yields C = 25.6 wt% PbThe temperature in ~ which the -( + Mg2Pb) phase border (Figure 9.18) has a value of 25.6 wt% Pb is about 360C (680F).9.62 Is it feasible to have actually an iron–carbon alloy for which the mass fractions of full ferrite and pearlite space 0.860 and also 0.969, respectively? Why or why not?9.62 This problem asks if it is possible to have actually an iron-carbon alloy because that which W = 0.860 and also Wp = 0.969. In order to make this determination, the is vital to collection up lever rule expressions for these 2 mass fountain in regards to the alloy composition, then to solve for the alloy composition of each; if both alloy ingredient values space equal, then such an alloy is possible. The expression for the mass portion of full ferrite isW =CFe C  C o 3 CFe C  C 3=6.70  Co 6.70  0.022= 0.860Solving because that this Co yields Co = 0.95 wt% C. Equation (9.20) asNow because that Wp we utilizeWp =6.70  C1\" 5.94= 0.969This expression leads to C1\" = 0.95 wt% C. Because Co = C1\" , this alloy is possible.9.63 Compute the maximum mass fraction of eutectoid cementite in an iron–carbon alloy that contain 1.00 wt% C.9.63 This difficulty asks that we compute the mass portion of eutectoid cementite in one iron-carbon alloy that consists of 1.00 wt% C. In stimulate to fix this problem it is essential to compute mass fountain of total and also proeutectoid cementites, and also then to subtract the last from the former. To calculate the mass portion of total cementite, that is important to usage the lever rule and a tie line that extends across the whole  + Fe3C phase ar asCo  C 1.00  0.022 WFe C = = = 0.146 CFe C  C 6.70  0.022 3 3Now, because that the mass portion of proeutectoid cementite we use Equation (9.21)WFe C\" = 3C1\"  0.76 5.94=1.00  0.76 = 0.040 5.94And, finally, the mass fraction of eutectoid cementite WFe C\"\" is simply 3 WFe C  WFe C  WFe C   0.146  0.040  0.106 3 3 39.65 The mass fraction of eutectoid ferrite in one iron–carbon alloy is 0.71. On the communication of this information, is it feasible to identify the ingredient of the alloy? If so, what is its composition? If this is not possible, define why.9.65 This difficulty asks whether or not it is possible to identify the ingredient of one iron-carbon alloy because that which the mass fraction ofeutectoid ferrite is 0.71; and also if so, to calculate the composition. Yes, that is possible to determine the alloy composition; and, in fact, there room two possible answers. For the first, the eutectoid ferrite exists in addition to proeutectoid ferrite. Because that this instance the mass portion of eutectoid ferrite (W\"\") is simply the difference between total ferrite and proeutectoid ferrite fixed fractions; the is W \"\" = W  - W \" Now, the is feasible to compose expressions because that W and also W\" in regards to Co, the alloy composition. Thus, W\"\" ==CFe C  Co 3 CFe C  C  30.76  C o 0.746.70  C o 0.76  C o  = 0.71 6.70  0.022 0.74And, addressing for Co returns Co = 0.61 wt% C. Because that the second possibility, we have actually a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite.

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Thus, that is crucial to set up a lever rule expression inside the mass portion of full ferrite is 0.71. Therefore, W =CFe C  C o 3 CFe C  C 3=6.70  Co 6.70  0.022And, fixing for Co yields Co = 1.96 wt% C.= 0.71