9.2 At 500C (930F), what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu? Solution (a) From Figure 9.7, the maximum solubility of Cu in Ag at 500C corresponds to the position of the –( + ) phase boundary at this temperature, or to about 2 wt% Cu. (b) From this same figure, the maximum solubility of Ag in Cu corresponds to the position of the –( + ) phase boundary at this temperature, or about 1.5 wt% Ag. 9.5 Consider a specimen of ice that is at 210C and 1 atm pressure. Using Figure 9.2, the pressure–temperature phase diagram for H2O, determine the pressure to which the specimen must be raised or lowered to cause it (a) to melt, and (b) to sublime. Solution The figure below shows the pressure-temperature phase diagram for H2O, Figure 10.2; a vertical line has been constructed at -10C, and the location on this line at 1 atm pressure (point B) is also noted. (a) Melting occurs, (by changing pressure) as, moving vertically (upward) at this temperature, we cross the Ice-Liquid phase boundary. This occurs at approximately 570 atm; thus, the pressure of the specimen must be raised from 1 to 570 atm. (b) In order to determine the pressure at which sublimation occurs at this temperature, we move vertically downward from 1 atm until we cross the Ice-Vapor phase boundary. occurs at approximately 0.0023 atm. This intersection 9.7 Given here are the solidus and liquidus temperatures for the germanium-silicon system. Construct the phase diagram for this system and label each region. Composition Solidus Liquidus (wt% Si) Temperature Temperature (°C) (°C) 0 938 938 10 1005 1147 20 1065 1226 30 1123 1278 40 1178 1315 50 1232 1346 60 1282 1367 70 1326 1385 80 1359 1397 90 1390 1408 100 1414 1414 Solution The germanium-silicon phase diagram is constructed below. 9.15 A 1.5-kg specimen of a 90 wt% Pb–10 wt% Sn alloy is heated to 250C (480F); temperature it is entirely an -phase solid solution (Figure 9.8). at this The alloy is to be melted to the extent that 50% of the specimen is liquid, the remainder being the phase. This may be accomplished either by heating the alloy or changing its composition while holding the temperature constant. (a) To what temperature must the specimen be heated? (b) How much tin must be added to the 1.5-kg specimen at 250C to achieve this state? Solution (a) Probably the easiest way to solve this part of the problem is by trial and error--that is, on the Pb-Sn phase diagram (Figure 9.8), moving vertically at the given composition, through the + L region until the tie-line lengths on both sides of the given composition are the same. This occurs at approximately 295C (560F). (b) We can also produce a 50% liquid solution at 250C, by adding Sn to the alloy. At 250C and within the + L phase region C = 14 wt% Sn-86 wt% Pb CL = 34 wt% Sn-66 wt% Pb Let C0 be the new alloy composition to give W = WL = 0.5. Then, W = 0.5 = CL C0 34 C0 = CL C 34 14 And solving for C0 gives 24 wt% Sn. Now, let mSn be the mass of Sn added to the alloy to achieve this new composition. The amount of Sn in the original alloy is (0.10)(1.5 kg) = 0.15 kg Then, using a modified form of Equation 4.3 0.15 kg m Sn 100 = 24 1.5 kg mSn And, solving for mSn (the mass of tin to be added), yields mSn = 0.276 kg. 9.27 A 45 wt% Pb–55 wt% Mg alloy is rapidly quenched to room temperature from an elevated temperature in such a way that the high-temperature microstructure is preserved. This microstructure is found to consist of the α phase and Mg 2Pb, having respective mass fractions of 0.65 and 0.35. Determine the approximate temperature from which the alloy was quenched. Solution We are asked to determine the approximate temperature from which a 45 wt% Pb-55 wt% Mg alloy was quenched, given the mass fractions of and Mg2Pb phases. We can write a lever-rule expression for the mass fraction of the phase as W = 0.65 = CMg 2Pb C0 CMg 2Pb C The value of C0 is stated as 45 wt% Pb-55 wt% Mg, and CMg Pb is 81 wt% Pb-19 wt% Mg, which is 2 independent of temperature (Figure 9.20); thus, 0.65 = 81 45 81 C which yields C = 25.6 wt% Pb The temperature at which the –( + Mg2Pb) phase boundary (Figure 9.20) has a value of 25.6 wt% Pb is about 360C (680F). 9.28 Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases. Solution Upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum.