Chapter 10 homework solutions (part 1).pdf

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Department
Materials Science & Engineering
Course
MSE101H1
Professor
Scott Ramsay
Semester
Spring

Description
CHAPTER 10 PHASE DIAGRAMS PROBLEM SOLUTIONS Solubility Limit 10.1 Consider the sugar–water phase diagram of Figure 10.1. (a) How much sugar will dissolve in 1000 g of water at 80°C (176°F)? (b) If the saturated liquid solution in part (a) is cooled to 20°C (68°F), some of the sugar will precipitate out as a solid. What will be the composition of the saturated liquid solution (in wt% sugar) at 20°C? (c) How much of the solid sugar will come out of solution upon cooling to 20°C? Solution (a) We are asked to determine how much sugar will dissolve in 1000 g of water at 80C. From the solubility limit curve in Figure 10.1, at 80C the maximum concentration of sugar in the syrup is about 74 wt%. It is now possible to calculate the mass of sugar using Equation 5.6 as m sugar C sugarwt%) = m  m  100 sugar water m  74 wt% = sugar  100 m sugar 1000 g Solving for sugaryields msugar= 2846 g  (b) Again using this same plot, at 20C the solubility limit (or the concentration of the saturated solution) is about 64 wt% sugar. (c) The mass of sugar in this saturated solution at 20sugar) may also be calculated using Equation 5.6 as follows: mÕ 64 wt% = sugar  100 mÕ  1000 g sugar  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. which yields a value for m' of 1778 g. Subtracting the latter from the former of these sugar concentrations sugar yields the amount of sugar that precipitated out of the solution upon coolim" ; that is sugar  m"sugar = msugar  m'sugar = 2846 g  1778 g = 1068 g  One-Component (or Unary) Phase Diagrams  10.3 Consider a specimen of ice that is at –15°C and 10 atm pressure. Using Figure 10.2, the pressure– temperature phase diagram for H O, determine the pressure to which the specimen must be raised or lowered to 2 cause it (a) to melt, and (b) to sublime. Solution The figure below shows the pressure-temperature phase diagram for H O, Figure 10.2; a vertical line has 2 been constructed at -15C, and the location on this line at 10 atm pressure (point B) is also noted. (a) Melting occurs, (by changing pressure) as, moving vertically (upward) along this line, we cross the Solid-Liquid phase boundary. This occurs at approximately 1,000 atm; thus, the pressure of the specimen must be raised from 10 to 1,000 atm. (b) In order to determine the pressure at which sublimation occurs at this temperature, we move vertically downward along this line from 10 atm until we cross the Solid-Vapor phase boundary. This intersection occurs at approximately 0.003 atm. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Interpretation of Phase Diagrams (Binary Isomorphous Systems) (Binary Eutectic Systems) (Equilibrium Diagrams Having Intermediate Phases or Compounds) 10.5 Cite the phases that are present and the phase compositions for the following alloys: (a) 25 wt% Pb–75 wt% Mg at 425°C (800°F) (b) 55 wt% Zn–45 wt% Cu at 600°C (1110°F) (c) 7.6 mb Cu and 144.4mlb Zn at 600°C (1110°F) (d) 4.2 mol Cu and 1.1 mol Ag at 900°C (1650°F) Solution (a) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425C, from Figure 10.20, only the  phase is present; its composition is 25 wt% Pb-75 wt% Mg. (b) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600C, from Figure 10.19,  and  phases are present, and C = 51 wt% Zn-49 wt% Cu  C= 58 wt% Zn-42 wt% Cu (c) For an alloy composed of 7.m lCu and 144.4 lm Zn and at 600C, we must first determine the Cu and Zn concentrations (using Equation 5.6), as 7.6 lb C = m  100 = 5.0 wt% Cu 7.6 lm  144.4 lbm C = 144.4 lbm  100 = 95.0 wt%  Zn 7.6 lb  144.4 lb m m From Figure 10.19, only the L phase is present; its composition is 95.0 wt% Zn-5.0 wt% Cu  (d) 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 5.7: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. m ' = n A = (4.2 mol)(63.55 g/mol) = 266.9 g Cu mCu Cu '  mAg = nmAg AAg = (1.1 mol)(107.87 g/mol) = 118.7 g Now, using Equation 5.6, concentrations of Cu and Ag are determined as follows:  266.9 g C = 100 = 69.2 wt% Cu 266.9 g  118.7 g 118.7 g CAg = 100 = 30.8 wt%  266.9 g  118.7 g From Figure 10.7,  a liquid phases are present; and C = 8 wt% Ag-92 w% Cu  C L 45 wt% Ag-55 wt% Cu Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 10.7 A 50 wt% Ni–50 wt% Cu alloy is slowly cooled from 1400°C (2550°F) to 1200°C (2190°F). (a) At what temperature does the first solid phase form? (b) What is the composition of this solid phase? (c) At what temperature does the liquid solidify? (d) What is the composition of this last remaining liquid phase? Solution Shown below is the Cu-Ni phase diagram (Figure 10.3a) and a vertical line constructed at a composition of 50 wt% Ni-50 wt% Cu. (a) Upon cooling form 1400C, the first solid phase forms at the temperature at which this vertical line intersects the L–( + L) phase boundary--i.e., at about 1320C. (b) The composition of this solid phase corresponds to the intersection with the L–( + L) phase boundary, of a tie line constructed across the  + L phase region at 1320C--i.e., C = 62 wt% Ni-38 wt% Cu.  (c) Complete solidification of the alloy occurs at the intersection of this same vertical line at 50 wt% Ni with the (+ L)– phase boundary--i.e., at about 1270C. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (d) The composition of the last liquid phase remaining prior to complete solidification corresponds to the intersection with the L–(+ L) boundary, of the tie line constructed across the + L phase region at 1270C--i.e., C is about 37 wt% Ni-63 wt% Cu. L Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 10.8 Determine the relative amounts (in terms of mas
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