ConceptCheck18

ConceptCheck18 - 674 Concept Check 18.1 Question If a...

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674 Concept Check 18.1 Question: If a metallic material is cooled through its melting temperature at an extremely rapid rate, it will form a noncrystalline solid (i.e., a metallic glass). Will the electrical conductivity of the noncrystalline metal be greater or less than its crystalline counterpart? Why? Answer: The electrical conductivity for a metallic glass will be less than for its crystalline counterpart. The glass will have virtually no periodic atomic structure, and, as a result, electrons that are involved in the conduction process will experience frequent and repeated scattering. (There is no electron scattering in a perfect crystal lattice of atoms.)
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676 Concept Check 18.2 Question: The room-temperature electrical resistivities of pure lead and pure tin are 2.06 x 10 -7 and 1.11 x 10 -7 Ω -m, respectively. (a) Make a schematic graph of the room-temperature electrical resistivity versus composition for all compositions between pure lead and pure tin. (b) On this same graph schematically plot electrical resistivity versus composition at 150 ° C. (c) Explain the shapes of these two curves, as well as any differences between them. Hin t: You may want to consult the lead-tin phase diagram, Figure 9.8 Answers: (a) and (b) Below is shown the electrical resistivity versus composition for lead-tin alloys at both room temperature and 150 ° C. (c) Upon consultation of the Pb-Sn phase diagram (Figure 9.8) we note upon extrapolation of the two solvus lines to at room temperature (e.g., 20 ° C), that the single phase α phase solid solution exists between pure lead and a composition of about 2 wt% of Sn-98 wt% Pb. In addition, the composition range over which the β phase is stable is between approximately 99 wt% Sn-1 wt% Pb and pure tin. Within both of these composition regions the resistivity increases in accordance with Equation 18.11; also, in the above plot, the resistivity of pure Pb is represented (schematically) as being greater than that for pure Sn, per the problem statement.
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Furthermore, for compositions between these extremes, both α and β phases coexist, and alloy resistivity will be a function of the resisitivities the individual phases and their volume fractions, as described by Equation 18.12. Also, mass fractions of the α and β phases within the two-phase region of Figure 9.8 change linearly with changing composition (according to the lever rule). There is a reasonable disparity between the densities of Pb and Sn (11.35 g/cm 3 versus 7.3 g/cm 3 ). Thus, according to Equation 9.6 phase volume fractions will not exactly equal mass
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ConceptCheck18 - 674 Concept Check 18.1 Question If a...

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