# E90mtm - precipitation becomes thermodynamically possible(a...

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Engineering 45 The Structure and Properties of Materials Midterm Exam October 31, 1990 Problem 1: (a) Describe how the Cu 3 Au, NaCl and ß-ZnS structures are derived from FCC. (b) Binary compounds with the Cu 3 Au structure are invariably metals, those with the ß-ZnS structure are usually semiconductors, and those with the NaCl structure are usually insulators. Why might you expect this behavior? Problem 2: The Debye temperature of copper is about 315 K, while that of diamond is nearly 2000 K. (a) At room temperature the specific heat of copper is much greater than that of diamond. Why? (b) At room temperature the thermal conductivity of diamond is greater than that of copper. Why? Problem 3: The kinetics of precipitation of a phase ß from a polygranular α phase are described by the kinetic diagram given below, where τ is the time required to initiate the transformation and T is the undercooling below the temperature at which ß

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Unformatted text preview: precipitation becomes thermodynamically possible. (a) If the material is cooled as indicated by the upper arrow, the final microstructure contains nuclei of ß almost exclusively in the grain boundaries of the α grains. Why might you expect this? Page 1 Engineering 45: Fall, 1990 Midterm Exam (b) If the material is cooled and then heated, as indicated by the lower path in the figure, the final microstructure consists of a dense distribution of ß precipitates in the interiors of the α grains. Why? Problem 4: Explain the following observations concerning the martensitic transformation in a particular steel: (a) The martensite transformation initiates at a given value of the temperature, M s , irrespective of the cooling rate. (b) The martensite transformation is not completed until the steel is cooled to a much lower temperature, M f , irrespective of the cooling rate. Page 2...
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E90mtm - precipitation becomes thermodynamically possible(a...

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