ENGR2000_chapter19_problems

ENGR2000_chapter19_problems - 1496T_c19_W1-W18 3/1/06 11:52...

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Questions and Problems W15 Thermal Stresses Thermal stresses, which are introduced in a body as a consequence of temperature changes, may lead to fracture or undesirable plastic deformation. The two prime sources of thermal stresses are restrained thermal expansion (or contraction) and temperature gradients established during heating or cooling. Thermal shock is the fracture of a body resulting from thermal stresses induced by rapid temperature changes. Because ceramic materials are brittle, they are especially susceptible to this type of failure. The thermal shock resistance of many materials is proportional to the fracture strength and thermal conductivity, and inversely proportional to both the modulus of elasticity and the coefficient of ther- mal expansion. IMPORTANT TERMS AND CONCEPTS Heat capacity Linear coefficient of thermal expansion Phonon Specific heat Thermal conductivity Thermal shock Thermal stress REFERENCES Kingery, W. D., H. K. Bowen, and D. R. Uhlmann, Introduction to Ceramics, 2nd edition, Wiley, New York, 1976. Chapters 12 and 16. Rose, R. M., L. A. Shepard, and J. Wulff, The Struc- ture and Properties of Materials, Vol. IV, Electronic Properties, Wiley, New York, 1966. Chapters 3 and 8. Ziman, J., “The Thermal Properties of Materials,” Scientific American, Vol. 217, No. 3, September 1967, pp. 180–188. QUESTIONS AND PROBLEMS Heat Capacity 19.1 Estimate the energy required to raise the temperature of 5 kg (11.0 lb m ) of the fol- lowing materials from 20 to (68 to ): aluminum, brass, aluminum oxide (alumina), and polypropylene. 19.2 To what temperature would 10 lb m of a brass specimen at be raised if 65 Btu of heat is supplied? 19.3 (a) Determine the room temperature heat capacities at constant pressure for the fol- lowing materials: copper, iron, gold, and nickel. (b) How do these values compare with one another? How do you explain this? 19.4 For copper, the heat capacity at constant volume at 20 K is 0.38 J/mol-K, and the Debye temperature is 340 K. Estimate the specific heat (a) at 40 K and (b) at 400 K. C v 25 ± C (77 ± F) 300 ± F 150 ± C 19.5 The constant A in Equation 19.2 is where R is the gas constant and is the Debye temperature (K). Estimate for aluminum, given that the specific heat is 4.60 J/kg-K at 15 K. 19.6 (a) Briefly explain why rises with in- creasing temperature at temperatures near 0K . (b) Briefly explain why becomes vir- tually independent of temperature at tem- peratures far removed from 0 K.
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This note was uploaded on 10/24/2009 for the course ENGR 2000 taught by Professor Goeser during the Fall '08 term at Armstrong State University.

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ENGR2000_chapter19_problems - 1496T_c19_W1-W18 3/1/06 11:52...

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