ENGR2000_chapter19_problems - 1496T_c19_W1-W18 11:52 Page...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 CONCEPT S 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. QUES TIONS 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 0 K. (b) Briefly explain why becomes vir- tually independent of temperature at tem- peratures far removed from 0 K.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern