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.