THERMAL PROPERTIES
PROBLEMS
19.4 For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/mol-K, and the Debye
temperature is 375 K. Estimate the specific heat (a) at 50 K and (b) at 425 K.
Solution
(a) For aluminum, Cv at 50 K may be approximated by Equation 19.2, since this temperature is
significantly below the Debye temperature (375 K). The value of Cv at 30 K is given, and thus, we may
compute the constant A as
A =
Cv
0.81 J / mol - K
=
= 3.00  10 -5 J/mol - K 4
(30 K) 3
T3
Therefore, at 50 K
C v = AT 3 = (3.00  10 -5 J/mol - K 4 ) (50 K) 3 = 3.75 J/mol - K
and
cv = (3.75 J/mol - K)(1 mol/26.98 g)(1000 g/kg) = 139 J/kg - K
(b) Since 425 K is above the Debye temperature, a good approximation for Cv is
C v = 3R
= (3)(8.31 J/mol - K) = 24.9 J/mol - K
And, converting this to specific heat
cv = (24.9 J/mol - K)(1 mol/26.98 g)(1000 g/kg) = 923 J/kg - K
19.12 When a metal is heated its density decreases. There are two sources that give rise to this
diminishment of ρ: (1) the thermal expansion of the solid, and (2) the formation of vacancies (Section 4.2).
Consider a specimen of copper at room temperature (20°C) that has a density of 8.940 g/cm3. (a)
Determine its density upon heating to 1000°C when only thermal expansion is considered. (b) Repeat the
calculation when the introduction of vacancies is taken into account. Assume that the energy of vacancy
formation is 0.90 eV/atom, and that the volume coefficient of thermal expansion, αv is equal to 3αl.
Solution
(a) In this portion of the problem we are asked to determine the density of copper at 1000C on
the basis of thermal expansion considerations. The basis for this determination will be 1 cm3 of material at
20C; this volume of copper has a mass of 8.940 g, which mass is assumed to remain constant upon
heating to the 1000C. Let us first compute the volume expansion of this cubic centimeter of copper as it is
heated to 1000C. According to Equation 19.4 volume expansion is equal to
V
=  v T
V0
where v, the volume coefficient of thermal expansion, as stipulated in the problem statement, is equal to
3l. The value of l given in Table 19.1 for copper is 17.0  10-6 (C)-1. Therefore, the volume of this
specimen of Cu at 1000C (V) is equal to
V = V0 + V = V0 + V0 v T  V0 (1   v T )


 V0 (1  3 l T )  V0 
1  3 l (T f  T0 )





= (1 cm3) 1  (3) 17.0  106 (C)1 (1000C  20C)
= 1.04998 cm3
Thus, the density is just the 8.940 g divided by this new volume—i.e.,
 =
8.940 g
= 8.514 g/cm3
1.04998 cm3
(b) Now we are asked to compute the density at 1000C taking into consideration the creation of
vacancies which will further lower the density. To begin, this determination requires that we calculate the
number of vacancies using Equation 4.1. But it first becomes necessary to compute the number of Cu
atoms per cubic centimeter (NCu) at 1000C using Equation 4.2. Thus,
N A Cu
ACu
N Cu =
=
(6.022  10 23 atoms / mol)(8.514
g/cm3)
63.55 g/mol
= 8.07  10 22 atoms/cm3
Now, from Equation 4.1, the total number of vacancies, Nv, is computed as
 Q 
N v = N Cu exp  v 
 kT 


0.90 eV / atom
= (8.07  10 22 atoms/cm3 ) exp 

 (8.62  105 eV / K) (1000  273 K) 
= 2.212  1019 vacancies/cm3
We now want to determine the number of vacancies per unit cell, which is possible if the unit cell volume
is multiplied by Nv. The unit cell volume (VC) may be calculated using Equation 3.5 taking n = 4 inasmuch
as Cu has the FCC crystal structure. Thus, from a rearranged form of Equation 3.5
VC =
=
nACu
Cu N A
(4 atoms/unit cell)(63.55 g/mol)
(8.514 g/cm3)(6.022  10 23 atoms / mol)
= 4.958  10 -23 cm3/unit cell
Now, the number of vacancies per unit cell, nv, is just
nv = N vVC
= (2.212  1019 vacancies/cm3 )(4.958  10 -23 cm3/unit cell)
= 0.001097 vacancies/unit cell
What this means is that instead of there being 4.0000 atoms per unit cell, there are only 4.0000 – 0.001097
= 3.998903 atoms per unit cell. And, finally, the density may be computed using Equation 3.5 taking n =
3.998903; thus
Cu =
nACu
VC N A
(3.998903 atoms/unit cell)(63.55 g/mol)
(4.958  1023 cm3/unit cell)(6.022  10 23 atoms / mol)
=
= 8.512 g/cm3
Thus, the influence of the vacancies is almost insignificant--their presence reduces the density by only
0.002 g/cm3 (from 8.514 g/cm3 to 8.512 g/cm3).
19.26 (a) If a rod of 1025 steel 0.5 m (19.7 in.) long is heated from 20 to 80°C (68 to 176°F) while its ends
are maintained rigid, determine the type and magnitude of stress that develops. Assume that at 20°C the
rod is stress free. (b) What will be the stress magnitude if a rod 1 m (39.4 in.) long is used? (c) If the rod in
part (a) is cooled from 20 to -10°C (68 to 14°F), what type and magnitude of stress will result?
Solution
(a) We are asked to compute the magnitude of the stress within a 1025 steel rod that is heated
while its ends are maintained rigid. To do this we employ Equation 19.8, using a value of 207 GPa for the
modulus of elasticity of steel (Table 6.1), and a value of 12.0  10-6 (C)-1 for l (Table 19.1). Therefore
 = E l (T0  T f )


= (207  10 3 MPa) 12.0  106 (C)1 (20C  80C)
= –150 MPa (–21,800 psi)
The stress will be compressive since its sign is negative.
(b) The stress will be the same [–150 MPa (–21,800 psi )], since stress is independent of bar
length.
(c) Upon cooling the indicated amount, the stress becomes
 = E l (T0  T f

)

= (207  10 3 MPa) 12.0  106 (C)1 (20C  (10C)
= +74.5 MPa (+10,900 psi)
This stress will be tensile since its sign is positive.