Practice Questions - Set # 6
MECH 221
November 2013
th
Note: These questions will be discussed during the tutorial sessions on November the 29 ..
Question 1
A cylindrical metal wire 2 mm in diameter is required to carry a current of 10 A with a minimum of 0.03
V drop per 300 mm of wire. Which of the metals and alloys listed in Table 18.1 are possible candidates?
Solution
Using Equations 18.3 and 18.4, let us determine the minimum conductivity required, and then select from
Table 18.1, those metals that have conductivities greater than this value. Combining Equations 18.3 and
18.4, the minimum conductivity is just:
σ=
=
1 Il
=
=
ρ VA
Il
 d 2
Vπ  
2
(10 A)(300 × 10−3 m)
 2 × 10−3 m 2
(0.03 V) (π) 

2


= 3.2 × 10 7 (Ω - m) -1
Thus, from Table 18.1, only aluminum, gold, copper, and silver are candidates.
Question 2
For each of the following pairs of semiconductors, decide which will have the smaller band gap energy,
Eg, and then cite the reason for your choice. (a) ZnS and CdSe, (b) Si and C (diamond), (c) Al2O3 and
ZnTe, (d) InSb and ZnSe, and (e) GaAs and AlP.
Solution
(a) Cadmium selenide will have a smaller band gap energy than zinc sulfide. Both are II-VI compounds,
and Cd and Se are both lower vertically in the periodic table (Figure 2.6) than Zn and S. In moving from
top to bottom down the periodic table, Eg decreases.
(b) Silicon will have a smaller band gap energy than diamond since Si is lower in column IVA of the
periodic table than is C.
(c) Zinc telluride will have a smaller band gap energy that aluminum oxide. There is a greater disparity
between the electronegativities for aluminum and oxygen [1.5 versus 3.5 (Figure 2.7)] than for zinc and
tellurium (1.6 and 2.1). For binary compounds, the larger the difference between the electronegativities
of the elements, the greater the band gap energy.
(d) Indium antimonide will have a smaller band gap energy than zinc selenide. These materials are III-V
and II-VI compounds, respectively; Thus, in the periodic table, In and Sb are closer together horizontally
than are Zn and Se. Furthermore, both In and Sb reside below Zn and Se in the periodic table.
(e) Gallium arsenide will have a smaller band gap energy than aluminum phosphide. Both are III-V
compounds, and Ga and As are both lower vertically in the periodic table than Al and P.
Question 3
Germanium to which 5 × 1022 m-3 Sb atoms have been added is an extrinsic semiconductor at room
temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier
exists for each Sb atom). (a) Is this material n-type or p-type? (b) Calculate the electrical conductivity of
this material, assuming electron and hole mobilities of 0.1 and 0.05 m2/V-s, respectively.
Solution
(a) This germanium material to which has been added 5 × 1022 m-3 Sb atoms is n-type since Sb is a donor
in Ge. (Antimony is from group VA of the periodic table--Ge is from group IVA.)
(b) Since this material is n-type extrinsic, Equation 18.16 is valid. Furthermore, each Sb will donate a
single electron, or the electron concentration is equal to the Sb concentration since all of the Sb atoms are
ionized at room temperature; that is n=5×1022 m-3 and as given in the problem statement, µe = 0.1 m2/V-s.
Thus:
𝜎 = 𝑛|𝑒|𝜇𝑒 = (5 × 1022 𝑚−3 )(1.62 × 10−19 𝐶)(0.1 𝑚2 ⁄𝑉𝑠) = 800 (Ω. 𝑚)−1
Question 4
A 0.1 m (3.9 in.) rod of a metal elongates 0.2 mm (0.0079 in.) on heating from 20 to 100°C (68 to 212°F).
Determine the value of the linear coefficient of thermal expansion for this material.
Solution
The linear coefficient of thermal expansion for this material may be determined using a rearranged form
of Equation 19.3b as
αl =
∆l
∆l
0.2 × 10−3 m
=
=
l0 ∆T
l0 (T f − T0 ) (0.1 m)(100°C − 20°C)
= 25.0 × 10 -6 (°C) -1
Question 5
To what temperature must a cylindrical rod of tungsten 10.000 mm in diameter and a plate of 316
stainless steel having a circular hole 9.988 mm in diameter have to be heated for the rod to just fit into the
hole? Assume that the initial temperature is 25°C.
Solution
This problem asks for us to determine the temperature to which a cylindrical rod of tungsten 10.000 mm
in diameter must be heated in order for it of just fit into a 9.988 mm diameter circular hole in a plate of
316 stainless steel, assuming that the initial temperature is 25°C. This requires the use of Equation 19.3a,
which is applied to the diameters of both the rod and hole. That is
d f − d0
= α l (T f − T0 )
d0
Solving this expression for df yields


d f = d0 1 + α l (T f −T0 )


Now all we need do is to establish expressions for df(316 stainless) and df (W), set them equal to one
another, and solve for Tf. According to Table 19.1, αl(316 stainless) = 16.0 × 10-6 (°C)-1 and αl (W) =
4.5 × 10-6 (°C)-1. Thus
d f (316 stainless) = d f (W)
[ {16.0 × 10−6 (°C)−1} (T f − 25°C)]
= (10.000 mm) [1 + {4.5 × 10−6 (°C)−1} (T f − 25°C)]
(9.988 mm) 1 +
Now solving for Tf gives Tf = 129.5°C
Question 6
(a) Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric materials,
rendering them more thermally insulative. (b) Briefly explain how the degree of crystallinity affects the
thermal conductivity of polymeric materials and why.
Solution
(a) Porosity decreases the thermal conductivity of ceramic and polymeric materials because the thermal
conductivity of a gas phase that occupies pore space is extremely small relative to that of the solid
material. Furthermore, contributions from gaseous convection are generally insignificant.
(b) Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal
conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal
transport when a crystalline structure prevails.
Question 7
For each of the following pairs of materials, decide which has the larger thermal conductivity. Justify your
choices.
(a) Pure copper; aluminum bronze (95 wt% Cu-5 wt% Al).
(b) Fused silica; quartz.
(c) Linear polyethylene; branched polyethylene.
(d) Random poly(styrene-butadiene) copolymer; alternating poly(styrene-butadiene) copolymer.
Solution
This question asks for us to decide, for each of several pairs of materials, which has the larger thermal
conductivity and why.
(a) Pure copper will have a larger conductivity than aluminum bronze because the impurity atoms in the
latter will lead to a greater degree of free electron scattering.
(b) Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline (whereas
quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline materials.
(c) The linear polyethylene will have the larger conductivity than the branched polyethylene because the
former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity
than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the
coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the
greater the thermal conductivity.
(d) The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random
copolymer; alternating copolymers crystallize more easily than random ones. The influence of
crystallinity on conductivity is explained in part (c).
Question 8
A copper wire is stretched with a stress of 70 MPa (10,000 psi) at 20°C (68°F). If the length is held
constant, to what temperature must the wire be heated to reduce the stress to 35 MPa (5000 psi)?
Solution
We want to heat the copper wire in order to reduce the stress level from 70 MPa to 35 MPa; in doing so,
we reduce the stress in the wire by 70 MPa – 35 MPa = 35 MPa, which stress will be a compressive one
(i.e., σ = –35 MPa). Solving for Tf from Equation 19.8 [and using values for E and αl of 110 GPa (Table
6.1) and 17.0 × 10-6 (°C)-1 (Table 19.1), respectively] yields
T f = T0 −
= 20°C −
σ
Eα l
− 35 MPa
(110 ×
10 3
[
MPa) 17.0 × 10−6 (°C)−1
]
= 20°C + 19°C = 39°C (101°F)
Question 9
Briefly explain what determines the characteristic color of (a) a metal and (b) a transparent nonmetal.
Solution
(a) The characteristic color of a metal is determined by the distribution of wavelengths of the
nonabsorbed light radiation that is reflected.
(b) The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of
the nonabsorbed light radiation that is transmitted through the material.
Question 10
Zinc telluride has a band gap of 2.26 eV. Over what range of wavelengths of visible light is it transparent?
Solution
Only photons having energies of 2.26 eV or greater are absorbed by valence-band-to-conduction-band
electron transitions. Thus, photons having energies less than 2.26 eV are not absorbed; the minimum
photon energy for visible light is 1.8 eV (Equation 21.16b), which corresponds to a wavelength of 0.7 µm.
From Equation 21.3, the wavelength of a photon having an energy of 2.26 eV (i.e., the band-gap energy)
is just
λ =
hc
=
E
(4.13 × 10−15 eV - s)(3
× 10 8 m /s)
2.26 eV
= 5.5 × 10 -7 m = 0.55 µm
Thus, pure ZnTe is transparent to visible light having wavelengths between 0.55 and 0.7 µm.