Chapter 4

Question 1

Define density as a material property

Answer 1

weight per unit of volume

Question 2

What is the difference in melting characteristics between pure metal elements and alloy metal

Answer 2

A pure metal element melts at one temperature (melting point) while an alloy begins melting at the solidus and finally becomes molten at the liquiduls. Between those two, the metal is a mix of solid and liquied

Question 3

Describe the melting cahraceristics of a noncrystalline material

Answer 3

THe material begins to soften as temperature increases I gradual transition) it finally ocnverts to a liquid at the melting point

Question 4

Define specific heat as a material property

Answer 4

Quantity of heat required to raise the temperature of a unit of mass of the material by one degree

Question 5

What is thermal conductivity as a material property

Answer 5

It is the capacity of a material to transfer heat energy through itself by thermal movement only

Question 6

Define thermal diffusivity

Answer 6

Thermal conductivity divided by the volumetric specific heat (heat energy required to raise the temperature of a unit of volume of a material by 1 degree)

Question 7

What are the important variables that affect mass diffusion?

Answer 7

depends on the diffusion coefficient (increases with temperature, so it can also be considered an important variable) concentration gradient, contact area, and time

Question 8

Define resistivity as a material property

Answer 8

A material’s capacity of resisnting electric flow of current

Question 9

Why are metals better conductors of electricity than ceramics and polymers

Answer 9

Because of the metalic bond

Question 10

Dielectric strength

Answer 10

Electrical potential requiered to break down the insulator per unit of thickness

Question 11

What is an electrolyte

Answer 11

an ionized solution capable of conducting electric current by movement of ions

Question 12

Which of the following metals has the lowest density (a) aluminum, (b) copper, (c) magnesium,
or (d) tin?

Answer 12

c magnesium

Question 13

Thermal expansion properties of polymers are generally (a) greater than, (b) less than, or (c) the
same as those of metals?

Answer 13

a greater

Question 14

In heating of most metal alloys, melting begins at a certain temperature and concludes at a higher temperature. In these cases which of the following marks the beginning of the melting (a)
liquidus or (b) solidus?

Answer 14

b solidus

Question 15

Which one of the following has the highest specific heat (a) aluminum, (b) concrete, (c)
polyethylene, or (d) water?

Answer 15

d water

Question 16

Copper is generally considered easy to weld because of its high thermal conductivity (a) true or (b)
false?

Answer 16

b false. Thigh high thermal conductivity of copper makes it difficult to weld because the heat flows away from the joint rather than being concentrated to permit melting of the metal

Question 17

The mass diffusion rate dm/dt across a boundary between two different metals is a function of which of the following variables (four best answers): (a) concentration gradient dc/dx, (b) contact area, (c)
density, (d) melting point, (e) thermal expansion, (f) temperature, and (g) time?

Answer 17

a concentration gradient, b contact area, f temperature, g time:

Included in the equation dm = -D(dc/dt)A dt

Question 18

Which of the following pure metals is the best conductor of electricity (a) aluminum, (b) copper, (c)
gold, or (d) silver?

Answer 18

d silver

Question 19

A superconductor is characterized by which of the following (one best answer): (a) high conductivity,
(b) resistivity properties between those of conductors and semiconductors, (c) very low resistivity, or
(d) zero resistivity

Answer 19

d zero resitivity (at certain temperatures)

Question 20

In an electrolytic cell, the anode is the electrode that is

Answer 20

positive

Question 21

The starting diameter of a shaft is 25.00 mm. This shaft is to be inserted into a hole in an expansion
fit assembly operation. To be readily inserted, the shaft must be reduced in diameter by cooling.
Determine the temperature to which the shaft must be reduced from room temperature (20°C) in
order to reduce its diameter to 24.98 mm. Refer to Table 4.1. (steel)

Answer 21

T2 = -46.67°C

Question 22

A bridge built with steel girders is 500 m in length and 12 m in width. Expansion joints are provided
to compensate for the change in length in the support girders as the temperature fluctuates. Each
expansion joint can compensate for a maximum of 40 mm of change in length. From historical
records it is estimated that the minimum and maximum temperatures in the region will be -35°C and
38°C, respectively. What is the minimum number of expansion joints required?

Answer 22

Solution: Assume L1 =500 m at -35 C, α=12x10-6
/C
L2 – L1 = αL1(T2 – T1)
L2 – L1 = 12x10-6
(500)(38 – (-35))
L2 – L1 = 0.42 m
Each expansion joint will control 40 mm = 0.04 m of expansion.
10 joints will provide 0.40 m of expansion. 11 joints will provide 0.45 m of expansion. Therefore, a
minimum of 11 joints are needed for coverage of the total length. Each bridge section will be 500/11
= 45.5 m long.

Question 23

Aluminum has a density of 2.70 g/cm3 at room temperature (20°C). Determine its density at 650°C,
using data in Table 4.1 as a reference.

Answer 23

Solution: Assume a 1 cm3 cube, 1 cm on each side.
From Table 4.1, α = 24(10-6
) mm/mm/°C
L2 - L1 = αL1 (T2 - T2).
L2 = 1.0 + 24(10-6
)(1.0)(650 - 20) = 1.01512 cm
(L2 )
3 = (1.01512)3 = 1.04605 cm3
Assume weight remains the same; thus ρ at 650°C = 2.70/1.04605 = 2.581 g/cm3

Question 24

With reference to Table 4.1, determine the increase in length of a steel bar whose length = 10.0 in, if
the bar is heated from room temperature of 70°F to 500°F.

Answer 24

Solution: Increase = (6.7 x 10-6 in/in/F)(10.0 in)(500°F - 70°F) = 0.0288 in

Question 25

With reference to Table 4.2, determine the quantity of heat required to increase the temperature of an
aluminum block that is 10 cm x 10 cm x 10 cm from room temperature (21°C) to 300°C.

Answer 25

Solution. Heat = (0.21 cal/g-°C)(103 cm3
)(2.70 g/cm3
)(300°C - 21°C) = 158,193 cal.
Conversion: 1.0 cal = 4.184J, so heat = 662,196 J.

Question 26

What is the resistance R of a length of copper wire whose length = 10 m and whose diameter = 0.10
mm? Use Table 4.3 as a reference.

Answer 26

Solution: R = rL/A, A = π(0.1)2
/4 = 0.007854 mm2 = 0.007854(10-6
) m2
From Table 4.3, r = 1.7 x 10-8 Ω-m2
/m

Question 27

A 16 gage nickel wire (0.0508-in diameter) connects a solenoid to a control circuit that is 32.8 ft
away. (a) What is the resistance of the wire? Use Table 4.3 as a reference. (b) If a current was passed
through the wire, it would heat up. How does this affect the resistance?

Answer 27

Solution: (a) L = 32.8 ft = 393.6 in
Area A = π(0.0508)2
/4 = 0.00203 in2
R = r (L/A) = 6.8 x 10-8 (39.4)(393.6/0.00203) = 0.520 ohm
(b) If a current is passed through the wire causing the wire to heat up, the resistivity of the wire would
change. Since nickel is a metal, the resistivity would increase, causing the resistance to increase. This,
in turn, would cause slightly more heat to be generated.

Question 28

Aluminum wiring was used in many homes in the 1960s due to the high cost of copper at the time.
Aluminum wire that was 12 gauge (a measure of cross-sectional area) was rated at 15 A of current.
If copper wire of the same gauge were used to replace the aluminum wire, what current should the
wire be capable of carrying if all factors except resistivity are considered equal? Assume that the
resistance of the wire is the primary factor that determines the current it can carry and the crosssectional area and length are the same for the aluminum and copper wires.

Answer 28

Solution: The area and length are constant between the types of wires. The overall change in
resistance is due to the change in resistivity of the materials. From Table 4.3:
For Aluminum r = 2.8 x 10-8
For Copper r = 1.7 x 10-8
The resistance will reduce by 1.7x10-8
/2.8x10-8 = 0.61
Since I = E/R and Rcu=0.61(Ral), then Icu = 1/0.61*Ial = 15/0.61= 25 A
Note that the code value is actually 20 A due to several factors including heat dissipation and
rounding down to the nearest 5 amp value.