CHAPTER 21 THERMAL EXPANSION
EXERCISE 112, Page 238
1. A length of lead piping is 50.0 m long at a temperature of 16C. When hot water flows through it
the temperature of the pipe rises to 80C. Determine the length of the hot pipe if the coefficient
of linear expansion of lead is 29  10 6 K 1 .
Length L 1 = 50.0 m, temperature t 1 = 16C, t 2 = 80C and  = 29  10 6 K 1
Length of pipe at 80C is given by:
L 2 = L 1 [1 + (t 2 - t 1 )] = 50.0[1 + (29  10 6 )(80 - 16)]
= 50.0[1 + 0.001856]
= 50.0[1.001856] = 50.0928 m
i.e. an increase in length of 0.0928 m or 92.28 mm
2. A rod of metal is measured at 285 K and is 3.521 m long. At 373 K the rod is 3.523 m long.
Determine the value of the coefficient of linear expansion for the metal.
Length L 1 = 3.521 m, L 2 = 3.523 m, temperature t 1 = 285 K and temperature t 2 = 373 K
Length L 2 = L 1 [1 + (t 2 - t 1 )]
i.e.
3.523 = 3.521[1 + (373 - 285)]
3.523 = 3.521 + (3.521)()(88)
i.e.
3.523 – 3.521 = (3.521)()(88)
Hence, the coefficient of linear expansion,  =
i.e.
0.002
= 0.00000645
(3.521)(88)
coefficient of linear expansion,  = 6.45  10  6 K 1
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3. A copper overhead transmission line has a length of 40.0 m between its supports at 20C.
Determine the increase in length at 50C if the coefficient of linear expansion of copper is
17  10 6 K 1 .
Length L 2 = L 1 [1 + (t 2 - t 1 )]
= L 1 + L 1 (t 2 - t 1 )
Hence, increase in length = L 1 (t 2 - t 1 )
= (40.0 m)(17  10 6 K 1 )(50 – 20)C
= (40.0)(17  10 6 )(30)
= 0.0204 m or 20.4 mm
4. A brass measuring tape measures 2.10 m at a temperature of 15C. Determine
(a) the increase in length when the temperature has increased to 40C
(b) the percentage error in measurement at 40C. Assume the coefficient of linear expansion of
brass to be 18  10 6 K 1 .
Length L 1 = 2.10 m, temperature t 1 = 15C, t 2 = 40C and  = 18  10 6 K 1
(a)
Length L 2 = L 1 [1 + (t 2 - t 1 )] = L 1 + L 1 (t 2 - t 1 )
Hence, increase in length = L 1 (t 2 - t 1 )
= (2.10 m)(18  10 6 K 1 )(40 – 15)C
= (2.10)(18  10 6 )(25)
= 0.000945 m or 0.945 mm
(b) Percentage error in measurement at 40C =
increase in length 0.000945

100%
original length
2.10
= 0.045%
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5. A pendulum of a ‘grandfather’ clock is 2.0 m long and made of steel. Determine the change in
length of the pendulum if the temperature rises by 15 K. Assume the coefficient of linear
expansion of steel to be 15  10 6 K 1 .
Length L 2 = L 1 [1 + (t 2 - t 1 )] = L 1 + L 1 (t 2 - t 1 )
Hence, increase in length = L 1 (t 2 - t 1 )
= (2.0 m)(15  10 6 K 1 )(15 K)
= (2.0)(15  10 6 )(15)
= 0.00045 m or 0.45 mm
6. A temperature control system is operated by the expansion of a zinc rod which is 200 mm long
at 15C. If the system is set so that the source of heat supply is cut off when the rod has
expanded by 0.20 mm, determine the temperature to which the system is limited. Assume the
coefficient of linear expansion of zinc to be 31  10 6 K 1 .
Length L 1 = 200 mm = 0.20 m, L 2 = 200 + 0.20 mm = 200.2 mm = 0.2002, temperature t 1 = 15C
Length L 2 = L 1 [1 + (t 2 - t 1 )] = L 1 + L 1 (t 2 - t 1 )
Hence,
i.e.
increase in length = L 1 (t 2 - t 1 )
0.2002 – 0.20 = (0.20)(31  10 6 )( t 2 - 15)
0.0002 = (0.20)( 31  10 6 )( t 2 - 15)
i.e.
( t 2 - 15) =
0.0002
= 32.26C
(0.20)(31106 )
i.e. the temperature to which the system is limited, t 2 = 32.26 + 15 = 47.26C
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7. A length of steel railway line is 30.0 m long when the temperature is 288 K. Determine the
increase in length of the line when the temperature is raised to 303 K. Assume the coefficient of
linear expansion of steel to be 15  10 6 K 1 .
Length L 2 = L 1 [1 + (t 2 - t 1 )] = L 1 + L 1 (t 2 - t 1 )
Hence, increase in length = L 1 (t 2 - t 1 )
= (30.0 m)(15  10 6 K 1 )(303 - 288)K
= (30.0)(15  10 6 )(15)
= 0.00675 m or 6.75 mm
8. A brass shaft is 15.02 mm in diameter and has to be inserted in a hole of diameter 15.0 mm.
Determine by how much the shaft must be cooled to make this possible, without using force.
Take the coefficient of linear expansion of brass as 18  10 6 K 1 .
Length L 1 = 15.02 mm = 0.01502 m, L 2 = 15 mm = 0.015 m
Length L 2 = L 1 [1 + (t 2 - t 1 )]
i.e.
0.015 = 0.01502[1 + (18  10 6 K 1 )(t 2 - t 1 )]
0.015 = 0.01502 + (0.01502)( 18  10 6 )( t 2 - t 1 )
i.e.
Hence,
0.015 – 0.01502 = (0.01502)( 18  10 6 )( t 2 - t 1 )
(t 2 - t 1 ) =
 0.00002
= - 73.98 K
(0.01502) 18 106 
i.e. the shaft must be cooled by 74 K
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EXERCISE 113, Page 240
1. A silver plate has an area of 800 mm 2 at 15C. Determine the increase in the area of the plate
when the temperature is raised to 100C. Assume the coefficient of linear expansion of silver to
be 19  10 6 K 1 .
A 2 = A 1 [1 + (t 2 - t 1 )]
i.e.
A 2 = A 1 [1 + 2(t 2 - t 1 )]
i.e.
A 2 = A 1 + A 1 2(t 2 - t 1 )
since  = 2, to a very close approximation
Hence, area increase = A 1 2(t 2 - t 1 )
=  800 10 6 m 2  2 19 10 6 K 1  100 15  C
= 800  10 6  2 19  10 6  85
= 2.584 10 6 m 2 or 2.584 mm 2
2. At 283 K a thermometer contains 440 mm 3 of alcohol. Determine the temperature at which the
volume is 480 mm 3 assuming that the coefficient of cubic expansion of the alcohol is
12  10 4 K 1 .
V 2 = V 1 [1 + (t 2 - t 1 )]
480 10 9 = 440 10 9 [1 + (12  10 4 )(t 2 - 283)]
i.e.
from which,
480 – 440 = 440(12  10 4 )(t 2 - 283)
and
from which,
and
480 = 440 + 440(12  10 4 )(t 2 - 283)
(t 2 - 283) =
40
= 75.76 K
440 12 104
temperature, t 2 = 75.76 + 283 = 358.8 K
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3. A zinc sphere has a radius of 30.0 mm at a temperature of 20C. If the temperature of the sphere
is raised to 420C, determine the increase in: (a) the radius, (b) the surface area, (c) the volume
of the sphere. Assume the coefficient of linear expansion for zinc to be 31  10 6 K 1 .
(a) Initial radius, L 1 = 30.0 mm, initial temperature, t 1 = 20 + 273 = 293 K,
final temperature, t 2 = 420 + 273 = 693 K and  = 31  10 6 K 1 .
New radius at 693 K is given by:
L 2 = L 1 [1 + (t 2 - t 1 )]
i.e.
L 2 = 30.0[1 + (31  10 6 )(693 - 293)]
= 30.0[1 + 0.0124] = 30.372 mm
Hence the increase in the radius is 0.372 mm
(b) Initial surface area of sphere, A 1 = 4r 2 = 4  30.0  = 3600 mm 2
2
New surface area at 693 K is given by:
A 2 = A 1 [1 + (t 2 - t 1 )]
since  = 2, to a very close approximation
i.e.
A 2 = A 1 [1 + 2(t 2 - t 1 )]
Thus
A 2 = 3600[1 + 2(31  10 6 )(400)]
= 3600[1 + 0.0248] = 3600 + 3600(0.0248)
Hence increase in surface area = 3600(0.0248) = 280.5 mm 2
(c) Initial volume of sphere, V 1 =
4 3 4
3
r =   30.0  mm 3
3
3
New volume at 693 K is given by:
V 2 = V 1 [1 + (t 2 - t 1 )]
i.e.
V 2 = V 1 [1 + 3(t 2 - t 1 )]
since  = 3, to a very close approximation
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Thus
V2 =
=
4
(30.0) 3 [1 + 3(31  10 6 )(400)]
3
4
4
4
(30.0) 3 [1 + 0.0372] = (30.0) 3 + (30.0) 3 (0.0372)
3
3
3
Hence, the increase in volume =
4
(30.0) 3 (0.0372) = 4207 mm 3
3
4. A block of cast iron has dimensions of 50 mm by 30 mm by 10 mm at 15C. Determine the
increase in volume when the temperature of the block is raised to 75C. Assume the coefficient
of linear expansion of cast iron to be 11  10 6 K 1 .
Initial volume of sphere, V 1 = 50  30  10 = 15000 mm 3
New volume at 75C is given by:
V 2 = V 1 [1 + (t 2 - t 1 )]
since  = 3, to a very close approximation
i.e.
V 2 = V 1 [1 + 3(t 2 - t 1 )]
Thus
V 2 = 15000 [1 + 3(11  10 6 )(75 - 15)]
= 15000[1 + 0.00198] = 15000 + 15000 (0.00198)
Hence, the increase in volume = 15000 (0.00198) = 29.7 mm 3
5. Two litres of water, initially at 20C, is heated to 40C. Determine the volume of water at 40C
if the coefficient of volumetric expansion of water within this range is 30  10 5 K 1 .
New volume at 40C is given by:
V 2 = V 1 [1 + (t 2 - t 1 )]
= 2[1 + (30  10 5 )(40 – 20)]
= 2[1 + 0.006] = 2[1.006]
= 2.012 litres
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6. Determine the increase in volume, in litres, of 3 m 3 of water when heated from 293 K to boiling
point if the coefficient of cubic expansion is 2.1  10 4 K 1 (1 litre  10 3 m 3 ).
Initial volume of sphere, V 1 = 3  103 = 3000 litres
New volume at boiling point (i.e. 373 K) is given by:
V 2 = V 1 [1 + (t 2 - t 1 )]
Thus
V 2 = 3000 [1 + (2.1  10 4 )(373 - 293)]
= 3000[1 + 0.0168] = 3000 + 3000 (0.0168)
Hence, the increase in volume = 3000 (0.0168) = 50.4 litres
7. Determine the reduction in volume when the temperature of 0.5 litre of ethyl alcohol is reduced
from 40C to - 15C. Take the coefficient of cubic expansion for ethyl alcohol as 1.1  10 3 K 1 .
New volume at - 15C is given by:
V 2 = V 1 [1 + (t 2 - t 1 )]
Thus
V 2 = 0.5 [1 + (1.1  10 3 )(- 15 - 40)]
= 0.5 [1 + (1.1  10 3 )(- 55)]
= 0.5 + (0.5)(1.1  10 3 )(- 55)
Hence, the reduction in volume = (0.5)(1.1  10 3 )(55) = 0.03025 litres
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EXERCISE 114, Page 241
Answers found from within the text of the chapter, pages 235 to 240.
EXERCISE 115, Page 241
1. (b) 2. (c) 3. (a) 4. (d) 5. (b) 6. (c) 7. (c) 8. (a) 9. (c) 10. (b)
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