CHAPTER 8: PHYSICS OF MATTERS

8.4 Thermal expansion

LEARNING OUTCOME:
At the end of this chapter, students should be able to:
a) Define coefficient of linear expansion, α, area expansion, β and volume expansion, γ

b) Solve problems related to thermal expansion of linear, area and volume.
*include expansion of liquid in a container
Δ = α oΔ , Δ = β oΔ , Δ = γ oΔ , β = 2α , γ = 3α

8.4 Thermal expansion

 Thermal expansion is defined as the change in dimensions of a body
accompanying a change in temperature.

 3 types of thermal expansion :
- Linear expansion
- Area expansion
- Volume expansion

 In solid, all types of thermal expansion are occurred.
 In liquid and gas, only volume expansion is occurred.

 At the same temperature, the gas expands greater than liquid and solid.

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Linear expansion

 Consider a thin rod of initial length, l0 at temperature,T0 is heated to a new
uniform temperature, T and acquires length, l as shown in figure below.

 If ΔT is not too large (< 100o C) l0

l  T and l  l0 l

l  l0T l

l : change in length  l  l0 52
T : change in temperature  T  T0

 : coefficient of linear expansion

l  l0T   l  Unit of  is C-1 or K-1.
l0 T

 The coefficient of linear expansion,  is defined as the change in length of a solid

per unit length per unit rise change in temperature.

 If the length of the object at a temperature T is l,

l  l  lo
l  l  lo
l  loT  lo
l  lo (T 1)

 For many materials, every linear dimension changes according to both equations

above. Thus, l could be the length of a rod, the side length of a square plate or the

diameter (radius) of a hole. 53

 For example, as a metal washer is heated, all dimensions including the radius of the
hole increase as shown in figure below.

r1 r1  r1 At T0  ΔT 
r2
r2  r2
At T0

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Area expansion

 This expansion involving the expansion of a surface area of an object.

 Consider a plate with initial area, A0 at temperature T0 is heated to a new uniform

temperature, T and expands by A, as shown in figure below.

 If ΔT is not too large (< 100o C)

A  A0 and A  T 55

A  A0T

A : change in area  A  A0
T : change in temperature  T T0

 : coefficient of area expansion

A  A0T   A  Unit of  is C-1 or K-1
A0 T

 The coefficient of area expansion,  is defined as the change in area of a solid

surface per unit area per unit rise in temperature.

 The area of the of the surface of object at a temperature T can be written as,

A  A0 1  T 

 For isotropic material (solid) , the area expansion is uniform in all direction, thus the

relationship between  and  is given by

  2

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Volume expansion
 Consider a metal cube with side length, l0 is heated and expands uniformly.

 If ΔT is not too large (< 100o C)

V  V0 and V  T
V   V0T

V : change in volume  V V0
T : change in temperature  T  T0

 : coefficient of volume expansion

  V  Unit of  is C-1 or K-1.

V0T

 The coefficient of volume expansion,  is defined as the change in volume of a solid

per unit volume per unit rise in temperature. 57

 The volume of an object at a temperature T can be written as,

V  V0 1  T 

 For isotropic material (solid), the volume expansion is uniform in all

direction, thus the relationship between  and  is given by

  3

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l  l0T A  A0T V  V0T

  l   A   V

loT AoT V0T

The coefficient of linear   2   3

expansion,  is defined as the The coefficient of area The coefficient of volume

change in length of a solid expansion,  is defined as expansion,  is defined as the
per unit length per unit rise
change in temperature. the change in area of a solid change in volume of a solid per
surface per unit area per unit volume per unit rise in
l  lo  l unit rise in temperature. temperature.

l  lo  loT A  Ao  A V  Vo  V

l  l0 1T  A  Ao  AoT V  Vo  VoT

A  A0 1 T  V  V0 1 T 

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Example 8.4.1
A 1.2 m steel rod is attached to a fixed wall as shown in figure below. At 30 C, a gap
between the rod and the opposite wall is 3.0 mm. By neglecting the expansion of the
walls, at what temperature will the gap close?
The coefficient of linear expansion for steel is 12  10-6 C1.
[3080C ]

1.2 m

gap = 3.0 mm (the scale is exaggerated)

Solution 8.4.1

1.2 m

gap = 3.0 mm (the scale is exaggerated)

l  l0T

 3103  12106 1.2T 

T  208.3 oC

T = T+ 30
= 208.3 +30
= 238.3 oC

Example 8.4.2

An iron sphere is 30.00 cm in diameter. (α iron =1.2  10-5 K-1). If it is heated from 20o C
to 220 o C, calculate the

(i) final diameter [74.5 cm3]
(ii) change in volume.

Solution 8.4.2

(i) final diameter

From d  d0T

   1.2105 30.00102 220  20

 7.2104 m

Final diameter, d  d  d0  30.00102  7.2104  30.072 102 m  30.072 cm

(ii) change in volume.

Initial volume , V0  4 r 3  4   d 3  4   30.00 102 3  0.09425 m3
3 3 2  3 2


Change in volume , V  V0T
 3V0T

 31.2105 0.09425220  20

 6.786 104 m3

 6.786 cm3

Example 8.4.3

A glass flask with volume 150 cm3 is filled to the brim with the mercury at 20 o C.
(α glass =0.40  10-5 K-1, γ mercury=18  10-5 K-1)
(i) How much mercury overflows when the temperature of the system is raised to 90 o C?

(ii) Why mercury overflows when the temperature is increased?

Solution 8.4.3

(α glass =0.40  10-5 K-1, γ mercury=18  10-5 K-1)

To= 20 o C , T= 90 o C
Vinitial for mercury = Vinitial for glass Vo =150 cm3

i) The volume of mercury overflows = Vmercury – Vglass
= (Vmercury+ Vo) – (Vglass+ Vo)
= Vmercury - Vglass

 V0T  3V0T

   3 V0T 

  18105  30.40105 15090  20

 1.764 m3

(ii) Why mercury overflows when the temperature is increased?

The coefficient of volume expansion of mercury is greater that that of glass so the
change in volume is greater for mercury.

Exercise 8

1. A load of 6.0 kg hangs from a steel wire with a length of 4.50 m and cross - sectional

area of 1.20 mm2. Young ' s modulus of steel is 2.00  1011 N m-2. Calculate

(a) the stress

(b) the elongation of the wire [4.91 107 Pa, 1.10 × 10-3 m, 2.44 × 10-4]
(c) the strain

2. Find the force required to stretch a 1.0 m steel wire of diameter 3.2 mm by 10.0 mm.

Given the Young ' s modulus of steel is 1.90  1011 N m-2. [1.53  104 N ]

3. A support cable on a bridge has an area of cross-section of 0.0085 m2 and a length of

35 m. It is made of high tensile steel whose Young’s modulus is 2.8 1011 Pa. The

tension in the cable is 7.20  105 N. Calculate [0.0106 m; 3.81 kJ]
(a) the extension of the cable.
(b) the strain energy stored in the cable.

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4. A wire made of a particular material is loaded with a load of 500 N. The diameter of the
wire is 1.0 mm. The length of the wire is 2.5 m, and it stretches 8.0 mm when under
load.
(a) What is the Young’s modulus of this material?
(b) What is the elastic strain energy per unit volume for the wire ?

[1.99 × 1011 Pa, 1.02 × 106 J m-3]

5. A 4.0 m iron rod with cross-sectional area 0.5 cm2 extends by 1.0 mm when a 225 kg

mass is suspended from one of its ends. Calculate the
(a) Young’s modulus of the rod.

(b) elastic energy stored in the rod. [Ans : 1.77 × 1011 Pa, 1.11 J]

6. A cylindrical brass rod with Young’s modulus 9.7 x 1010 Pa and original diameter

10.0 mm experiences only elastic deformation when a tensile load of 240 N is applied.

(a) Calculate the stress that produces the deformation.

(b) If the original length of the rod is 0.20 m, calculate the change in the length of

the rod. [3.1 × 106 N m-2, 6.3 × 10-6 m]

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7. A 2.0-m-long steel rod has a cross-sectional area of 0.30 cm2. The rod is a part of a
vertical support that holds a heavy 550-kg platform that hangs attached to the rod’s

lower end. Ignoring the weight of the rod, what is the tensile stress in the rod and the
elongation of the rod under the stress? Young’s modulus for steel is Y=2.0×1011Pa.

[1.8  108 N m-2,, 1.8  10-3 ]

8. A copper wire LM is fused at one end, M to an iron wire MN as shown in figure below.

LM F
N

The copper wire has length 0.90 m and cross-section area 0.90 × 10−6 m2. The

iron has length 1.40 m and cross-section area 1.30 × 10−6 m2. The compound wire is

stretched and the total length increases by 0.01 m. Determine

(a) the ratio of the extension of copper wire to the extension of iron wire,

(b) the extension of each wire,

(c) the applied force to the compound wire.

(Given Y iron = 2.10 × 1011 Pa , Y copper = 1.30 × 1011 Pa ) [1.5, 4.0 mm, 6.0 mm, 780 N]

9. A metal plate 5.0 cm thick has a cross sectional area of 300 cm2. One of its face is

maintained at 100 o C by placing it in contact with steam and another face is maintained

at 30 o C by placing it in contact with water flow. Assume the heat flow is steady and no

heat is lost to the surrounding. Determine the thermal conductivity of the metal plate if

the rate of heat flow through the plate is 9 kW. [214 W m-1 K-1 ]

10. A rod 1.300 m long consists of a 0.800 m length of aluminum joined end to end to a
0.500 m of brass. The free end of the aluminum section is maintained at 150.0 o C and
the free end of the brass piece is maintained at 20.0 o C. No heat is lost through the sides

of the rod. Calculate the temperature of the point where two metals are joined metal.
(Given k of aluminum =205 W m-1 o C-1 and k of brass =109 W m-1 o C-1 )

[90.2 o C ]

11. Figure below shows part of a wall which consists of two layers of material.

concrete wood

300 oC 27 oC

The first layer is concrete of thermal conductivity 0.80 W/m. K and thickness 15.0 cm,

whereas the second layer is wood of thermal conductivity 0.60 W/m. K and thickness

1.5 cm. The temperature at the external surface of the concrete is 300 o C and the

temperature of the external surface of the wood is 27 o C. Calculate

(a) the temperature at the boundary of the concrete and the wood. [59.1 o C]

(b) the rate of heat flow if the area is 200 cm2. [25.7 J s-1]

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12. The length of metal rod is 30.000 cm at 20 o C and 30.019 cm at 45 o C, respectively.

Calculate the coefficient of linear expansion for the rod. [2.53  10-5 o C-1]

13. Determine the change in volume of block of cast iron 5.0 cm  10 cm  6.0 cm, when
the temperature changes from 15 o C to 47 o C. (cast iron =0.000010 oC-1) [0.29 cm3]

14. The length of a copper rod is 2.001 m and the length of a wolfram rod is 2.003 m at

the same temperature. Calculate the change in temperature so that the two rods have

the same length where the final temperature for both rods is equal.

(copper= 0.000017 o C-1 and wolfram= 0.0000043 o C-1) [78.72 o C]

15. A metal sphere with radius of 9.0 cm at 30.0 o C is heated until the temperature of

100.0 o C. Determine the change in volume for the sphere.

(metal spehere= 0.000051 oC-1) [1.09  10-5 cm-3]