Linear expansion or longitudinal expansion refers to the increase in length of a solid material when its temperature rises. This phenomenon occurs due to the increased thermal energy within the material, causing its constituent particles to vibrate more vigorously, thereby increasing the average distance between them. Understanding the linear expansion is crucial for predicting and engineering the thermal behaviour of materials in various applications, including construction, manufacturing, and thermal management systems.

LIST OF SUB-TOPICS:

• Applications of Linear Expansion of Solids
• Applications of Superficial Expansion of Solids
• Applications of Cubical Expansion of Solids

Applications of Linear Expansion of Solids:

A gap is kept between two successive rails

The rails of a railway expand in summer and contract in winter. Therefore gaps are kept between successive rails to allow for their expansion. The gap allows the rails to expand without exerting excessive force on the structure.

If there are no gaps, the increase in temperature will cause the rails to expand and they will overlap one another or dislodge from the position. This will be dangerous to the trains and may result in a severe accident. If the rails were tightly connected without any gap, thermal expansion could cause bending, buckling, or even structural damage.

To put an iron tyre on a wooden wheel of bullock cart, the iron tyre is heated first

Putting an iron tyre on a wooden wheel of a bullock cart involves a process known as “shrinking” or “setting” the tyre onto the wheel.

The diameter of the iron tyre is always slightly less than the wooden wheel of the bullock cart. Heating the iron tyre causes it to expand due to thermal expansion. As the temperature increases, the diameter of iron tyre. This expansion makes the iron tire slightly larger in diameter. Hence the tyre easily slides over the wooden wheel of the bullock cart. Then water is poured on the red-hot tyre. It contracts and grips the wheel firmly.

The clocks regulated with pendulum require adjustment during summer and winter.

The pendulums of clocks expand in summer and contract in winter, therefore, they lose time in summer and gain time in winter. In order that the clocks should give correct time, the pendulums are made from invar which is an alloy having a very small coefficient of linear expansion in some clocks, compensating pendulums are used.

A compensating pendulum is made of a number of iron and brass rods joined in such a way that the length of the pendulums remains constant even if there is a change in temperature. Such clocks show accurate time. If this is not possible the clocks require the adjustment in summer and winter.

Bimetallic strips are used as temperature controlling devices.

The bending of the bimetallic strip is due to the difference in the coefficient of linear expansion of two different metals used in the bimetallic strip.

A bimetallic strip consists two strips of different metals fixed to each other lengthwise An increase in temperature causes bending of the strip in such a way that the metal of greater linear coefficient of expansion lies on the outer side. A lowering of temperature again bends the strip, but with the metal of smaller linear coefficient of expansion on the outer side. Such strips can be used in electric iron, electric oven, refrigerators to control the temperature.

When the hot glass of a lamp is touched with a cold knife, it sometimes cracks.

When the hot glass of a lamp is touched with a cold knife,  the part of the glass, which is in contact with the knife contracts. Due to this local contraction, While remaining portion remains in the expanded condition. Due to uneven expansion in the glass, the glass cracks. This phenomenon is known as thermal shock.

When hot milk is poured into a thick-walled glass vessel it cracks.

When milk is poured into a thick-walled glass vessel at room temperature, the inner surface of the vessel gets expanded, while the outer surface remains at room temperature. Thus there is no expansion on the outer surface. Due to uneven expansion in the vessel, the glass cracks. This phenomenon is known as thermal shock.

The unit of the coefficient of expansion the same for linear, superficial as well as a cubical expansion:

The coefficient of expansion is the ratio of two similar quantities divided by temperature. Therefore, its unit is the same as the reciprocal of the unit of temperature for any kind of expansion.

If the temperature is measured on the Celsius scale the unit is per degree Celsius and if the temperature is measured on the Kelvin scale, the unit is per degree Kelvin.

Other Applications:

• Telephone and electric wires are arranged little bit loosely between any of the two poles. It is simply because with the decrease in temperature in winter season, the wire contracts and if they were arranged tightly that will become further tight and they may break. To avoid this problem, they were arranged loosely.

• Iron or steel girders are used in the construction of a bridge. One end of the girder is rigidly fixed with bricks and concrete. The other end is not fixed. Instead, the end is set on a roller over the support. When there is rise or fall in temperature due to seasonal changes, girders may expand or contract, without developing any thermal stress.

• In construction and engineering, materials like concrete, steel, and asphalt expand and contract with temperature changes. Expansion joints are used to allow for these movements without causing damage to structures. For example, bridges, buildings, and pipelines incorporate expansion joints to accommodate thermal expansion and contraction.

Applications of Superficial Expansion of Solids:

Removing Tight Lids of Glass Jar:

To open the lid of a glass jar that is tight enough, it is immersed in hot water for a minute or so. Metal cap expands and becomes loose. It would now be easy to turn it to open.

The high-temperature water causes the metal lid and glass jar to expand. But the glass has a low coefficient of expansion than the material of the lid. Hence the lid expands more than the glass jar and the lid can be easily removed.

To Pass a Nail Through Hole in Metal Plate:

To pass nail through a metal plate having a hole of a diameter slightly less than that of the nail, the plate is heated. so that the diameter of hole increases and the nail can easily pass through it.

Applications of Volumetric Expansion of Solids:

Use of Mercury or Alcohol in Thermometer:

A thermometer measures temperature by measuring a temperature-dependent property. Expansion of liquid is the temperature-dependent property and they are in direct proportion. Thus measuring the change in volume of mercury we can find the change in temperature.

Mercury has a high boiling point, and a highly predictable and shows a uniform response to changes in temperature. Mercury has a very high coefficient of volumetric expansion than the glass. Hence expands at a faster rate than glass. The expansion of the glass is negligible.

In a typical mercury thermometer, mercury is placed in a long, narrow sealed tube called a capillary. Because it expands at a much faster rate than the glass capillary, the mercury rises and falls with the temperature. A thermometer is calibrated with one of the temperature scales.

Riveting of two Metal Plates:

Two steel plates can be jointly tightly together by a process called riveting. Rivets are heated to red hot condition and are forced through coaxial holes in the two plates. The end of hot rivets is then hammered and shaped. On cooling, the rivets contract and bring the plates tightly gripped to each other.

Design of Air Craft:

The aircraft expands by 15-25centimetress during its flight due to the increase in temperature on account of heat created by friction with the air. Designers used rollers (separators) to isolate the cabin and passenger area from the body of the aircraft so that the expansion does not rip the plane apart.

Overflow Tanks of Coolant in Automobiles:

The efficiency of an internal combustion engine depends on heat rejected to the surroundings. Thus efficient cooling of the engine ensures the efficiency of the engine. Coolants are used to cool the engine. But during the process coolant itself undergoes volumetric expansion. If overflow tanks are not provided, there is a possibility of the bursting of coolant line. Hence automobiles have coolant overflow tanks.

Use of Thick Bottles for Soft Drinks:

To avoid bursting of soft drink bottles containing gas, due to thermal expansion, their walls are made very thick.

Conclusion:

The expansion of solids, which refers to the increase in size or volume of a material in response to changes in temperature, has several practical applications across various fields. Understanding the principles of thermal expansion allows engineers and designers to develop systems and structures that can accommodate temperature changes and operate reliably in various environments.

For More Topics in Thermal Properties of Matter and Thermodynamics Click Here

For More Topics in Physics Click Here

The post Applications of Expansion of Solids appeared first on The Fact Factor.

In this article, we shall study to solve problems to calculate the coefficient of linear expansion of solid, final length and the change in temperature.

Example – 01:

A metal scale is graduated at 0 ^oC. What would be the true length of an object which when measured with the scale at 25 ^oC, reads 50 cm? α for metal is 18 x 10^-6 /^oC.

Given: Initial temperature = t₁ = 0 ^oC, final temperature = t₂ = 25 ^oC, measured length = l₁ = 50 cm, coefficient of linear expansion = α = 18 x 10^-6 /^oC.

To Find: Actual length = l₂ =?

Solution:

l ₂ = l ₁ (1 + α (t₂ – t₁))

∴  l ₂ = 50 x (1 + 18 x 10^-6  (25 – 0))

∴  l ₂ = 50 x (1 + 18 x 10^-6  x 25)

∴  l ₂ = 50 x (1 + 450 x 10^-6)

∴  l ₂ = 50 x (1 + 0. 000450)

∴  l ₂ = 50 x 1. 000450

∴  l ₂ = 50.225 cm

Ans: the true length of object is 50.225 cm

Example – 02:

A metal rod is 64.522 cm long at 12 ^oC and 64.576 cm at 90 ^oC. Find the coefficient of linear expansion of its material.

Given: Initial temperature = t₁ = 12 ^oC, final temperature = t₂ = 90 ^oC, initial length = l₁ = 64.522 cm, final length = l ₂ = 64.576 cm.

To Find: coefficient of linear expansion = α =?

Solution:

l ₂ = l ₁ (1 + α (t₂ – t₁))

∴  l ₂ = l ₁ + α l₁ (t₂ – t₁)

∴  l ₂ – l ₁ = α l₁ (t₂ – t₁)

∴  64.576  – 64.522  = α x 64.522 x  (90 – 12)

∴  0.054  = α x 64.522 x  78

∴  α = (0.054)/(64.522 x  78) = 1.073 x 10^-5 /^oC.

Ans: The coefficient of linear expansion is 1.073 x 10^-5 /^oC.

Example – 03:

A metal bar measures 60 cm at 10 ^oC. What would be its length at 110 ^oC, α = 1.5 x 10^-5 /^oC.

Given: Initial temperature = t₁ = 10 ^oC, final temperature = t₂ = 110 ^oC, initial length = l₁ = 60 cm, coefficient of linear expansion = α = 1.5 x 10^-5 /^oC.

To Find: final length = l₂ =?

Solution:

l ₂ = l ₁ (1 + α (t₂ – t₁))

∴  l ₂ = 60 x (1 + 1.5 x 10^-5  (110 – 10))

∴  l ₂ = 60 x (1 + 1.5 x 10^-5  x 100)

∴  l ₂ = 60 x (1 + 1.5 x 10^-3)

∴  l ₂ = 60 x (1 + 0. 0015)

∴  l ₂ = 60 x 1. 0015

∴  l ₂ = 60.09 cm

Ans: the length at 110 ^oC will be 60.09 cm

Example – 04:

A rod is found to be 0.04 cm longer at 30 ^oC than it is at 10 ^oC. Calculate its length at 0 ^oC if coefficient of linear expansion = α = 2 x 10^-5 /^oC.

Given: Difference in lengths = l₂ – l ₁ = 0.04 cm, Initial temperature = t₁ = 10 ^oC, final temperature = t₂ = 30 ^oC, coefficient of linear expansion = α = 2 x 10^-5 /^oC.

To find: Initial length = l _o = ?

Solution:

l ₁ = l _o (1 + αt₁)

∴  l ₁ = l _o (1 + 2 x 10^-5 x 10)

∴  l ₁ = l _o (1 + 2 x 10^-5)   ……….  (1)

Now, l ₂ = l _o (1 + αt₂)

∴  l ₂ = l _o (1 + 2 x 10^-5 x 30)

∴  l ₂ = ll _o (1 + 60 x 10^-5)   ……….  (2)

From equations (1) and (2)

l ₂ – l ₁ = l _o (1 + 60 x 10^-5) – l _o (1 + 20 x 10^-5)

∴  0.04  = l _o (1 + 60 x 10^-5 – 1 – 20 x 10^-5)

∴  0.04  = l_o x  40 x 10^-5

∴  l _o = 0.04 / (40 x 10^-5) = 100 cm

Ans: The length of rod at 0 ^oC is 100 cm

Example – 05:

The length of iron rod at 100 ^oC is 300.36 cm and at 150 ^oC is 300.54 cm. Calculate its length at 0 ^oC and coefficient of linear expansion of iron.

Given: initial length = l₁ = 300.36 cm, final length = l ₂ = 300.54 cm, Initial temperature = t₁ = 100 ^oC, final temperature = t₂ = 150 ^oC,

To find: Initial length = l _o = ? and coefficient of linear expansion = α =?

Solution:

l ₁ = l _o (1 + αt₁)

∴  300.36 = l _o (1 + 100α)   ……….  (1)

l ₂ = l _o (1 + αt₂)

∴  300.54 = l _o (1 + 150α)   ……….  (2)

Dividing equation (2) by (1)

∴  300.54/300.36 = l _o (1 + 150α)/ l _o (1 + 100α)

∴  1.0006 = (1 + 150α)/(1 + 100α)

∴  1.0006(1 + 100α) = (1 + 150α)

∴  1.0006 + 100.06α =  1 + 150α

∴  1.0006 – 1 =  150α – 100.06α

∴  0.0006  =  49.94α

∴  α = 0.0006/49.94

∴  α = 1.2 x 10^-5 /^oC

From equation (1)

300.36 = l _o (1 + 100α)

∴  300.36 = l _o (1 + 100 x 1.2 x 10^-5)

∴  300.36 = l _o (1 + 1.2 x 10^-3)

∴  300.36 = l _o (1 + 0.0012)

∴  300.36 = l _o (1 .0012)

∴  l _o = 300.36/1 .0012 = 300 cm

Ans: The length of rod at 0 ^oC is 300 cm and coefficient of linear expansion is 1.2 x 10^-5 /^oC

Example – 06:

By how much will a steel rod 1 m long expand when heated from 25 ^oC to 55 ^oC? The coefficient of volume expansion of steel is 3 x 10^-5 /^oC.

Given: Initial temperature = t₁ = 25 ^oC, final temperature = t₂ = 55 ^oC, initial length = l₁ =  1m, coefficient of volume expansion = γ = 3 x 10^-5 /^oC.

To Find: Increase in length = l ₂ – l ₁ =?

Solution:

coefficient of volume expansion (γ) = 3 x coefficient of linear expansion (α)

3 x 10^-5  = 3 x α

∴  α = 1 x 10^-5 /^oC

l ₂ = l ₁ (1 + α (t₂ – t₁))

∴  l ₂ = l ₁ + α l ₁ (t₂ – t₁)

∴  l ₂ – l ₁ = α l ₁ (t₂ – t₁)

∴  l ₂ – l ₁ = 1 x 10^-5  x 1 x  (55 – 25)

∴  l ₂ – l ₁ = 10^-5 x  30

∴  l ₂ – l ₁ = 3 x 10^-4 m

∴  l ₂ – l ₁ = 0.3 x 10^-3 m = 0.3 mm

Ans: The rod will expand by 0.3 mm.

Example – 07:

A brass rod and an iron rod are each 1m in length at 0 ^oC. Find the difference in their lengths at 110 ^oC. α for brass is 19 x 10^-6 /^oC and α for iron is 10 x 10^-6 /^oC.

Given: Initial temperature = t₁ = 0 ^oC, final temperature = t₂ = 110 ^oC, initial length = l_Br1 = l_Fe1 =1m, coefficient of linear expansion for brass = α_Br = 19 x 10^-6 /^oC. Coefficient of linear expansion for iron = α_Fe = 10 x 10^-6 /^oC.

To Find: Difference in length = l_Br2 – l_Fe2 =?

Solution:

For brass rod

l_Br2 = l _Br1 (1 + α _Br (t₂ -t₁))

For iron rod

l_Fe2 = l _Fe1 (1 + α _Fe (t₂ -t₁))  ……….  (2)

Subtracting equation (2) from (1)

∴  l_Br2 – l_Fe2 = l _Br1 (1 + α _Br (t₂ -t₁)) – l _Fe1 (1 + α _Fe (t₂ -t₁))

∴  l_Br2 – l_Fe2  = 1 (1 + 19 x 10^-6 (110 – 10)) – 1 (1 + 19 x 10^-6 (110 – 10))

∴  l_Br2 – l_Fe2  = 1 + 19 x 10^-6 x 100 – 1 – 10 x 10^-6 x 100

∴  l_Br2 – l_Fe2  = 9  x 10^-6 x 100 = 900 x 10^-6 m

∴  l_Br2 – l_Fe2  = 0.9 x 10^-3 m = 0.9 mm

Ans: The difference in their lengths is 0.9 mm

Example – 08:

A brass rod and an iron rod are each 1m in length at 20 ^oC. At what temperature the difference in their lengths is 1.4 mm. α for brass is 18.92 x 10^-6 /^oC and α for iron is 11.92 x 10^-6 /^oC.

Given: Initial temperature = t₁ = 20 ^oC, initial length = l_Br1 = l_Fe1 =1m, coefficient of linear expansion for brass = α_Br = 18.92 x 10^-6 /^oC, coefficient of linear expansion for iron = α_Fe = 11.92 x 10^-6 /^oC. Difference in length = l_Br2 – l_Fe2 = 1.4 mm = 1.4 x 10^-3 m.

To Find: final temperature = t₂ =?

Solution:

For brass rod

l_Br2 = l _Br1 (1 + α _Br (t₂ -t₁))

For iron rod

l_Fe2 = l _Fe1 (1 + α _Fe (t₂ -t₁))  ……….  (2)

Subtracting equation (2) from (1)

∴  l_Br2 – l_Fe2 = l _Br1 (1 + α _Br (t₂ -t₁)) – l _Fe1 (1 + α_Fe (t₂ -t₁))

∴  1.4 x 10^-3 = 1 (1 + 18.92 x 10^-6 (t₂ -20)) – 1 (1 + 11.92 x 10^-6 (t₂ -20))

∴  1.4 x 10^-3 = 1 + 18.92 x 10^-6 (t₂ -20) – 1 – 11.92 x 10^-6 (t₂ -20)

∴  1.4 x 10^-3 = 7 x 10^-6 (t₂ -20)

∴  (t₂ -20) =1.4 x 10^-3/7 x 10^-6 = 200

∴  t₂ = 200 + 20 = 220 ^oC

Ans: At 220 ^oC the difference in their lengths is 1.4 mm.

Example – 09:

A rod A and a rod B are of equal length at 0 ^oC. If at 100 ^oC they differ by 1mm find their lengths at 0 ^oc, α_A = 8 x 10^-6 /^oC, α_B = 12 x 10^-6 /^oC.

Given: Initial temperature = t₁ = 0 ^oC, final temperature = t₂ = 100 ^oC, initial length = l_A1 = l_B1, coefficient of linear expansion for rod A = α_A = 8 x 10^-6 /^oC, coefficient of linear expansion for rod B = α_B = 12 x 10^-6 /^oC. Difference in length = l_B2 – l_A2 = 1 mm = 1 x 10^-3 m.

To Find: Initial lengths of rod at 0 ^oC

Solution:

For rod A

l_A2 = l _A1 (1 + α _A (t₂ -t₁))

For rod B

l_B2 = l _B1 (1 + α _B (t₂ -t₁))  ……….  (2)

Subtracting equation (2) from (1)

∴  l_B2 – l_A2 = l _B1 (1 + α _B (t₂ -t₁)) – l _A1 (1 + α_A (t₂ -t₁))

∴  1 x 10^-3 = l _B1 (1 + 12 x 10^-6 (100 – 0)) – l _A1 (1 + 8 x 10^-6 (100 – 0))

∴  1 x 10^-3 = l _B1 (1 + 12 x 10^-6 (100)) – l _A1 (1 + 8 x 10^-6 (100))

Now, given l_A1 = l_B1

∴  1 x 10^-3 = l _A1 (1 + 12 x 10^-4  – l _A1 (1 + 8 x 10^-4)

∴  1 x 10^-3 = l _A1 (1 + 12 x 10^-4  – 1 – 8 x 10^-4)

∴  1 x 10^-3 = l _A1 x 4 x 10^-4

∴  l _A1  = 1 x 10^-3 /  4 x 10^-4= 2.5 m

∴  l_A1  = l_B1 = 2.5 m

Ans: At 0 ^oC the lengths of the two rods is 2.5 m.

Example – 10:

A brass rod and an iron rod are 4 m and 4.01 m respectively at 20 ^oC. At what temperature the two rods have the same length. α for brass is 18.92 x 10^-6 /^oC and α for iron is 11.92 x 10^-6 /^oC.

Given: Initial temperature = t₁ = 20 ^oC, initial length = l_Br1 = l_Fe1 =1m, coefficient of linear expansion for brass = α_Br = 18 x 10^-6 /^oC, coefficient of linear expansion for iron = α_Fe = 12 x 10^-6 /^oC, l_Br2 = l_Fe2

To Find: final temperature = t₂ =?

Solution:

For brass rod

l_Br2 = l _Br1 (1 + α _Br (t₂ -t₁))

For iron rod

l_Fe2 = l _Fe1 (1 + α _Fe (t₂ -t₁))  ……….  (2)

Given l_Br2 = l_Fe2

∴  l _Br1 (1 + α _Br (t₂ -t₁)) = l _Fe1 (1 + α_Fe (t₂ -t₁))

∴  4 (1 + 18 x 10^-6 (t₂ -20)) =  4.01(1 + 12 x 10^-6 (t₂ -20))

∴  4 + 72 x 10^-6 (t₂ -20) =  4.01 + 48.12 x 10^-6 (t₂ -20)

∴   72 x 10^-6 (t₂ -20) –  48.12 x 10^-6 (t₂ -20) =  4.01 – 4

∴  23.88 x 10^-6(t₂ -20) = 0.01

∴  (t₂ -20) = 0.01/(23.88 x 10^-6)

∴  t₂  – 20 = 418.8

t₂ = 418.8 + 20 = 438.8 ^oC

Ans: At 438.8 ^oC the rods will have the same length

Example – 11:

The difference in lengths of rod A and a rod B is 60 cm at all temperatures. Find their lengths at 0 ^oC, α_A = 18 x 10^-6 /^oC, α_B = 27 x 10^-6 /^oC.

Given: Initial temperature = t₁ = 0 ^oC, final temperature = t₂ ^oC, coefficient of linear expansion for rod A = α_A = 18 x 10^-6 /^oC, coefficient of linear expansion for rod B = α_B = 27 x 10^-6 /^oC. Difference in length = l_B2 – l_A2 = 60 cm = 60 x 10^-2 m, (t₂ -t₁) is same for both rod.

To Find:Initial lengths of rod at 0 ^oC

Solution:

For rod A

l_A2 = l _A1 (1 + α _A (t₂ -t₁))

l_A2 = l _A1 +  l_A1α _A (t₂ -t₁))

l_A2 – l _A1 =  l _A1α _A (t₂ -t₁))  ……….  (1)

For rod B

l_B2 – l _B1 =  l _B1α _A (t₂ -t₁))  ……….  (1)

As the difference between the two rods is always 60 cm their expansion should be equal

l_A2 – l _A1 = l_B2 – l _B1

l _A1α _A (t₂ -t₁)) = l _B1α _A (t₂ -t₁))

∴ l _A1 α_A=  l _B1α _B

∴ l _A1 /l _B1 =  α _B/ α_A = 27 x 10^-6/18 x 10^-6  = 3/2

∴ l _A1  = (3/2) l _B1

Length of rod A is greater than rod B

Now (l _A1 – l _B1) = 60

∴ ( (3/2) l _B1 – l _B1) = 60

∴ (1/2) l _B1 = 60

∴ l_B1 = 120 cm

l _A1  = (3/2)l x 120 = 180 cm

Ans: At 0 ^oC the length of rod A is 180 cm and that of rod B is 120 cm

Example – 12:

The difference in lengths of rod A and a rod B is 5 cm at all temperatures. Find their lengths at 0 ^oC, α_A = 12 x 10^-6 /^oC, α_B = 18 x 10^-6 /^oC.

Given: Initial temperature = t₁ = 0 ^oC, final temperature = t₂ ^oC, coefficient of linear expansion for rod A = α_A = 12 x 10^-6 /^oC, coefficient of linear expansion for rod B = α_B = 18 x 10^-6 /^oC. Difference in length = l_B2 – l_A2 = 5 cm = 5 x 10^-2 m, (t₂ -t₁) is same for both rod.

To Find:Initial lengths of rod at 0 ^oC

Solution:

Assuming length of rod B is greater than rod B

For rod A

l_A2 = l _A1 (1 + α _A (t₂ -t₁))

For rod B

l_B2 = l _B1 (1 + α _B (t₂ -t₁))  ……….  (2)

Subtracting equation (2) from (1)

∴  l_B2 – l_A2 = l _B1 (1 + α _B (t₂ -t₁)) – l _A1 (1 + α_A (t₂ -t₁))

∴  l_B2 – l_A2 = l _B1 + l _B1α _B (t₂ -t₁) – l _A1  – l _A1 α_A (t₂ -t₁)

∴  l_B2 – l_A2 = (l _B1 – l_A1) + (l _B1α _B  – l _A1 α_A)(t₂ -t₁)

∴ 60 x 10^-2= 60 x 10^-2 + (l _B1α _B  – l _A1 α_A)(t₂ -t₁)

∴ 0=  (l _B1α _B  – l _A1 α_A)(t₂ -t₁)

∴ 0=  (l _B1α _B  – l _A1 α_A)

∴ l_A1 α_A=  l_B1α _B

∴ l_A1 /l _B1 =  α _B/ α_A = 18 x 10^-6/12 x 10^-6  = 3/2

∴ l_A1  = (3/2)l _B1

Length of rod A is greater than rod B

Now (l _A1 – l _B1) = 5

∴ ((3/2) l _B1 – l _B1) = 5

∴ (1/2) l _B1 = 5

∴ l_B1 = 10 cm

l_A1  = (3/2)l x 10 = 15 cm

Ans: At 0 ^oC the length of rod A is 15 cm and that of rod B is 10 cm

Science > Physics > Expansion of Solids > Numerical Problems on Linear Expansion of Solids

The post Numerical Problems on Linear Expansion of Solids appeared first on The Fact Factor.

Whenever there is an increase in the dimensions of a body due to heating, then the body is said to be expanded and the phenomenon is known as expansion of solids. Solids undergo three types of expansions a) Linear (Longitudinal) expansions, b) Superficial expansions (Arial) and c) Cubical expansions (Volumetric)

LIST OF SUB-TOPICS:

• Linear Expansion of Solids
• Characteristics of Coefficient of Linear Expansion
• Factors Influencing the Coefficient of Linear Expansion
• Coefficient of Linear Expansion for Different Materials
• Superficial Expansion of Solids
• Cubical Expansion of Solids
• Relation Between α and β
• Relation Between α and γ
• Relation Between α, β and γ
• Molecular Explanation of Expansion of Solids:

Linear Expansion of Solid:

Linear expansion or longitudinal expansion refers to the increase in length of a solid material when its temperature rises. This phenomenon occurs due to the increased thermal energy within the material, causing its constituent particles to vibrate more vigorously, thereby increasing the average distance between them.

Expression for the Coefficient of Linear Expansion of Solids:

Consider a metal rod of length ‘l₀’ at temperature 0 °C. Let the rod be heated to some higher temperature say t °C. Let ‘l’ be the length of the rod at temperature t °C.

∴ Change in temperature = t₂ – t₁ = t – 0 = t

and Change in length = l – l₀

Experimentally it is found that the change in length ( l – l₀) is

Directly proportional to the original length (l₀)

l – l₀   ∝  l₀   ………………. (1)

Directly proportional to the change in temperature (t)

l – l₀   ∝  t …………… (1)

Dependent upon the material of the rod.

From equation (1) and (2)

l – l₀   ∝  l₀ t

∴   l – l₀   =  α l₀ t    …………… (3)

Where ‘α’ is a constant called a coefficient of linear expansion

This is an expression for the coefficient of linear expansion of a solid.

The coefficient of linear-expansion is defined as the increase in length per unit original length at 0⁰c per unit rise in temperature.

From equation (3) we get

∴   l   = l₀ + α l₀ t

∴   l   = l₀ (1 + α t) ………….. (4)

This is an expression for the length of rod at t °C

Note: The magnitude of the coefficient of linear expansion is so small that it is not necessary to take the initial temperature at 0 °C.

Consider a metal rod of length ‘l₁’ at temperature t₁0 °C. Let the rod be heated to some higher temperature say t °C. Let ‘l₂’ be the length of the rod at temperature t₂ °C. Let l₀’ be the length of the rod at the temperature of 0 °C. Let α be the coefficient of linear expansion, then we have

l₁ = l₀ (1 + α t₁) ………….. (2)

l₂ = l₀ (1 + α t₂) ………….. (2)

Dividing equation (2) by (1) we get

The coefficient of linear expansion is different for different material

Back to List of Sub-Topics

Characteristics of Coefficient of Linear Expansion:

The coefficient of linear expansion is a material property that describes how much a material will expand or contract in length for a given change in temperature. The change in length of the material is directly proportional to the original length, the change in temperature, and the coefficient of linear expansion. Key characteristics of the coefficient of linear expansion are as follows:

• Material Specific: The coefficient of linear expansion varies from one material to another. Different materials have different rates of expansion or contraction for the same change in temperature.
• Temperature Dependence: While the coefficient of linear expansion is often treated as a constant over a small temperature range, it can vary slightly with temperature. However, for most materials and within normal temperature ranges, this variation is negligible for practical purposes. Coefficient of linear expansion generally decreases with the increase in temperature.
• Directionality: The coefficient of linear expansion is directional and applies along a specific axis of the material. For instance, in an anisotropic material like wood, the coefficient of linear expansion can be different along different axes.
• Units: The coefficient of linear expansion is typically expressed in units of per degree Celsius or per Kelvin
• Thermal Stress Inducer: Differences in the coefficients of linear expansion between materials can lead to thermal stresses when they are bonded together. These stresses can cause mechanical failure or deformation in structures or devices.
• Measurement: Coefficients of linear expansion can be determined experimentally through techniques such as dilatometry, which involves measuring the change in length of a material for known changes in temperature.

Understanding the coefficient of linear expansion is crucial in engineering and construction to design structures that can accommodate thermal expansion and contraction without causing damage or failure. The coefficient of linear expansion is an essential parameter for understanding how materials respond to temperature changes and for designing systems that can withstand such thermal effects.

Back to List of Sub-Topics

Factors Influencing the Coefficient of Linear Expansion:

The coefficient of linear expansion of a material is influenced by several factors, including:

• Chemical Composition: The chemical composition of a material significantly affects its coefficient of linear expansion. Different materials have different atomic and molecular structures, which lead to variations in their expansion behaviour.
• Crystal Structure: The crystal structure of a material can influence its coefficient of linear expansion. For example, crystalline materials may have different expansion characteristics along different crystallographic directions.
• Temperature Range: The coefficient of linear expansion can vary depending on the temperature range over which it is measured. In some materials, the coefficient of linear expansion may change with temperature, especially at extreme temperatures.
• Impurities and Alloying Elements: The presence of impurities or alloying elements in a material can alter its coefficient of linear expansion. Alloying elements may introduce lattice distortions or changes in the bonding behaviour, affecting the material’s expansion properties.
• Phase Transitions: Phase transitions, such as melting or solid-state transformations, can affect the coefficient of linear expansion. Different phases of a material may exhibit different expansion behaviours.
• Anisotropy: Some materials exhibit anisotropic behaviour, meaning their properties vary depending on the direction of measurement. Anisotropy can result from the material’s crystal structure or processing methods and can lead to different coefficients of linear expansion along different axes.
• Microstructure: The microstructure of a material, including factors such as grain size, grain boundaries, and defects, can influence its coefficient of linear expansion. Grain boundaries and defects may act as obstacles to atomic movement and affect the material’s expansion behaviour.
• External Stress and Strain: The coefficient of linear expansion may change under the influence of external stress or strain. Mechanical deformation or stress can alter the material’s atomic arrangement and affect its expansion properties.
• Environmental Conditions: Environmental factors such as humidity, pressure, and the presence of reactive gases can influence the coefficient of linear expansion. Changes in these environmental conditions may affect the material’s structure and expansion behaviour.

Understanding the factors that influence the coefficient of linear expansion is crucial for predicting and engineering the thermal behaviour of materials in various applications, including construction, manufacturing, and thermal management systems.

Back to List of Sub-Topics

Coefficient of Linear Expansion for Different Materials:

Back to List of Sub-Topics

Superficial Expansion of Solids:

Whenever there is an increase in the area of a solid body due to heating then the expansion is called superficial or Arial expansion.

Expression for the Coefficient of Superficial Expansion of Solids:

Consider a thin metal plate of area ‘A₀’ at temperature 0 °C. Let the plate be heated to some higher temperature say t °C. Let ‘A’ be the area of the plate at temperature t °C.

∴ Change in temperature = t₂ – t₁ = t – 0 = t

and Change in area = A- A₀

Experimentally it is found that the change in the area (A- A₀) is

Directly proportional to the original area (A₀)

A – A₀   ∝ A₀   ………………. (1)

Directly proportional to the change in temperature (t)

A – A₀   ∝ t ………….. (1)

Dependent upon the material of the plate

From equation (1) and (2)

A – A₀   ∝ A₀ t

∴   A – A₀   = β A₀ t    …………… (3)

Where ‘β’ is a constant called a coefficient of superficial expansion

This is an expression for the coefficient of superficial expansion of a solid.

The coefficient of superficial expansion is defined as the increase in area per unit original area at 0⁰c per unit rise in temperature.

From equation (3) we get

∴ A   = A₀ + β A₀ t

∴ A   = A₀ (1 + β t) ………….. (4)

This is an expression for the area of the plate at t °C

Note: The magnitude of the coefficient of superficial expansion is so small that it is not necessary to take the initial temperature as 0 °C.

Consider a thin metal plate of area ‘A₁’ at temperature t₁0 °C. Let the plate be heated to some higher temperature say t °C. Let ‘A₂’ be the area of the plate at temperature t₂ °C. Let ‘A₀’ be the area of the plate at a temperature of 0 °C. Let β be the coefficient of superficial expansion, then we have

A₁ = A₀ (1 +    β t₁) ………….. (2)

A₂ = A₀ (1 +    β t₂) ………….. (2)

Dividing equation (2) by (1) we get

The coefficient of superficial expansion is different for different material

Back to List of Sub-Topics

Cubical Expansion of Solids:

Whenever there is an increase in the volume of the body due to heating the expansion is called cubical or volumetric expansion.

Expression for the Coefficient of Cubical Expansion of Solids:

Consider a solid body of volume ‘V₀’ at temperature 0 °C. Let the body be heated to some higher temperature say t °C. Let ‘V’ be the volume of the body at temperature t °C.

∴ Change in temperature = t₂ – t₁ = t – 0 = t

and Change in volume = V – V₀

Experimentally it is found that the change in the volume (V – V₀) is

Directly proportional to the original volume (V₀)

V – V₀  ∝  V₀   ………………. (1)

Directly proportional to the change in temperature (t)

V – V₀  ∝  t………………. (1)

Dependent upon the material of the body.

From equation (1) and (2)

V – V₀  ∝  V₀ t

∴ V – V₀   =  γ V₀ t    …………… (3)

Where ‘γ’ is a constant called a coefficient of cubical expansion

This is an expression for the coefficient of cubical expansion of a solid.

The coefficient cubical expansion is defined as an increase in volume per unit original volume at 0⁰c per unit rise in temperature.

From equation (3) we get

∴   V = V₀  +  γ V₀ t

∴   V   = V₀ (1 +    γ t) ………….. (4)

This is an expression for the volume of the body at t °C

Note: The magnitude of the coefficient of cubical expansion is so small that it is not necessary to take the initial temperature as 0 °C.

Consider a solid body of volume ‘V₁’ at temperature t₁0 °C. Let the body be heated to some higher temperature say t °C. Let ‘V₂’ be the volume of the body at temperature t₂ °C. Let ‘V0’ be the volume of the body at the temperature of 0 °C. Let γ be the coefficient of cubical-expansion, then we have

V₁  = V₀ (1 +   γ t₁) ………….. (2)

V₂  = V₀ (1 +  γ t₂) ………….. (2)

Dividing equation (2) by (1) we get

Back to List of Sub-Topics

Relation Between α and β:

Consider a thin metal plate of length, breadth, and area l₀, b₀, and A₀ at temperature 0 °C. Let the plate be heated to some higher temperature say t °C. Let l, b and A be the length, breadth, and area of the plate at temperature t °C.

Then original area   =   A₀ =   l₀ b₀ ……………..  (1)

Consider linear expansion

Length,   l =   l₀ (1+ αt)

Breadth, b = b₀ (1 + αt)

where α = coefficient of linear expansion

Final area   =    A    = l  b  =  l₀ (1+ αt) × b₀ (1 + αt)

∴   A    = l₀ b₀  (1+ 2 αt + α²t²)

Now α is very small hence α² is still small, hence quantity α²t² can be neglected

∴   A    = A₀ (1+ 2 αt)  …………  (2)

Consider superficial expansion of the plate area.

A = A₀( 1+ βt) …………  (3)

From (2) and (3)

β = 2α

Thus the coefficient of superficial expansion is twice the coefficient of linear expansion.

Back to List of Sub-Topics

Relation Between α and γ:

Consider a thin rectangular parallelopiped solid of length, breadth, height, and volume l₀, b₀, h₀, and V₀ at temperature 0 °C. Let the solid be heated to some higher temperature say t °C. Let l, b, h and V be the length, breadth, height, and volume of the solid at temperature t °C.

Then original volume    =   V₀ =   l₀ b₀ h₀   ………… (1)

Consider linear expansion

Length,   l =   l₀ (1+ αt)

Breadth, b = b₀ (1 + αt)

Height   h  = h₀ (1 + αt)

where α = coefficient of linear expansion

Final volume    =    V    =  l  b h  =  l₀ (1+ αt) × b₀ (1 + αt)× h₀ (1 + αt)

∴   V    =  l₀ b₀  h₀ (1+ 3 αt + 3 α²t² + α³t³ )

Now α is very small hence α² is still small, hence quantity α²t², α³t³ can be neglected

∴   V    =  V₀ (1+ 3 αt)  ,,,,,,,,,,,,,,,,,,  (2)

Consider cubical expansion of the solid.

V = V₀( 1+ γt) ,,,,,,,,,,,,,,,,,  (3)

From (2) and (3)

γ = 3α

Thus the coefficient of cubical expansion is the thrice coefficient of linear expansion.

Back to List of Sub-Topics

Notes: 

We have β = 2α   hence α = β/2   ………………… (1)

We have γ = 3α   hence α = γ/3   ………………… (2)

From relations (1) and (2) we get

α = β/2 = γ/3

Hence 6 α = 3 β   = 2γ

Back to List of Sub-Topics

Molecular Explanation of Expansion of Solids:

The expansion of solids at the molecular level can be understood through the principles of thermal expansion and the behaviour of atoms and molecules within the solid lattice structure.

• Vibrational Motion: Atoms and molecules within a solid are constantly in motion due to thermal energy. As the temperature of the solid increases, the average kinetic energy of the atoms and molecules also increases. This increased thermal energy causes the atoms and molecules to vibrate more vigorously about their equilibrium positions within the solid lattice.
• Increased Average Distance: The increased vibrational motion of atoms and molecules leads to an increase in the average distance between them. This phenomenon occurs because the atoms and molecules push against each other as they vibrate, causing the overall dimensions of the solid to expand.
• Interatomic Forces: The expansion of solids is influenced by the strength and nature of the interatomic or intermolecular forces within the material. In solids held together by stronger bonds, such as metallic bonds or covalent bonds, the expansion may be less pronounced compared to materials with weaker bonds, such as those held together by van der Waals forces.
• Anisotropic Expansion: Some solids exhibit anisotropic expansion, meaning they expand unequally along different crystallographic directions. This behaviour arises from the anisotropic arrangement of atoms or molecules within the crystal lattice structure. For example, in materials with layered or fibrous structures, expansion may be greater along certain directions than others.
• Phase Transitions: Phase transitions, such as melting or solid-state transformations, can also affect the expansion behaviour of solids. During phase transitions, the arrangement of atoms or molecules within the solid lattice undergoes significant changes, leading to alterations in the material’s expansion properties.
• Coefficient of Thermal Expansion: The coefficient of thermal expansion (CTE) quantifies the extent to which a material expands or contracts in response to changes in temperature. The CTE is a material-specific property that depends on factors such as the material’s composition, crystal structure, and bonding characteristics.

Thus, the expansion of solids at the molecular level is a result of the increased vibrational motion of atoms and molecules within the solid lattice structure as the temperature rises. Understanding the molecular mechanisms underlying thermal expansion is essential for various applications in materials science, engineering, and thermal management.

Back to List of Sub-Topics

For More Topics in Thermal Properties of Matter and Thermodynamics Click Here

For More Topics in Physics Click Here

The post Expansion of Solids appeared first on The Fact Factor.