Thermal Expansion And Its Type (Board Level)

Video Lecture

Theory For Notes Making

Thermal Expansion

When matter is heated without any change in it’s state, it usually expands (Except few . Solids can expand in one dimension (linear expansion), two dimension (superficial expansion) and three dimension (volume expansion) while liquids and gases usually suffers change in volume only.Hence there are three types of thermal expansion

Linear expansion

Superficial expansion

Volumetric expansion

Linear expansion

Consider a linear object of length l at a temperature t₁.Now it is heated through a small temperature dtas a result the fractional change in length (dl/l) is directly proportional to change in temperature `dt’

so $\frac{{d\,l}}{l}\propto dt$

in place of proportionality sign we put a constant a.wherea is known as the thermal coefficient of linear expansion and it depends on the material of the rod.

therefore $\frac{{d\,l}}{l}=\alpha dt$ .

$\frac{{d\,l}}{l}=\alpha dt$

$\Rightarrow dl=l\alpha dt$ For a small change in temperature we can integrate dl to get final length after heating.

$\Rightarrow \int\limits_{{{{l}_{{}}}}}^{L}{{dl}}=l\alpha \int\limits_{{{{t}_{1}}}}^{{{{t}_{2}}}}{{dt}}$

$\Rightarrow L-l=l\alpha ({{t}_{2}}-{{t}_{1}})$

hence $\Rightarrow L=l[1+\alpha ({{t}_{2}}-{{t}_{1}})]$ or $L=l[1+\alpha \Delta t]$

here we have assumed that change in length varies linearly with change in temperature and the change in length is also very small .But it happens only for a small change in temperature. The graph between the length `L’ and $\Delta t$ is a straight line as shown below

Similar expressions can be written for change in area for a two-dimensional object and change in volume for a three dimensional object.

S = S₀ (1 + βΔt ) $\Rightarrow$ $\Delta S=\beta {{S}_{o}}\,\Delta t$

& V = V₀ (1 + γΔt) $\Rightarrow$ $\Delta V=\gamma {{V}_{o}}\,\Delta t$

whereb and gare coefficients of superficial expansion and coefficient of cubical expansion respectively.

The relation between a, β and γ for an isotropic solid is a : β : γ = 1 : 2 : 3.

The three coefficients of expansion are not constant for a given solid. Their values depend on the temperature range in which they are measured. The values of a , β , γ are independent of the units of length, area and volume respectively.

For anisotropic solids $\gamma ={{\alpha }_{x}}+{{\alpha }_{y}}+{{\alpha }_{z}}$ where a_x, a_y, and a_zrepresent the mean coefficients of linear expansion along three mutually perpendicular directions.

(For the derivation of relation between a, b and g please watch the video lectures)

Thermal Stress

When a rod is held between two fixed supports and its temperature is increased, the fixed supports do not allow the rod to expand, which results in a stress, which is called thermal stress.

Let a rod of length l is held between two fixed supports and its temperature is increased by ΔT, then change in length of the rod

Δl = laΔT

where a is thermal coefficient of linear expansion for the material of the rod.

If Y is the Young’s modulus for the material of the rod and A is the area of cross-section of the rod then mechanical compression in the rod.

Δl’ = $\left( {\frac{F}{A}} \right)\,\,\frac{l}{Y}$

Since support is rigid Δ l + Δl’ = 0, which gives $\frac{F}{A}\,=\,-\,\alpha \,Y\Delta T$

This is thermal stress, – ve sign indicates that thermal stress is compressive in nature.

Some other examples of Thermal Expansion in Solids

(1) Bi-metallic strip

(2) Effect of temperature on the time period of a simple pendulum

(3) Error in scale reading due to expansion or contraction

(5) Expansion of cavity

(6) When rails are laid down on the ground, space is left between the ends of two rails.

(7) The transmission cable are not tightly fixed to the poles.

(8) Test tubes, beakers and crucibles are made of pyrex-glass or silica because they have very low value of coefficient of linear expansion.

(9) The iron rim to be put on a cart wheel is always of slightly smaller diameter than that of wheel.

For detailed explanation of first five examples, watch the video lectures

Variation of Density with Temperature

Most substances (solid and liquid) expand when they are heated, i.e., volume of a given mass of a substance increases on heating, so the density should decrease $\left( {\text{as }\rho =\frac{m}{V}} \right)$ ( where m is mass ).

Let the density be $\rho$ at temperature t₁ and $\rho ‘$ at temperature t₂. Then $\displaystyle \rho =\frac{m}{V}$ and $\displaystyle \rho ‘=\frac{m}{{V’}}$

Þ $\frac{{{\rho }’}}{\rho }=\frac{V}{{{V}’}}=\frac{V}{{V+\Delta V}}=\frac{V}{{V+\gamma V\Delta t}}=\frac{1}{{1+\gamma \,\Delta t}}$

Þ ${\rho }’=\frac{\rho }{{1+\gamma \,\Delta t}}$

If $\gamma \,\Delta t$ is very small binomial approximation can be used

So $\rho ‘=\rho \,{{(1+\gamma \,\Delta t)}^{{-1}}}$

hence $\rho ‘=\rho \,(1-\gamma \,\Delta t)$ so with increase in temperature the density decreases.

Illustration

The design of some physical instrument requires that there be a constant difference in length of 10 cm between an iron rod and copper rod laid side by side at all temperatures. Find their lengths.

(α_Fe = 11 × 10^–6 C^–1, α_Cu = 17 × 10^–6 °C^–1)

Solution

Since the α_CU>α_Fe so, length of iron rod should be greater than the length of copper rod.

Let the initial lengths of iron and copper rods be l₁ and l₂ , then

l₁–l₂ = 10 cm … (i)

also since the difference has to be constant at all the temperatures, so

Δl = l₁α_FeΔT = l₂α_CuΔT

$\frac{{{{l}_{1}}}}{{{{l}_{2}}}}\,=\,\frac{{{{\alpha }_{{Cu}}}}}{{{{\alpha }_{{Fe}}}}}$ … (ii)

Solving (i) & (ii) we get l₁ =28.3 cm and l₂ = 18.3 cm

Illustration

The height of mercury column measured at t°C with a brass scale which gives correct reading H₁ at 0°C. What height H₀ will the mercury column have at 0°C? The coefficient of linear expansion of brass is α and the coefficient of volume expansion of mercury is g.

Solution

H₁ at t°C = H₁ (1 + αt) actually since pressure is same, so

ρ₀gH₀ = ρ_tgH₁ (1 + αt)

H₀ = H₁ (1 + αt) $\frac{{{{\rho }_{t}}}}{{{{\rho }_{0}}}}$ = H₁ (1 + αt) (1 + gt)^–1

H₀ = H₁ [1 + (α- g) t]

Objective Assignment

Two rods of lengths ${{l}_{1}}$ and ${{l}_{2}}$ are made of materials whose coefficients of linear expansion are ${{\alpha }_{1}}$ and ${{\alpha }_{2}}$ respectively. If the difference between the two lengths is independent of temperature, then

(a) $\frac{{{{l}_{1}}}}{{{{l}_{2}}}}=\frac{{{{\alpha }_{2}}}}{{{{\alpha }_{1}}}}$

(b) $\frac{{{{l}_{1}}}}{{{{l}_{2}}}}=\frac{{{{\alpha }_{1}}}}{{{{\alpha }_{2}}}}$

(d) $\alpha _{2}^{2}{{l}_{1}}=\alpha _{1}^{2}{{l}_{2}}$

Ans (a)

A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically in the hand, such that iron strip is towards the left hand and copper strip is towards the right hand side. This bimetal is then heated by flame. The bimetal strip will

(a) remain straight

(b) bend towards right

(d) no change

Ans (b)

Subjective Assignment

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QUIZ

Created on August 05, 2022 By

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Thermal Expansion And Its Types (Basic Level)

1 / 10

A piece of glass is heated to a high temperature and then allowed to cool. If it cracks, a probable reason for this is the following property of glass

Low thermal conductivity

High thermal conductivity

High specific heat

High melting point

2 / 10

If the radius and length of a copper rod are both doubled, the rate of flow of heat along the rod increases

4 times

2 times

8 times

16 times

3 / 10

Consider two hot bodies ${{B}_{1}}$ and ${{B}_{2}}$ which have temperatures ${{100}^{o}}C$ and ${{80}^{o}}C$ respectively at $t=0$ . The temperature of the surroundings is ${{40}^{o}}C$ . The ratio of the respective rates of cooling ${{R}_{1}}$ and ${{R}_{2}}$ of these two bodies at $t=0$ will be

{{R}_{1}}:{{R}_{2}}=3:2

{{R}_{1}}:{{R}_{2}}=5:4

{{R}_{1}}:{{R}_{2}}=2:3

None Of these

4 / 10

A bucket full of hot water cools from ${{75}^{o}}C$ to ${{70}^{0}}C$ in time ${{T}_{1}}$ , from ${{70}^{o}}C$ to ${{65}^{o}}C$ in time ${{T}_{2}}$ and from ${{65}^{o}}C$ to ${{60}^{o}}C$ in time ${{T}_{3}}$ , then

{{T}_{1}}={{T}_{2}}={{T}_{3}}

{{T}_{1}}&gt;{{T}_{2}}&gt;{{T}_{3}}

{{T}_{1}}&lt;{{T}_{2}}&lt;{{T}_{3}}

{{T}_{1}}&gt;{{T}_{2}}&lt;{{T}_{3}}

5 / 10

Hot water cools from ${{60}^{o}}C$ to ${{50}^{o}}C$ in the first 10 minutes and to ${{42}^{0}}C$ in the next 10 minutes. The temperature of the surrounding is

{{5}^{o}}C

{{10}^{o}}C

{{15}^{o}}C

{{20}^{o}}C

6 / 10

A metal ball of surface area $200\ c{{m}^{2}}$ and temperature ${{527}^{o}}C$ is surrounded by a vessel at ${{27}^{o}}C$ . If the emissivity of the metal is 0.4, then the rate of loss of heat from the ball is $(\sigma =5.67\times {{10}^{{-8}}}J/{{m}^{2}}\ -s-{{k}^{4}})$

108 joules approx

182 joules approx

168 joules approx

138 joules approx

7 / 10

According to Wein's law

{{\lambda }_{m}}T

= constant

\frac{{{{\lambda }_{m}}}}{T}

= constant

\frac{T}{{{{\lambda }_{m}}}}

= constant

T+{{\lambda }_{m}}

= constant

8 / 10

The thickness of a metallic plate is 0.4 cm. The temperature between its two surfaces is ${{20}^{o}}C$ . The quantity of heat flowing per second is 50 calories from $5c{{m}^{2}}$ area. In CGS system, the coefficient of thermal conductivity will be0.4

0.4

0.6

0.2

0.5

9 / 10

In Searle's method for finding conductivity of metals, the temperature gradient along the bar

Is greater nearer the hot end

Is greater nearer to the cold end

Is the same at all points along the bar

Increases as we go from hot end to cold end

10 / 10

The dimensions of thermal resistance are

{{M}^{{-1}}}{{L}^{{-2}}}{{T}^{3}}K

M{{L}^{2}}{{T}^{{-2}}}{{K}^{{-1}}}

M{{L}^{2}}{{T}^{{-3}}}K

M{{L}^{2}}{{T}^{{-2}}}{{K}^{{-2}}}

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Thermal Expansion And Its Type (Board Level)

Theory For Notes Making

Objective Assignment

Subjective Assignment

On Line Test

Video Lecture

Theory For Notes Making

Objective Assignment

Subjective Assignment

QUIZ