Thermal Expansion And Its Type (NEET LEVEL)

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

Expansion of liquids

Liquids like solids expand on heating. The liquids have no definite shape and they experience volume expansion only. As the liquid is contained in a vessel, the observed expansion of liquid is less than the real expansion, because the capacity of vessel containing liquid increases with temperature. If {{\gamma }_{r}} is the coefficient of real expansion of liquid and {{\gamma }_{a}} the coefficient of apparent expansion of liquid and {{\gamma }_{g}}  coefficient of volume expansion of solid container then

{{\gamma }_{r}}={{\gamma }_{a}}+{{\gamma }_{g}}

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 tis 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.