# Variation in Density with temperature

#### Variation in Density with Temperature

In this article we are going to discuss about Variation in Density with temperature of solids and liquids.

We know that on heating a substance, its volume increases but mass remains the same.

Hence we conclude that :-

Heating of bodies is accompanied by an increase in volume and decrease in density.

Let

M = mass of a substance

V0 = initial volume,   ρ0 = initial density

ΔT = change in temperature

γ = coefficient of volume expansion of the substance

V = final volume,    ρ = final density

Since mass remains the same ⇒  M = V0ρ0 = Vρ

Now final volume is given by  V = V0(1 + γΔT)

Dividing both sides by M, we get

$\displaystyle \frac{V}{M}=\frac{{{V}_{0}}}{M}(1+\gamma \Delta T)$

$\displaystyle \Rightarrow \frac{1}{\rho }=\frac{1}{{{\rho }_{0}}}(1+\gamma \Delta T)$

$\displaystyle \Rightarrow {{\rho }_{0}}=\rho (1+\gamma \Delta T)$

$\displaystyle \Rightarrow \rho =\frac{{{\rho }_{0}}}{1+\gamma \Delta T}$

$\displaystyle \Rightarrow \rho ={{\rho }_{0}}{{(1+\gamma \Delta T)}^{-1}}$

Now as γΔT<<<1, hence

$\displaystyle \Rightarrow \rho ={{\rho }_{0}}(1-\gamma \Delta T)$

This is the expression to find final density of a material after an increase in temperature of ΔT.

Example. A sphere of diameter 7 cm and mass 266.5 gm floats in a bath of a liquid. As the temperature rises, the sphere begins to sink & at a temperature of 35ºC the sphere gets completely submerged. If the density of a liquid at is 1.527 gm/cc, then neglecting the expansion of the sphere, find the coefficient of cubical(volume) expansion of the liquid.

Solution.  Volume of the sphere = $\displaystyle \frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\times \frac{22}{7}\times {{(7)}^{3}}=179.59c{{m}^{3}}$

∴ Density of sphere = 266.5/179.59 = 1.484 g/cc

The density of the material of the sphere and the density of the liquid becomes equal at the moment when the sphere is just submerged in the liquid.

$\displaystyle \Rightarrow {{\rho }_{L}}={{\rho }_{S}}$

$\displaystyle \Rightarrow \frac{{{\rho }_{0}}}{1+\gamma \Delta T}={{\rho }_{S}}$

$\displaystyle \Rightarrow \frac{1.527}{1+\gamma \times 35}=1.484$

γ = 8.27×10-4 ºC-1