Variation in Density with temperature

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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

 

About the author

Manoj Kumar Verma

Hi, I'm Manoj Kumar Verma, a physics faculty having 7 years of teaching experience. I have done B.Tech (E.E.). I am also a YouTuber and Blogger. This blog is dedicated to help students learn the physics concepts easily.

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