Question

The density of mercury is $13.6\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right.$ at $0^{°}C$ and its coefficient of cubical expansion is $1.82\times {10}^{-4}°C^{-1}$. Calculate the density of mercury at $50°C$.

Answer 1

Step 1. Given data:

The density of mercury $\left(\rho \right)$ = $13.6\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right.$ at $0^{°}C$

The coefficient of cubical expansion is $\left(\alpha \right)$ $1.82\times {10}^{-4}°C^{-1}$

Temprature at which we have to find density of mercury $∆T$ = $50°C$

Step 2. Formula used:

Let the mass of mercury $m=13.6gm$

and initial volume be $V=1c{m}^3$ at $0^{°}C$

Volume thermal expansion coefficient formula:

Then change in volume, $∆V=V\alpha ∆T$

Step 3. Calculations:

Substituting the given values, we get.

$∆V=1\times 1.82\times {10}^{-4}\times 50=91\times {10}^{-4}c{m}^3$

New volume $$\begin{gathered} V_1=V+∆V \\ =1+0.0091=1.0091c{m}^3 \end{gathered}$$

New density $\left({\rho }_1\right)$=$=\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$V_1$}\right.=\raisebox{1ex}{$13.6$}\!\left/ \!\raisebox{-1ex}{$1.0091$}\right.=13.48\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right.$​

Thus, the density of mercury at $50°C$ will be $13.48\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$c{m}^3$}\right.$.