Question

A piece of metal weighs 46 g in air and 30 g in a liquid of density $1.24\times {10}^{3}kg/m^3$kept at ${27}^{o}C$. when the temperature of the liquid is raised to ${42}^{o}C$, the metal piece weighs 30.5 g. The density of the liquid at ${42}^{o}C$ is $1.20\times {10}^{3}kg/m^3$. Calculate the coefficient of linear expansion of the metal.

Answer 1

Loss of weight at ${27}^{0}C$ $46$

$\begin{array}{c}30=16=v_1\times 1.24rl\times g\\ \left(i\right)Lossofweightat{42}^{0}Cis\\ 30.5=15.5=V_2\times 1.2rl\times g\\ \left(ii\right)Nowdividing\left(i\right)by\left(ii\right),weget\\ \frac{16}{15.5}=\frac{V_1}{V_2}\times \frac{1.24}{1.2}\\ But\frac{V_2}{V_1}=1+3a\left(t_2?t_1\right)\\ =\frac{15.5\times 1.24}{16\times 1.2}=1.001042\\ 3a\left({42}^0?{27}^0\right)=0.001042\\ a=2.316\times 10?5{/}^{0}C\end{array}$