Question

12. A piece of metal weighs 46 g in air and 30 g in a liquid of density 1.24 x 10³ kg m^-3 kept at 27$°$C. When the temperature of the liquid is raised to 42$°$C, the metal piece weighs 30.5 g. The density of the liquid at 42$°$C is 1.20 x 10³ kg m ^-3. Calculate the coefficient of linear expansion of the metal.

Answer 1

Dear Student

$$\begin{gathered} Lossinmassofthemetal=massofliquiddisplaced=46-30=16g \\ Volumeofmetalat{27}^{o}C=Volumeofliquiddisplacedat{27}^{o}C \\ {V}_1=\frac{16}{densityofliquid}=\frac{16g}{1.24gc{m}^{-3}}=12.903c{m}^3 \\ Similarlyvolumeofmetalat{42}^{o}C \\ {V}_2=\frac{46-30.5}{1.2gc{m}^{-3}}=12.917c{m}^3 \\ Coefficientofcubicalexpansionofmetal \\ \gamma =\frac{V_2-V_1}{V_1\left(T_2-T_1\right)}=\frac{\left(12.917-12.903\right)}{12.903\left(42-27\right)}=7.23\times {10}^{5o}{C}^{-1} \\ Since\gamma =3\alpha \\ \alpha =\frac{1}{3}\gamma =\frac{1}{3}\left(7.23\times {10}^{5o}{C}^{-1}\right)=2.41\times {10}^{5o}{C}^{-1} \end{gathered}$$
Regards