Question

The cubical expansivities of benzene and wood are $1.2\times {10}^{-3}{}^{\circ }{C}^{-1}$ and $1.5\times {10}^{-4}{}^{\circ }{C}^{-1}$, their densities at $0^{\circ }C$ are $9\times {10}^{2}kg{m}^{-3}$ and $8.8\times {10}^{2}kg{m}^{-3}$ respectively. At what temperature will wood just sink in benzene?

Answer 1

Let the volume of wood be $V_W$ and mass be $m_W$.

So ${\rho }_w=\frac{m_W}{V_W}=8.8\times {10}^2$ ....(1)

Let the volume of benzene be $V_b$ and mass be $m_b$.

So ${\rho }_b=\frac{m_b}{V_b}=9\times {10}^2$ ....(1)

When the Temperature is increased by $\Delta T$,

Volume of wood be $V_{W1}=V_W(1+{\gamma }_{wood}\Delta T)$

and mass be $m_W$

So ${\rho }_{w1}=\frac{m_W}{V_W(1+{\gamma }_{wood}\Delta T)}=\frac{8.8\times {10}^2}{(1+{\gamma }_{wood}\Delta T)}$ ....(3)

Volume of benzene be $V_{b1}=V_b(1+{\gamma }_{benzene}\Delta T)$

and mass be $m_b$.

So ${\rho }_{b1}=\frac{m_b}{V_b(1+{\gamma }_{benzene}\Delta T)}=\frac{9\times {10}^2}{(1+{\gamma }_{benzene}\Delta T)}$ ..(4)

So if both densities become same, wood will start sinking i.e. ${\rho }_{w1}={\rho }_{b1}$

$\frac{8.8\times {10}^2}{(1+1.5\times {10}^{-4}\Delta T)}=\frac{9\times {10}^2}{(1+1.2\times {10}^{-3}\Delta T)}$

$\Delta T={21.70}^{0}C$