Question

The densities of wood and benzene at $0^{\circ }C$ are $880kg{m}^{-3}$ and $900kg{m}^{-3}$, respectively. The coefficient of volume expansion is $12\times 10^{-3}{C}^{-1\circ }$ for wood and $1.5\times 10^{-3\circ }{C}^{-1}$ for benzene. Then the temperature at which a piece of wood just sinks in benzene is

• A. $88^{\circ }C$
• B. $90^{\circ }C$
• C. $83.3^{\circ }C$
• D. $90.3^{\circ }C$

Answer 1

Answer: (C)

Given, density of wood at $0{}^{\circ }C,{\rho }_w=880Kg{m}^{-3}$
density of benzene at $0^{\circ }C,{\rho }_b=900kg{m}^{-3}$
coefficient of volume expansion of wood,
${\gamma }_w=1.2\times 10^{-3\circ }{C}^{-1}$
coefficient of volume expansion of benzene,
${\gamma }_b=1.5\times 10^{-3}{}^{\circ }{C}^{-1}$
and initial temperature, $T_1=0^{\circ }C$
Let $T_2$ be the temperature at which pieces of wood will just sink in benzene and $\Delta T=T_2-T_1$
The piece of wood begins to sink when it weight is equal to the weight of benzene displaced. Mass $=$ Volume $\times$ Density Therefore, $V{\rho }_{w}g=V{\rho }_{b}g$
$\therefore \frac{{\rho }_w}{1+{\gamma }_w\Delta T}=\frac{{\rho }_b}{1+{\gamma }_b\Delta T}$
$\frac{880}{1+1.2\times 10^{-3}\Delta T}=\frac{900}{1+1.5\times 10^{-3}\Delta T}$
$880+880\times 1.5\times 10^{-3}\Delta T=900+900\times 1.2\times 10^{-3}\Delta T$
$(1320-1080)\times 10^{-3}\Delta T=20$
$\Delta T=\frac{20}{240\times 10^{-3}}$
$\Delta T=83.3^{\circ }C$
$T_2-T_1=83.3^{\circ }C$
Hence, $T_2=83.3^{\circ }C$