Question

The densities of wood and benzene at $0^{\circ} C$ are $880 \,kg \,m ^{-3}$ and $900 \,kg m ^{-3}$, 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 $0{ }^{\circ} C , \rho_{w}=880\, Kg\, m ^{-3}$
density of benzene at $0^{\circ} C , \, \rho_{b}=900\, kg \, m ^{-3}$
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, $T_{2}=83.3^{\circ} C$