Question

The densities of wood and benzene at $0^{\circ }C$ are 880 kg $m^3$ and 900 kg $m^{-3}$ respectively. The coefficients of volume expansion are $1.2\times {10}^{-3}{}^{\circ }{C}^{-1}$ for wood and $1.5\times {10}^{-3}{}^{\circ }C$ for benzene. At what temperature will a piece of wood just sink in benzene ?

Answer 1

Given $f_w=880kg/m^3,$

$f_b=900kg/m^3$

$T_1=0^{\circ }C,$

${\gamma }_w=1.2\times {10}^{-3}/{}^{\circ }C,$

${\gamma }_b=1.5\times {10}^{-3}/{}^{\circ }C$

The sphere begins to sink when

$\Rightarrow (mg{)}_{sphere}$ = displaced water

$V{f}_w^{\prime }g=V{f}_b^{\prime }g$

$\Rightarrow \frac{f_w}{1+{\gamma }_w\Delta \theta }=\frac{f_b}{1+{\gamma }_b\Delta \theta }$

$\Rightarrow \frac{880}{1+1.2\times {10}^{-3}\Delta \theta }=\frac{900}{1+1.5\times {10}^{-3}\Delta \theta }$

$\Rightarrow 880+880\times 1.5\times {10}^{-3}\Delta \theta$

$=900+900\times 1.2\times {10}^{-3}\Delta \theta$

$\Rightarrow (880\times 1.5\times {10}^{-3}-900\times 1.2\times {10}^{-3})\Delta \theta =20$

$\Rightarrow (1320-1080)\times {10}^{-3}\Delta \theta =20$

$\Rightarrow \Delta \theta ={83.3}^{\circ }C$

$T_2-T_1=83\Rightarrow T_2-0^{\circ }=83$

$\Rightarrow T_2={83}^{\circ }C$