Question

The volume of thin brass and the volume of a solid cube made of same material is both equal to $1m^3$ at $0^{0}C$. What will be the ratio of change in volumes of vessel and the cube on being heated to ${25}^0$ C? Given $\alpha$ for brass = $1.9\times {10}^{-5}{/}^{{}^0}C$

Answer 1

A hollow cube and a solid cube is made of brass $(\alpha =1.9\times {10}^{-5}{k}^{-1})$

as volume $V_0=1m^3,$ side length = $L_0=1m$

and area of one surface, $A_0=1m^2$

on heating the hollow cube, it experiences Area expansion.

so, $A=A_0(1+2\alpha \Delta T)$

$\Delta T:$ change in temp = 25 k.

$\Rightarrow A=1(1+2\times 1.9\times {10}^{-5}\times 25)=1.001m^2$

so, new volume of hollow cube,

$V_1=A^{3/2}$ $[A=L^2,V=L^3$ given $V^2=A^3$ for a cube]

$=(1.001{)}^{3/2}$

$\Rightarrow V_1=1.0015m^3$

On the other hand, the solid cube experiences volumetric expansion,

$V_2=V_0(1+3\times AT)$

$=1(1+3\times 1.9\times {10}^{-5}\times 25)=\mathrm{1.001425.}{m}^3$

Thus ratio of change of volume,

$\frac{V_1-V_0}{V_2-V_0}=\frac{1.0015-1}{1.001425-1}=\frac{0.0015}{0.001425}=\frac{1500}{1425}=\frac{20}{19}$