Question

A brass cube, 10 cm on a side, is raised in temperature

Answer 1

Answer:

$\delta L\%=22.5\%$

Explanation:

Assuming that the thermal expansion of brass cube occurs isotropically (i.e. equal in all the directions).

Given:

• length of the cube, $l=10\ cm=0.1\ m$

• change in temperature of the cube, $\Deta T=200\^{\circ}C$

• coefficient of volume expansion, $\beta=57\times 10^{-6}\ ^{\circ}C^{-1}$

Hence volume of the cube:

$V=10^{-3}\ m^3$

Now the volume of the cube after expansion:

$\delta V=V.\beta.\Delta T$

$\delta V=10^{-3}\times 57\times 10^{-6}\times 200$

$\delta V=1.14\times 10^{-5}\ m^3$

Therefore,

$\delta L=0.0225\ m$

Now the percentage change in the edges of the cube:

$\delta L\%=\frac{\delta L}{L} \times 100$

$\delta L\%=\frac{0.0225}{0.1} \times 100$

$\delta L\%=22.5\%$