Question

A spherical brass shell has an interior volume of 1.84 x 10-3 m3. Within this interior volume is a solid steel ball that have a volume of 0.63 x 10-3 m3. The space between the steel ball and the inner surface of the brass shell is filled completely with mercury. A small hole is drilled through the brass, and the temperature of the arrangement in increased by 19 C°. What is the volume of mercury that spills out of the hole?

Answer 1

Coefficient of linear expansion of the mentioned materials in the problems are -

brass - 19 x 10^-6
steel - 11 x 10^-6
mercury - 60 x 10^-6

Now, as we know that the volumetric expansion is 3 times linear, so when heated by 19 deg. C brass volume would expand by 1083*10^-6 of initial volume,

so brass sphere volume will increase by 1083 x 10^-6 x 1.84 x 10^-3 = 1.993 x 10^-6 m^3.

For steel sphere volumetric expansion is 627 x 10^-6 and increase in volume is 627 x 10^-6 x 0.63 x 10^-3

= 0.395 x 10^-6 m^3.

Finally mercury volumetric expansion is 3.42 x 10^-3 with initial volume 1.84 x 10^-3 - 0.63 x 10^-3 = 1.21 x 10^-3

So, it give increase in volume 4.138 x 10^-6 m^3.

Therefore, we have excessive increase in volume = 4.138 x 10^-6 - 1.993 x 10^-6 m^3 + 0.395 x 10^-6 m^3

= 2.54 x 10^-6 m^3.

So, volume of mercury that spills out of the hole = 2.54 x 10^-6 m^3.