Question

A spherical brass shell has an interior volume of 1.30 x 10-3 m3. Within this interior volume is a solid steel ball that have a volume of 0.89 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 12 C°. What is the volume of mercury that spills out of the hole?

Answer 1

?V/V? = 3??T ------------------------> (i)

? = coefficient of linear expansion
for steel = 13 x 10^(-6) per °C or K
for brass = 19 x 10^(-6) per °C
for mercury = 181/3 x 10^(-6) per °C

brass shell interior volume V? = 1.30 x 10 ? ³ m³
steel ball volume = v? = 0.89 x 10 ? ³ m³
mercury volume = V'? = (1.30- 0.89) = 0.41 x 10 ? ³ m³

from (i) extra volume of brass due to expansion

=> ?V = V?(3??T) = (1.30 x 10 ? ³ ) [3x19 x 10^(-6)]12 = 8.892 x 10^(-7) m³

?v = v?(3??T) = (0.89 x 10 ? ³ ) [3x13 x 10^(-6)]12= 4.1652 x 10^(-7) m³

?V' = V'?(3??T) = (0.89 x 10 ? ³ ) [3x(181/3) x 10^(-6)]12 = 1.93308 x 10^(-6) m³

the mercury that will spill = 1.93308 x 10^(-6) m³ - [available extra volume after expansion]

the mercury that will spill = 1.93308 x 10^(-6) m³ - [8.892 x 10^(-7) m³ - 4.1652 x 10^(-7) m³]

= [1.4604 x 10^(-6)] m³