Question

A spherical brass shell has an interior volume of 1.55 × 10-3 m³. Within this interior volume is a solid steel ball that has a volume of 6.00 × 10-4 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 is increased by 14 °C. What is the volume of the mercury that spills out of the hole?

Answer in m^3. Answer is not 7.031 x 10^-7 m^3

Answer 1

ΔV/Vₒ = 3αΔT ---------------- (i)

α = 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 Vₒ = 1.55 x 10^-3 m^3

steel ball volume = vₒ = 0.6 x 10^-3 m^3

mercury volume = V'ₒ = (1.55 - 0.60) = 0.95 x 10^-3 m^3

from (i) extra volume of brass due to expansion

ΔV = Vₒ(3αΔT) = (1.55 x 10^-3 ) [ 3 x 19 x 10^-6]14 = 1.24 x 10^-6 m^3

Δv = vₒ(3αΔT) = (0.6 x 10^-3 ) [3 x 13 x 10^-6]14 = 3.276 x 10^-7 m^3

ΔV' = V'ₒ(3αΔT) = (0.95 x 10^-3 ) [3 x (181/3) x 10^(-6)]14 = 2.4 x 10^-6 m^3

the mercury that will spill = 2.4 x 10^-6 m^3 - [available extra volume after expansion]

the mercury that will spill = 2.4 x 10^-6 m^3 - [1.24 x 10^-6 - 3.276 x 10^-7]

= 1.16 x 10^-6 + 0.3276 x 10^-6

= 1.4876 x 10^-6 m^-3