A spherical brass shell has an interior volume of 1.27 x 10 -3 m 3 . Within this

Question

A spherical brass shell has an interior volume of 1.27 x 10-3 m3. Within this interior volume is a solid steel ball that have a volume of 0.61 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 15 C°. What is the volume of mercury that spills out of the hole?

So, I plugged in my numbers to a formula (which was a post on Chegg) and ended up with the final equation of: 1674.45 x 10^-9 – (1811.7 x 10^-9 – 356.85 x 10^-9)= 219.6 x 10^-9 or 2.2 x 10^-7 and both answers were wrong. I have one more chance to input an answer…Please help!

Explanation / Answer

initial volumes:

V10 = 1.27 e-3 m^3

V20 =0.61 e -3 m^3

V30 =0.66 e -3 m^3

after temp rise:

V1= V10 (1+31T)= 1.27 e -3*(1 + 3*18.7 e -6 *15) = 1.27106 e -3 m^2

V2=V20 (1+32T)= 0.61 e -3 * (1 + 3 * 12e-6 * 15) = 0.33001 e-3 m^3

V3=V30 (1+33T)= 0.66 e -3* (1 + 3 *61e-6* 15) = 0.661817e-3 m^3

new volume of space between the steel ball and brass shell

dV =V1-V2 =1.27106e -3 - 1.27e -3 = 1.06 e -6 m^3

volume of mercury overflow through hole

dV = 0.661817 e -3 - 1.06 e -6 = 0.6607 e -3 m^3

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