a spherical brass shell has an interior volume of 1.60 x 10^-3 m^3. Within this interior volume is a solid steel ball that has a volume of 0.70 x 10^-3 m^3. 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 12 degrees celsius. What is the volume of the mercury that spills out of the hole?
B(brass) = 57 x 10^-6 (degrees celsius)^-1
B(mercury) = 182 x 10^-6 (degrees celsius)^-1
B(steel) = 36 x 10^-6 (degrees celsius)^-1
My Approach:
Change in Velocity = mercury(BVchangeT) - steel(BVchangeT) - brass(BVchangeT)
= (1.6 x 10^-3 - 0.7 x 10^-3)m^3 (182 x 10^-6 celsius^-1)(12 celsius) + (36 x 10^-6 celsius^-1) (0.7 x 10 m^3)(12 celsius) - (1.60 x 10^-3 m^3)( 57 x 10^-6 celsius^-1)(12 celsius)
= 1.17 x 10^ -6 m^3 <----my answer, what do you think?
2006-11-15
12:09:18
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1 answers
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asked by
Anonymous
in
Science & Mathematics
➔ Physics
Change in Volume= mercury(BVchangeT) + steel(BVchangeT) - brass(BVchangeT)
^^edit
2006-11-15
12:10:05 ·
update #1