Volume Thermal Expansion Question ?

[cheechnchong]

Volume Thermal Expansion Question!??

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)

= 5.70 x 10^.7 m^3

Let Me know if this number is right. I really struggled on this one...kinda tricky for me!

 

[Doc Al]

My Approach:

Change in Velocity = mercury(BVchangeT) - steel(BVchangeT) - brass(BVchangeT)

I assume you meant volume, not velocity!

In any case, you want to compare the increase in volume of the mercury to the increase of volume of the space between steel ball and brass shell: spillover volume = change in mercury volume - change in interior volume.

That change in interior volume is given by: change in brass volume - change in steel volume. Rewrite your equation, being more careful with your signs.

 

[kyle8921]

I would first confirm that the given values for the thermal expansion coefficients (B) for brass, mercury, and steel are correct. Then, I would use the given formula for thermal expansion to calculate the change in volume of each material at a temperature increase of 12 degrees Celsius. This would give me the change in volume for each material, which I would then use to calculate the total change in volume of the arrangement.

Next, I would subtract the volume of the steel ball from the total change in volume to get the volume of the mercury that spilled out of the hole. This is because the steel ball is taking up space within the brass shell, so it would not contribute to the spilled mercury.

Finally, I would double-check my calculations and units to ensure accuracy. If the given values and formula were correct, then the calculated volume of spilled mercury should be accurate as well. If the number does not seem correct, I would go back and check my calculations and units to identify any errors.