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Help with Archimedes' Principle

  • Thread starter lmbiango
  • Start date
  • #1
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Homework Statement



A solid aluminum sphere has a radius of 1.84 m and a temperature of 77.5 °C. The sphere is then completely immersed in a pool of water whose temperature is 26.7 °C. The sphere cools, while the water temperature remains nearly at 26.7 °C, because the pool is very large. The sphere is weighed in the water immediately after being submerged (before it begins to cool) and then again after cooling to 26.7 °C. Use Archimedes' principle to find the magnitude of the difference between the weights.



Homework Equations



Volume of the sphere: (4/3) pi (R)^3
V=26.094

and change in volume: coefficient of volume expansion for aluminum (69 x 10^-6) * (Initial Volume: 26.094) * (Change in Temp: -50.8)
= - .0915

This is the other equation I have:
difference in weight = -(density)(gravity)(volume)
But the answer I get isn't right...


The Attempt at a Solution



I think everything I am doing is right up until the part about finding the difference in weights. I don't know what equation to use, or how to use it I guess. Any help would be appreciated.
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
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Remember the difference in volume leads to a different upthrust because of the different amount of water displaced - the mass of the Al sphere doesn't change.
 
  • #3
21
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Remember the difference in volume leads to a different upthrust because of the different amount of water displaced - the mass of the Al sphere doesn't change.
So, are you saying that the difference between the weights would be zero? I thought that would make sense, but I tried that also (although I thought it seemed too easy) and the homework tool said it wasn't right... all I know is the units are newtons ...
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,774
12
No the weight would be different because the shrunk ball would displace less water and so receive less upthrust and so weigh more.
But the difference in volume corresponds to a different amount of water - not a different amount of Al so it is the density of water not Al that you need.
 

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