Help with Archimedes' Principle

In summary, the problem involves a solid aluminum sphere with a radius of 1.84 m and a temperature of 77.5 °C being immersed in a pool of water at 26.7 °C. The difference in weight between the sphere before and after cooling to 26.7 °C can be found using Archimedes' principle. The volume of the sphere is calculated using its radius and the change in volume is determined using the coefficient of volume expansion for aluminum and the change in temperature. The difference in weight is then found by using the difference in volume and the density of water.
  • #1
lmbiango
21
0

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.
 
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  • #2
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
mgb_phys said:
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
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.
 

What is Archimedes' Principle?

Archimedes' Principle is a scientific law that states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. In simpler terms, it explains why objects float or sink in a liquid.

How can Archimedes' Principle be used?

Archimedes' Principle has many real-world applications, such as determining the volume of irregularly shaped objects, designing ships and submarines, and understanding the behavior of hot air balloons. It is also essential in understanding the concept of density and buoyancy.

What factors affect the buoyant force according to Archimedes' Principle?

The buoyant force is affected by two main factors: the density of the fluid and the volume of the object. The denser the fluid, the greater the buoyant force. Similarly, the larger the volume of the object, the greater the buoyant force.

What is the relation between density and buoyancy according to Archimedes' Principle?

Archimedes' Principle states that the buoyant force is equal to the weight of the fluid that the object displaces. This means that if an object is less dense than the fluid it is submerged in, it will experience a net upward force and float. If the object is more dense, it will experience a net downward force and sink.

Are there any exceptions to Archimedes' Principle?

Archimedes' Principle is a fundamental law of physics and holds true in most cases. However, there are a few exceptions, such as when the object is compressible or when the fluid is not at rest. In these cases, more complex equations are required to accurately calculate the buoyant force.

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