Find the magnitude of the difference between the weights (temperature question)

[physics1007]

Homework Statement

A solid aluminum sphere has a radius of 2.34 m and a temperature of 89.0 °C. The sphere is then completely immersed in a pool of water whose temperature is 22.3 °C. The sphere cools, while the water temperature remains nearly at 22.3 °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 22.3 °C. Use Archimedes' principle to find the magnitude of the difference between the weights.

Homework Equations

Weight of sphere in water before cooling W1 = W - ρ * g * V1 ρ = density of water
where W = actual weight of sphere and V1 the volume before cooling
Also, W2 = W - ρ * g * V2 Then W2 - W1 = -(V2 - V1 ) * ρ * g
V1 = 4/3 * π * r^3 V2 = 4/3 * π * (r + Δr)^3
Expanding (r + Δr)^3 and dropping any terms with Δr^3 or Δr^2 since these are very small
V2 = 4/3 * π * (r^3 + 3 * r * Δr)
V2 - V1 = 4 * π * r^2 * Δr
Δr = k * r * ΔT where k = coefficient of linear expansion for Al and ΔT the temperature change

This gives V2 - V1 = 4 * π * r^3 * k * ΔT
W2 - W1 = -4 * π * r^3 * k * ρ * g * ΔT

The Attempt at a Solution

i get -2420.678 and that's not right. any suggestions?

 

[OlderDan]

You seem to have the right idea. I did not verify your number, but the problem asks for the magnitude of the difference in weights, which should be a positive number.

Besides the sign, you appear to have a problem with your ΔV expression, but I think you fixed it. ΔV should be the area of the sphere times Δr. In one place you have just an rΔr instead of r²Δr. I think you used the correct expression in your calculation.

 

[kyle8921]

I appreciate your attempt at solving this problem. However, it seems that you may have made a mistake in your calculations. The magnitude of the difference between the weights can be found using Archimedes' principle, which states that the buoyant force (Fb) acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the fluid is water and the object is the aluminum sphere.

We can use the following equation to find the weight of the sphere in water before and after cooling:

W1 = ρ * g * V1
W2 = ρ * g * V2

Where W1 and W2 represent the weights of the sphere in water before and after cooling, respectively. ρ is the density of water, g is the acceleration due to gravity, and V1 and V2 are the volumes of the sphere before and after cooling, respectively.

Using the given values, we can calculate the volume of the sphere before and after cooling:

V1 = 4/3 * π * (2.34 m)^3 = 39.66 m^3
V2 = 4/3 * π * (2.34 m)^3 = 39.66 m^3

Since the sphere is completely submerged in water, the volume of water displaced by the sphere is equal to its volume. Therefore, V1 and V2 are equal and we can simplify the equation to:

W2 - W1 = ρ * g * V1 - ρ * g * V1 = 0

This means that there is no difference in the weight of the sphere before and after cooling. This makes sense because the buoyant force acting on the sphere is equal to its weight, and since the sphere is completely submerged, the buoyant force remains the same even as the sphere cools.

I hope this helps clarify the situation and leads you to the correct solution. If you are still having trouble, I suggest double checking your calculations and equations to ensure accuracy. Good luck!