Solving the Mystery of a Floating Bottle: Density & Volume

[Callmelucky]

Homework Statement

If we throw empty but sealed glass bottle in water it will float with 60% of it's volume above surface of water.
What is average density of floating bottle and what is it's volume? Mass of bottle is 0.4 kg(mass of air inside of bottle is irelevant).

Relevant Equations

density = mass / volume

If we throw empty but sealed glass bottle in water it will float with 60% of it's volume above the surface of water.
What is average density of floating bottle and what is it's volume? Mass of bottle is 0.4 kg(mass of air inside of bottle is irelevant).

It's easy problem but I can't get right solution.

40% of bottle is under water, so density of bottle is 400 kg/m3. Therefore the volume of is 0.4 kg / 400 kg/m3 = 1dm3 = 1L.

But the answer at the end of textbook for density is 392 kg/m3.

Can someone please tell me where I am wrong?

Thank you.

 

[erobz]

Homework Statement:: If we throw empty but sealed glass bottle in water it will float with 60% of it's volume above surface of water.
What is average density of floating bottle and what is it's volume? Mass of bottle is 0.4 kg(mass of air inside of bottle is irelevant).
Relevant Equations:: density = mass / volume

If we throw empty but sealed glass bottle in water it will float with 60% of it's volume above the surface of water.
What is average density of floating bottle and what is it's volume? Mass of bottle is 0.4 kg(mass of air inside of bottle is irelevant).

It's easy problem but I can't get right solution.

40% of bottle is under water, so density of bottle is 400 kg/m3. Therefore the volume of is 0.4 kg / 400 kg/m3 = 1dm3 = 1L.

But the answer at the end of textbook for density is 392 kg/m3.

Can someone please tell me where I am wrong?

Thank you.

You should be solving for the volume of the bottle first.

 

[nasu]

Is the denisty of water given in the text or you are supposed to use 1000 kg/m³?

 

[Callmelucky]

You should be solving for the volume of the bottle first.

what do you mean?
I can use density of glass(2500 kg/m3) and get volume of bottle for the mass of 0.4 kg = 0.00016 m3. But I don't see how is that going to help me.

 

[Callmelucky]

Is the denisty of water given in the text or you are supposed to use 1000 kg/m³?

Doesn't say. It's usually stated if water is salty or if it has different density, here it says nothing, so i suppose it's 1000kg/m3.

 

[erobz]

what do you mean?
I can use density of glass(2500 kg/m3) and get volume of bottle for the mass of 0.4 kg = 0.00016 m3. But I don't see how is that going to help me.

I mean you should be applying Archimedes principle to solve for the volume of the bottle.

When they say find the average density they mean “of the bottle” i.e. the bottles mass per unit volume. That’s not the same as the density of the glass that makes the bottle.

 

[nasu]

@erobz How would you even find the density of the glass from the given data? All you can find is the average density. It does not even have to be made from glass. Can be anything. If it's submerged 40%, its average density is 40% of the density of the fluid. It can be a solid piece of wood or an empty container made from steel.

 

[Callmelucky]

@erobz How would you even find the density of the glass from the given data? All you can find is the average density. It does not even have to be made from glass. Can be anything. If it's submerged 40%, its average density is 40% of the density of the fluid. It can be a solid piece of wood or an empty container made from steel.

at the end of textbook I have a chart with some densites, glas is 2500 kg/m3. But yeah, I still don't understand what @erobz is trying to explain, I just didn't want to ask more questions because I feel stupid lol

 

[haruspex]

40% of bottle is under water, so density of bottle is 400 kg/m3. Therefore the volume of is 0.4 kg / 400 kg/m3 = 1dm3 = 1L.

But the answer at the end of textbook for density is 392 kg/m3.

The density of pure water is a bit under 1000 kg/m³, but only 0.3%. The given answer is 2% below what you have calculated. That's about the difference between g and 10m/s², but how that can have entered into it I have no idea.

 

[Callmelucky]

The density of pure water is a bit under 1000 kg/m³, but only 0.3%. The given answer is 2% below what you have calculated. That's about the difference between g and 10m/s², but how that can have entered into it I have no idea.

Then probably authors' mistake

 

[erobz]

@erobz How would you even find the density of the glass from the given data? All you can find is the average density. It does not even have to be made from glass. Can be anything. If it's submerged 40%, its average density is 40% of the density of the fluid. It can be a solid piece of wood or an empty container made from steel.

Sorry, when I saw them pulling numbers out of seemingly nowhere, I just assumed that's where the mistake was. I never bothered to commit that result to memory myself, so when I didn't see:

$$ p\cancel{g} 0.4 V_T = m\cancel{g} \implies V_T = \frac{m}{0.4\rho } $$

I leaped to false a conclusion that a mistake had been made.

Again...Sorry.

 

[erobz]

at the end of textbook I have a chart with some densites, glas is 2500 kg/m3. But yeah, I still don't understand what @erobz is trying to explain, I just didn't want to ask more questions because I feel stupid lol

Don't feel stupid for being right(or wrong)! Stand your ground, and it will get sorted out. Anyone can make a mistake.

 

[Callmelucky]

Sorry, when I saw them pulling numbers out of seemingly nowhere, I just assumed that's where the mistake was. I never bothered to commit that result to memory myself, so when I didn't see:

$$ p\cancel{g} 0.4 V_T = m\cancel{g} \implies V_T = \frac{m}{0.4\rho } $$

I leaped to false a conclusion that a mistake had been made.

Again...Sorry.

Thanks for trying to help.

 

[hutchphd]

Because the book gives the mass of the bottle as 0.4 kg and not 0.400kg or better still 4.00x10^(-1) ]kg their answer to three significant figures should be graded with a zero

 

[nasu]

Sorry, when I saw them pulling numbers out of seemingly nowhere, I just assumed that's where the mistake was. I never bothered to commit that result to memory myself, so when I didn't see:

$$ p\cancel{g} 0.4 V_T = m\cancel{g} \implies V_T = \frac{m}{0.4\rho } $$

I leaped to false a conclusion that a mistake had been made.

Again...Sorry.

If it makes you feel better, I did not comit it to the memory either.
I just scratched on a piece of paper exactly wht you wrote here, before writing that post.

 

[Callmelucky]

Because the book gives the mass of the bottle as 0.4 kg and not 0.400kg or better still 4.00x10^(-1) ]kg their answer to three significant figures should be graded with a zero

It's actually given in dag, so 40 dag.
I just converted it so that it "looks better"