Object placed in a fluid, both fluid and object are different densities

[mbkau00]

Hi, I was wondering what basic formulae can be used to solve questions like this? I haven't encountered a problem using two different densities, and no height or displacement volume given.

Q.
A block of ice, 100g, density 900kg m-3, is placed in a 400ml of solution A, density 1100 kg m-3. Before any ice has melted, the level of solution A is closest to?
A: 490mL

Thanks in advance for any help.

 

[LawrenceC]

It would be better if you figured it out. We are not permitted to solve problems for you but we can assist. Let's see your reasoning/work.

 

[mbkau00]

The only formulae that I can think of is d=m/v. I have been given the answer. I just don't know where to start in solving it

 

[LawrenceC]

Have you heard of Archimedes principle?

 

[mbkau00]

Yes, I have, however I was under the impression that it had to do with the amound of water being displaced? As there is no height given to the ice block, nor a height or % of the ice block being submerged - i didnt think it was applicable?

As the answer is 490mL. I'm assuming that it is .1kg*900kg = 90g
Then add this 90g to the existing 400ml of solution = 490mL

I can't find a formula that relates mass*density, and without seeing the answer, i wouldn't know how to approach it.

Thanks for your help.

 

[LawrenceC]

The % of the ice block submerged can be deduced. If the densities were equal and the ice were gently placed in the solution, how much would be above or below the surface?

"As the answer is 490mL. I'm assuming that it is .1kg*900kg = 90g"

kg*kg=kg^2, so how can this be volume?
kg^2 is not milliliters.

 

[mbkau00]

I'm not really sure. I'll guess that it would almost be fully submerged; and as the ice block weighs less than the fluid it would float. By how much I don't know.

Yes I didn't look at that calculation like that. It isn't ^3, only ^2.

 

[LawrenceC]

If the density of the ice is the same as the density of the solution, the top of the ice would be right at the surface of the water if gently placed in the water. If the ice were placed under the water, it would neither sink nor rise. It would merely remain where you put it.

However, if the ice density is greater than the solution, it sinks to the bottom. The opposite is also true and that is what we have here.

The upward force on the ice exactly equals the amount of solution pushed out of the way (displaced) by the portion of the ice that is under water. When you gently place the ice on the surface of the solution, it sinks until the solution displaced equals the weight of the ice. So determine the weight of the ice, then determine how much solution needs to be pushed out of the way (displaced) to create a force balance.

You can also look at this problem from a pressure-depth standpoint if you assume the ice is a right circular cylinder or the shape of a brick with a flat surface parallel to the free surface of the liquid. This keeps the calculations straightforward.

 

[mbkau00]

Ok I am very new to physics (3 weeks) so i don't know or understand what calculations will keep this straight forward; as i still don't have a formula to work with.

nonetheless i'll try and workout what you mean:

- ice is .1kg
- how much solution needs to be disaplced to keep forces balanced? - no idea where to start with that one.

.1kg of ice with a density of 900kg, into 400ml with a density of 1100kg.
I still can't move from this point.

Surely the .1kg of ice and its density (900kg) must be manipulated so it can work with the 400ml of 1100kg solution?

As my original post, I still don't know how to manipulate these numbers. I don't know how to relate these numbers. I don't know what to multiply or divide by what, to get what needs to make sense of this situation.

 

[mbkau00]

Regarding pressure - depth standpoint: i understand that density will increase as the object decreases vertically. no change horizontally at a fixed depth.

thanks for your help.

 

[LawrenceC]

Regarding pressure - depth standpoint: i understand that density will increase as the object decreases vertically. no change horizontally at a fixed depth.

thanks for your help.

The density does not change. It is constant.

 

[HallsofIvy]

You said you knew Archimede's principle: the block of ice will displace its own mass in water. You know the mass of the block and density of water so can calculate the volume of water displaced.

 

[mbkau00]

Sorry, I meant pressure will vary, not density.

Ok if the ice block will displace it's own mass (100g) in a solution, it will displace 100ml if the densities were the same? As the density of the fluid is 1100kg m-3, whereas the iceblock is 900kg m-3, it won't be 100ml displaced.

As named in my opening post, the answer is 90ml displaced, or new volume = 490ml (400+90)

I still don't know a concrete formula to give me 90ml. Every example I have seen with the use of archimedes principle involves the same density.

 

[mbkau00]

am i right in saying this:

if the ice block is fully submerged, the Volume displaced will equal the volume of the ice block.

if the ice block is somewhat submerged, the volume displced will be less than the volume of the ice block

 

[LawrenceC]

am i right in saying this:

if the ice block is fully submerged, the Volume displaced will equal the volume of the ice block.

if the ice block is somewhat submerged, the volume displced will be less than the volume of the ice block

Both statements above are correct.

We are not going to supply you with a formula because you learn nothing by plugging numbers into a formula just to get an answer. You should be able to reason this out and come up with your own formula. For every situation there is some formula that can be used. It would be impossible to memorize them all. So, figure it out.

 

[mbkau00]

considering i am no more advanced in solving this problem than when I posted it; i think it is time to surrender and accept defeat!