Finding weight using Archimedes' principle

[kari82]

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids. The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, 54.2% is immersed in the oil and the balance in the water. In a separate experiment, and empty flask is weighed, 35.3 cm^3 of the lubricating oil is poured into the flask, and the flask is reweighed. If the scale reading was 124.8g in the first weighting, what would it be in the second weighting?

mblock=m of oil displaced

density graphite=2.26g/cm^3

I'm not sure if I am right but I assumed that volume of oil=0.542volume of block

ρ(block) x V(block)=ρ(oil) x (0.542Vb)

ρ(oil)=ρ(b) x V(b)/0.542Vb=2.26/0.542= 4.17 g/cm^3

Then we can calculate the mass of the poured oil

m(oil)=ρ x V(poured) = 4.17 x 35.3 = 147.2g

m(total)=m(o) + m(f) = 147.2g +124.8g = 272g

I feel like I am missing something. Can someone please tell me if my approach is not correct? Thank you for your help!

 

[NascentOxygen]

How do you justify not taking into account the effect of the water?

 

[kari82]

I cant.. And that was the main reason why I'm pretty sure I'm missing something. When I tried to include water into my calculations I was getting that the density of the oil is less than 1, which it does not make sense.

What I assumed was that the m(block)=m(oil displaced) and the m(block)=m(water displaced); therefore, the m(oil displaced)=m(water displaced)... Is that correct?

 

[Doc Al]

What I assumed was that the m(block)=m(oil displaced) and the m(block)=m(water displaced); therefore, the m(oil displaced)=m(water displaced)... Is that correct?

No. What's the total buoyant force acting on the floating block?

 

[kari82]

mf(oil)g + mf(water)g ?

 

[Doc Al]

mf(oil)g + mf(water)g ?

Yes. (Assuming that by mf(oil) you mean the mass of the displaced oil, etc.)

Express everything in terms of density and volume.

 

[kari82]

Thank you so much! So my correct equation will be:

ρ(block)V(block)=ρ(oil)0.542Vb + ρ(water)0.458Vb

Thank you again! :)

 

[Doc Al]

Looks good to me. ;-)