Determining the Density of Turpentine: Archimedes' Principle Explained

  • Thread starter Thread starter frozen7
  • Start date Start date
  • Tags Tags
    Archimedes
AI Thread Summary
The discussion centers on calculating the density of turpentine using Archimedes' Principle with a solid aluminum cylinder. The first method calculates the density as 887 kg/m³, while the lecturer's method yields a density of 2699.4 kg/m³, suggesting a nearly equal density to aluminum. It is noted that if the densities were indeed similar, the aluminum would displace its weight, resulting in a near-zero reading when immersed. The consensus is that the initial calculation is correct, as the professor's approach appears flawed. The correct density of turpentine is confirmed to be 887 kg/m³.
frozen7
Messages
163
Reaction score
0
A solid aluminium cylinder with density of 2700 has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

I do it in this way:
W = ( 0.067 - 0.045 ) (9.8)
= 0.2156

Volume of aluminium = 0.067 / 2700 = 2.48 x 10-5

0.2156 = (density of turpentine ) (2.48 x 10-5) (9.8)
density of turpentine = 887

However, my lecturer do it in this way:

W = (density of aluminium) (g)(volume of solid)
volume of solid = xxx

W = (density of turpentine) ( g ) ( volume of solid)
density of turpentine = 2699.4

Which solution is correct actually?
 
Physics news on Phys.org
If you think about it ...

By Archimedes' Principle, the upthrust on an object immersed in a fluid is equal to the weight of the fluid displaced.
According to your professor, turps and aluminium have approximately the same density (2699.4 vs 2700)
This would mean that the aluminium would displace approximately its own weight and so would register a weight of almost zero instead of 45 g.

You working out looks fine.

I wonder if your professor equated the weight of the aluminium cylinder with the weight of the turps displaced ?
 
Your solution is correct.

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top