Thermal/mechanical properties of matter

  • #1
1)Explain why a correction must be applied for displacement of air in accurate weighing with a common balance.

Calculate the percentage error which would arise through neglect of this correction in weighing water with platinum weights of density 2.15x10^4 in air density 1.22. (units are kg and m)

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I'm not really sure what to do with this question, is the first part involving the non-uniform distribution of air and the resulting varying pressure or something? hmm. I've scanned a textbook i have access to, it's not very well up on thermal stuff and I can't get to library for a few days. beh. As for the second part, hints are welcome :uhh:
 

Answers and Replies

  • #2
dextercioby
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Are u sure it's "thermal stuff" and not Archimede's buoyant force...?

Judging from the second problem,it would seem that way...

Daniel.
 
  • #3
yeah me too, thats just what the problem sheet was titled. makes it a little more confusing. I still don't get it though.. :|
 
  • #4
dextercioby
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You mean "thermal/mechanical properties of matter"...?Neglecting complete vagueness of the phrase,i think the "mechanical" part would account for Archimede's buoyant force...:wink:

How about asking the dude/chick who gave the problem what on Earth was he/she meaning...

Daniel.
 
  • #5
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I am itching to give it a shot....

subscripts a, w and p correspond to air, water and platinum resp.
Buoyant force act on both sides of the scale.


[tex] v_w \rho _w - v_w\rho_a = v_p\rho_p - v_p\rho_a [/tex]


[tex] \frac{v_w}{v_p} = \frac{(\rho_p - \rho_a)}{ \rho_w - \rho_a[/tex]

real weight = [tex] \rho_w v_w[/tex]

Apparent weight = [tex] v_w\rho_w - (v_w\rho_a - v_p\rho_a) [/tex]

% diff = [tex] \frac{\rho_a}{ \rho_w}\left( \frac { \rho_p - \rho_w}{\rho_p - \rho_a} \right)[/tex]


Does it sounds right?

edit: I was attempting to write the ratio of [tex] v_w/v_p[/tex] by rearranging the first equation. It shows up OK in my preview page.
But shows error after submitting.
 
Last edited:

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