Easy question but I get a crazy answer

  • Thread starter sai2020
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In summary, Mercury has a thermal expansion coefficient (volume) of about 1/5500 perKelvin. Estimate the inner diameter of the thermometer tube.
  • #1
sai2020
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The question is

Mercury has a thermal expansion coefficient (volume) of about 1/5500 perKelvin. Estimate the inner diameter of the thermometer tube.

Here's my attempt.

L [tex]\approx[/tex] 30cm
V = 2*pi*r^2*L
Change in volume at 0 C and 110 C approximately equal to V

=> 110/5500 = r^2(3*pi/5)
=> r = sqrt(1/30*pi) [tex]\approx[/tex] 0.10m

I am well known to make crazy errors but I can't find out where that is over here
 
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  • #2
Theres not enough information to determine an actual value for the diameter. You give no initial volume (Vo) and I don't know if you need to consider the expanding glass container as well. Given the assumptions that the volume at 0 C is unknown and the expansion of glass is negligible, I got a solution of

d = 0.2899*sqrt(Vo)
 
  • #3
Just a few tips to help you on your way;
Recheck how you determine the volume of a cylinder.
Where does (3*pi/5) come from?
What are your justifications for those assumptions in the problem? There's insufficient data to complete it with the question you provided.
Make sure you don't fluke up on units.
Remember that 100 cm^2 =/= 1 m^2. 100^2 cm^2 = 1 m^2
 
  • #4
Another clue: the point of a decent thermometer is to provide sufficient changes in length to read (say) a quarter-degree change in temperature.
 
  • #5
Cvan said:
Just a few tips to help you on your way;
Where does (3*pi/5) come from?

It comes from

L = 30*10^-2 = 0.3m
V = r^2(2*pi*L) = r^2(3*pi/5)

Topher925 said:
Theres not enough information to determine an actual value for the diameter. You give no initial volume (Vo) and I don't know if you need to consider the expanding glass container as well. Given the assumptions that the volume at 0 C is unknown and the expansion of glass is negligible, I got a solution of

d = 0.2899*sqrt(Vo)

The *Hint* says the initial volume is 1mm^3. and I think the glass expansion can be neglected cos this is an introductory thermal course and this is the first tutorial.

Can you please tell me how you got to that.
 
Last edited:
  • #6
*I made a mistake in my first post, that should be the radius, r not the diameter, d

I just used the equation

dV = B*dT*Vo

where:
dV = change in volume (0.03 x pi x r^2)
B = expansion coefficient (18x10^-5/C)
dT = change in temp (110C)
Vo = initial volume (1mm^3)

Plug all that in and solve for r and you get r = 0.2899*sqrt(Vo).
 
  • #7
OH! My definition of B was wrong.. Thanks a lot..
 

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