Thermal Expansion of lead rod question

[Intrusionv2]

A lead rod and a common glass rod both have the same length when at 21.0° C. The lead rod is heated to 50.0° C. To what temperature must the glass rod be heated so that they are again at the same length?

I did this:

L = a*L0*T
a (thermal expansion) of lead = (29e-6)
a (thermal expansion) of glass pyrex = (3.25e-6)

[a(50-21)]lead = [a(Tf-21)]glass
Tf = 280° C.

 

[gneill]

Do you happen to know what temperature they're looking for?

Although it doesn't make a difference to the solution to this problem, for future reference you might take note that the length of the material that has undergone a temperature change should be

$$L_{new} = L_{old} (1 + \alpha \Delta T)$$

This is because the coefficient of expansion is based upon

$$\frac{\Delta L}{L} = \alpha \Delta T$$

 

[Intrusionv2]

Hey thanks for the reply, I actually got it worked out.

They wanted a larger value for alpha of glass than I had.

I thought that pyrex glass would be fine to use as alpha for a glass rod, guess not. They wanted 9x10^-6.

But other than that my procedure is good :)

 

[gneill]

Hey thanks for the reply, I actually got it worked out.

They wanted a larger value for alpha of glass than I had.

I thought that pyrex glass would be fine to use as alpha for a glass rod, guess not. They wanted 9x10^-6.

But other than that my procedure is good :)

Ah yes. It might be of interest to note that Pyrex glass was developed to address thermal expansion issues -- it was designed to have a lower thermal expansion coefficient than "ordinary" glass. It also meant fewer breakages due to thermal stress in the lab as well as the kitchen!

 

[Intrusionv2]

Ah yes. It might be of interest to note that Pyrex glass was developed to address thermal expansion issues -- it was designed to have a lower thermal expansion coefficient than "ordinary" glass. It also meant fewer breakages due to thermal stress in the lab as well as the kitchen!

Makes sense, good to know!