# 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)

Tf = 280° C.

Last edited:

Mentor
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 :)

Mentor
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!