Solving for Temp Change of Epoxy Frames for Lenses Insertion

[KristinaMr]

Homework Statement

This is the problem.

A pair of eyeglass frames is made of epoxy plastic. At
room temperature (20.0°C), the frames have circular
lens holes 2.20 cm in radius. To what temperature must
the frames be heated if lenses 2.21 cm in radius are to
be inserted in them? The average coefficient of linear
expansion for epoxy is 1.30 x 10^-4 (°C)^-1.

Homework Equations

The Attempt at a Solution

I know how to solve this but I'm not very sure what the ℃ elevated to -1 means. Can someone explain what that means and why it is expressed this way?

 

[Phylosopher]

Homework Statement

This is the problem.

A pair of eyeglass frames is made of epoxy plastic. At
room temperature (20.0°C), the frames have circular
lens holes 2.20 cm in radius. To what temperature must
the frames be heated if lenses 2.21 cm in radius are to
be inserted in them? The average coefficient of linear
expansion for epoxy is 1.30 x 10^-4 (°C)^-1.

Homework Equations

The Attempt at a Solution

I know how to solve this but I'm not very sure what the ℃ elevated to -1 means. Can someone explain what that means and why it is expressed this way?

Look at wiki page, "Thermal expansion coefficients for various materials" section. It's just the inverse of the temperature unit.

Write down the equation of the thermal expansion, and weigh down the units for the thermal coefficient.

 

[KristinaMr]

α=(∆L/L)/T

So the units are ℃^-1 because the temperature in the equation is in the denominator (which means T ^-1) . Right?

 

[Phylosopher]

α=(∆L/L)/T

So the units are ℃^-1 because the temperature in the equation is in the denominator (which means T ^-1) . Right?

Yes. The physical unit for the length is irrelevant because you have (( ΔL/L )) which always cancels out.

The choice of the unit for the temperature influence the unit of your coefficient and vice versa. You can choose to use K (Kelvin) instead of °C (Celsius). In this scenario, α have a unit of K^-1

 

[KristinaMr]

Thank you for your help :)