Relationship between heat and resistivity of a wire

[sollinton]

Homework Statement

"A metal wire has a resistance of 10 ohms at a temperature of 20 degrees C. If the same wire has a resistance of 10.55 ohms at 90 degrees C, what is the resistance of this same wire when its temperature is -20 degrees C?"

Homework Equations

I am not exactly having trouble solving this problem. It's just that my teacher realized after she wrote the problem that she never taught us the relationship between heat and electrical resistivity. Therefore she gave us the problem as extra credit and told us to find out how to solve it on our own.

So all I am really looking for is an equation. I would like to do the calculations myself, but if someone could point me in the right direction, it would be greatly appreciated.

 

[Delphi51]

This temperature dependence of resistance seems to be quite complicated, but I think all you want in this case is the simplest approximation. In Wikipedia "Electrical Resistance" it says "Near room temperature, the electric resistance of a typical metal increases linearly with rising temperature". In other words R = Ro + c(delta T), where Ro is the resistance at room temperature, c is a constant and delta T is the temperature minus room temperature.

 

[sollinton]

Okay, so using your equation

[tex]R = R_{o} + c(\Delta T)[/tex]​

I solved for c

[tex]c = \frac{(R - Ro)}{\Delta T}[/tex]

[tex]c = \frac{(10.55 \Omega - 10 \Omega)}{70 C}[/tex]​

And found c to equal

[tex]0.0078 \frac{\Omega}{C}[/tex]​

Then plugging c back into the original equation and using the new [tex]\Delta[/tex]T:

[tex]R = 10\Omega + 0.0078 \frac{\Omega}{C}(-40C)[/tex]
[tex]R = 9.68 \Omega[/tex]​

This seems to make perfect sense, as 9.68 is about as far from 10 as 10.55 is, and the smaller change in temperature accounts for the smaller change in resistivity.

Thank you very much for your help!

P.S.
Sorry the equations are kind of sloppy, I don't have that program that puts out the symbols and such.

**EDIT**
Woo got the fancy equations working!

 

[Delphi51]

Looks good!
People are making these equations by putting "tex" before and "/tex" after, using square brackets instead of quotes. Then the website will translate cryptic commands - like "\theta" into the Greek letter theta. If you hold your mouse over a fancy formula you can see how it was done. Unfortunately it is quite a bit of work and you can't see what you are doing as you do it.