Resistance and resistivity problem

[jony529]

Hey I had a question on my quiz where it asked for the Resistance of aluminum at 600 deg C. But the formula rho-rho(i) = rho(i)a(T-T(i)) only finds the value of the resistivity at 600 deg C how do I convert it to resistance without more given info?


Question: A spool of aluminum wire has a resistance of 634 at 20o C. Find the resistance of the spool at 600o C

Alum = (rho(i)= 2.75x10^-8), (alpha = 4.4x10^-3)

My work
rho = (2.75x10^-8*4.4x10^-3*580)+(2.75x10^-8)

rho = (9.768x10^-8)

So I need to convert the resistivity to resistance from here but I just don't see how.

 

[mgb_phys]

Assuming nothing else about the dimensions of the wire changed then it's just a ratio question.
You could pick an arbitrary length and CSA and work out the resistance and then the resistance change but it would just cancel out.

 

[jony529]

Sorry but I didnt fully understand what you were trying to tell me. I know R = rho(L/A), So if I pick an arbitrary length say 10m. How do I find that A of the wire, and furthermore what's canceling out? Confused!

 

[jony529]

BTW the answer is 2250 ohms. Like I said originally it was a quiz online I just took. But how do I arrive at that answer??

 

[mgb_phys]

The reel has a certain length and area an a resistance of 634R if the resistivity doubles then the resistance of that piece must also double.
If the resistivity is proportional to the temperature then you just have to fin the c change in temperature and so the proportional change in resitivity and resistance.

Imagine an analogy, if you ha a block that weight 1kg and the density doubles then the new block weighs 2kg - whatever the size is.

 

[jony529]

The thrilling conclusion to my problem:


rho = rho(i)alpha(T-T(i)) + rho(i)
R = rho(L/A)

Find R.

{(L/A)} * {rho} = {(L/A)} * {rho(i)alpha(T-T(i)) + rho(i)}

which equals

R = R(i)alpha(T-T(i)) + R(i)

2250 ohms = 634(4.4x10^-3)580 + 634