Simple Resistance-Temp relationship

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The discussion centers on calculating the resistance of a copper coil at -131.3 °C, starting from a resistance of 0.541 Ω at room temperature. The formula used incorporates the temperature coefficient of resistance, with a noted value of 4.3E-3, although some participants suggest using 3.9E-3 instead. There are concerns about the accuracy of the homework program, as users suspect it may have coding errors or incorrect coefficients. Additionally, there is confusion regarding whether the initial resistance should be based on 0 °C instead of 20 °C. The conversation highlights the importance of significant figures and potential discrepancies in educational materials.
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Homework Statement



A copper coil has a resistance of 0.541 Ω at room temperature. What is its resistance when it is cooled to -131.3 °C?

Homework Equations



Ri=New resistance
R0=initial resistance
a=alpha

Ri=R0+R0(a(temp change))
temp change is positive when it goes up

alpha is 4.3E-3 in the book, so that is the right value
Room temp is 20 degC

The Attempt at a Solution



Ri=.541+.541(.0043)(-151.3)
Ri=.189031 ohms

I really feel like I'm doing this right and for all I know the homework program is wrong.
 
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PennStateFan1 said:
Ri=.189031 ohms

I really feel like I'm doing this right and for all I know the homework program is wrong.

Did you enter your result with that many significant figures? The program may be picky about that.
 
I tried fixing the sigfigs and its still wrong.
 
Looks good, but maybe the program is using a different value for the temperature coefficient of copper? I've seen it more commonly as 0.0039.
 
It's the right coefficient, because it told what table in the book to use. It's probably the coding for the program. It's a new book we're using and my teacher thinks there might be mistakes.
 
Look at the the R in your equation(0.541).This is the resistance at 20 degrees.Should it not be the resistance at 0 degrees?

R at 0 degrees=0.541/1+20 alpha
Dont forget that if alpha is referred to R at 0 degrees then the temperature drop is not 151.3
 
Last edited:
Okay, I don't know how late I am to answering your question, but the book I have says that the alpha is 3.9E-3, not 4.3E-3. Everything else is right in your problem because I just did a similar problem to yours in my homework and I got it right. :)
 

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