Temperature coefficient of resistance

[Philip Wong]

Homework Statement

ok we did a temperature coefficient of resistance at class, we need to calculate the coefficient of resistance (a) ourselves.
known variable:
R = resistance
ln R = log of resistance
T = tempreature
a = coefficient of resistance
c = the intercept

Homework Equations

ln R = aT+c
i.e. R= k exp(aT), where k=exp(c) is a constant.

The Attempt at a Solution

at first I thought I could calculate a as if it is gradient, and therefore worked out the intercept (c). But since a is the coefficient of resistance it didn't make sense to me to do it this way. So I'm lost
Anyways this is my working:
gradient = delta ln R/ delta T
= (lnR_1-lnR_2/T_1-T_2) = -.912185/44
= -0.02682897
therefore intercept (c) = -.912185 - (-.02682897 * 44)
= 10.8925618

put it back to the equation
lnR = a*T+10.893
given T = 64, its correspond ln R = 4.957 (we measured these ourselves, assume correct)

so:
4.957 = a*64+10.893
-5.936=a*64
a=-5.936/64
a=-0.0927

put everything back to original equation:
ln R = -0.0927 T + 10.893

is this even correct??
thanks

 

[Philip Wong]

if my above working was correct, here is the second part of my question.
during the experiment we only measure the resistance every 2C drop (from 100C to 30C). I've plot a ln R vs T graph, how can a determine the a at temperature of 65C?

 

[Philip Wong]

arha! I was right in general, but had made a small mistake! I forgot to convert temperature into kelvin before I calculated the coefficient. when I convert it, everything looks more reasonable and correct!