Unformatted text preview: 64. A least squares ﬁt of the data gives R = 537
5 + 1111
1750 T with T in degrees Celsius. (a) At T = 20◦ C, our expression gives R =
(b) At T = 0◦ C, our expression gives R = 21017
175 ≈ 120 Ω.
537
5 ≈ 107 Ω. (c) Deﬁning αR by
αR = R − R20
R20 (T − 20◦ C) then we are eﬀectively requiring αR R20 to equal the 1111 factor in our least squares ﬁt. This implies
1750
that αR = 1111/210170 = 0.00529/C◦ if R20 = 21017 ≈ 120 Ω is used as the reference.
175
(d) Now we deﬁne αR by
αR = R − R0
,
R0 (T − 0◦ C) which means we require αR R0 to equal the 1111 factor in our least squares ﬁt. In this case,
1750
αR = 1111/187950 = 0.00591/C◦ if R0 = 537 ≈ 107 Ω is used as the reference.
5
(e) Our least squares ﬁt expression predicts R = 96473/350 ≈ 276 Ω at T = 265◦ C. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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