Question

The resistance of a 1.50 m long copper wire, with a cross-sectional area of 7.06 x 10^-6 m², is measured at different temperatures starting at 20.0 ^oC and ending at 70.0 ^oC. The given table shows those resistances as a function of temperature differences from the initial temperature.

1. Generate an accurate plot of the resistance as a function of DT.
2. Determine the resistivity of copper from your plot and compare to the accepted value for copper’s resistivity.
3. Determine the temperature coefficient for the resistivity of copper from your plot. (Note: The resistivity, and therefore, the resistance, of a material is temperature dependent, in a similar format as the linear expansion of a material from thermodynamics.)

DT (± 0.1 oC)	R (± 1 x 105W)
0.0	0.00373
5.0	0.00377
10.0	0.00392
15.0	0.00401
20.0	0.00405
25.0	0.00407
30.0	0.00415
35.0	0.00430
40.0	0.00438
45.0	0.00446
50.0	0.00448

Answer 1

Let's plot the resistance vs the temperature difference: a. R vs AT 0.0047 y 1.558E-05x +3.730E-03 0.0045 0.0043 0.0041 0.0039 0.0037 0.0035 0 10 30 50 60 AT The model for the resistance and the change in temperature is: b. R Ro(1aAT) Then: R RoRoaAT The y-intercept is Ro = 3.73 x 10-3 (105) = 373 0 The slope Roa = 1.558 x 10-5 105 = 1.558 N°C-1 Remember that: PoL Ro A Po(1.5) 372 7.06 x 10-6 Po 1.75 x 10-8 Nm The accepted value is: 1.7 x 10-8 m 1.75-1.7 The percentage of error is: %E The temperature coefficient is 1.558 100 2.94% 1.7 c. 1.558 Ro 373 a 0.00418°C-1 40 20 sOT)