An engineering intern is studying the effect of temperature on the resistance of a current carrying wire. she applies a voltage to a aluminum wire at a temperature of 61.0°c and notes that it produces a current of 1.40 a. if she then applies the same voltage to the same wire at −88.0°c, what current should she expect (in a)? the temperature coefficient of resistivity for aluminum is 3.90 ✕ 10−3 (°c)−1. (assume that the reference temperature is 20°c.)

I=2.80 A

Explanation:

We Know that    R =R₀(1+∝ ΔT)

                           R=R₀ (1+3.9*10⁻³ *(61-20))

                           R=R₀ (1.1599)

                           I=V/R=V/(R₀ (1.1599)

                           1.4 = V/(R₀ (1.1599)                       ∵ equation 1

                     We have to calculate I when T=-88°

                              R =R₀(1+∝ ΔT)

                             R=R₀ (1+3.9*10⁻³ *(-88-20))

                             R=R₀ (0.5788)

                             I=V/(R₀ (0.5788)                          ∵equation 2

Dividing equation 2 by equation 1

                            I = 2.80 A

this is a good question! i'll do some research and come back when i have the answer!

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.