78. (a) Let ∆Tbe the change in temperature and κbe the coefficient of linear expansion for copper. Then ∆L = L ∆Tand ∆∆LLT==×°=×−−(./..17 1017 1055K)(1.0 C)This is equivalent to 0.0017%. Since a change in Celsius is equivalent to a change on the Kelvin temperature scale, the value of used in this calculation is not inconsistent with the other units involved. Incorporating a factor of 2 for the two-dimensional nature of A, the fractional change in area is ∆∆AAT×°=×22 17 1034 10/.K)(1.0 C)which is 0.0034%. For small changes in the resistivity ρ, length L, and area Aof a wire, the change in the resistance is given by ∆∆∆∆RRRLLRAA=∂∂+∂∂+∂∂.Since R = L/A, ∂R/∂= L/A = R/, ∂R/∂L= /A = R/L, and ∂R/∂A = –L/A2–R/A. Furthermore, ∆/= α∆T, where is the temperature coefficient of resistivity for copper
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