## Materials Science and Engineering: An Introduction

$Δl = -0.012 m \approx -12.50 mm (-0.49 in.)$
Given: copper wire of 15 m (49.2 ft) $T_{0}$ = 40°C (104°F) $T_{f}$ = -9°C (15°F) Required: change in length (C) Solution: Rearranging Equation 19.3b, and using $α_{l}$ value from Table 19.1 for Copper: $Δl = l_{0}α_{l} (T_{f} - T_{0}) = (15 m) (17.0 \times 10^{-6 } (°C)^{-1})(-9°C - 40°C) = -0.012 m \approx -12.50 mm$