Chapter 10 - Temperature and Kinetic Theory - Learning Path Questions and Exercises - Exercises - Page 384: 36

(a) $-0.54mm$ (b) $0.3mm$

(a) We know that $\Delta L=L\alpha \Delta T$ We plug in the known values to obtain: $\Delta L=12\times 10^{-6}(-45)$ This simplifies to: $\Delta L=-0.54mm$ (b) As $\Delta L=L\alpha \Delta T$ We plug in the known values to obtain: $\Delta L=12\times 10^{-6} \times 25$ This simplifies to: $\Delta L= 0.3mm$