Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ∆T, then the rod’s length changes by ∆L=αL0∆T, where α is the thermal expansion coefficient. For steel, α=1.24x10^-5°C.
Express length L as a function of T if L0=65 in. at T0=100°C.

Question

Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ∆T, then the rod’s length changes by ∆L=αL0∆T, where α is the thermal expansion coefficient. For steel, α=1.24x10^-5°C.
Express length L as a function of T if L0=65 in. at T0=100°C.

anonymous · Answer

well, first recognize the total length is given by $L(T)=L_0+\Delta L$

anonymous · Answer

we're told $\Delta L=\alpha L_0\Delta T=\alpha L_0(T-T_0)$ since $\Delta T=T-T_0$ hence:$$L(T)=L_0+\alpha L_0(T-T_0)$$

anonymous · Answer

How did you realize that L(T)=L0+ΔL?

anonymous · Answer

@OddlyWeird the length at temperature $T$ is the original length $L_0$ plus the change in length $\Delta L$ at that temperature