$A$ thin rod having length $L_0$ at $0\,^{\circ}C$ and coefficient of linear expansion $\alpha$ has its two ends maintained at temperatures $\theta_1$ and $\theta_2$,respectively. Find its new length.

(N/A) In thermal steady state,the temperature gradient along the rod is constant. The temperature varies linearly from $\theta_1$ to $\theta_2$. The average temperature of the rod is $\theta_{avg} = \frac{\theta_1 + \theta_2}{2}$.
The change in length $\Delta L$ is given by $\Delta L = L_0 \alpha \Delta T$,where $\Delta T$ is the change in temperature from the reference temperature $(0\,^{\circ}C)$.
Thus,the new length $L$ is:
$L = L_0(1 + \alpha \theta_{avg})$
$L = L_0 \left( 1 + \alpha \left( \frac{\theta_1 + \theta_2}{2} \right) \right)$