L=L₁+L₂=L₀₁+ΔL₁ +L₀₂ + ΔL ₂=
= L₀₁+α₁•L₀₁•ΔT +L₀₂ + α₂•L₀₂•ΔT=
=1+ 23.1•10⁻⁶(450-25) +1+ 12•10⁻⁶(450-25) =
=2.015 m
A 1.0-m thin rod of aluminum is laid end-to-end with a 1.0-m thin rod of steel. The rods
are initially at 25 oC. What is the total length of both rods when heated to 450 oC?
1 answer
1 answer