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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?

  • physics -

    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

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