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Answer:
α = 2,857 10⁻⁵ ºC⁻¹
Explanation:
The thermal expansion of materials is described by the expression
ΔL = α Lo ΔT
α = [tex]\frac{\Delta L}{L_o \ \Delta T}[/tex]
in the case of the bar the expansion is
ΔL = L_f - L₀
ΔL= 1.002 -1
ΔL = 0.002 m
the temperature variation is
ΔT = 100 - 30
ΔT = 70º C
we calculate
α = 0.002 / 1 70
α = 2,857 10⁻⁵ ºC⁻¹