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Δl = 6 x 1,7 x 10⁻⁵ x (75 - 25)
Δl = 5,1 x 10⁻³ = 0,0051
l₀ = 6
α = 1,7×10⁻⁵/K
t₁ = 25°C
t₂ = 75°C
Δt = t₂-t₁
Δt = 50
Dit = Δl
Jwb =
Δl = l₀.α.Δt
Δl = 6.1,7×10⁻⁵.50
Δl = 5,1×10⁻³
Δl = 0,0051