To what temperature should an aluminum bar be heated in order to increase its length by 1% in relative to length at O°C? alpha is =2.4× 10⁵ 1°C​

Answer:Explanation:Given:ΔL / L₀   = 0.01t₀ = 0°Cα = 2.4·10⁻⁵  1/°C_______________t - ?L = L₀· (1 + α·Δt)L = L₀ + L₀·α·ΔtΔL = L₀·α·Δtα·Δt = ΔL / L₀Δt = (ΔL/L₀)  / αΔt = 0.01 / 2.4·10⁻⁵ ≈ 417°CΔt = t - t₀t = t₀ + Δt = 0 + 417 = 417°C