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9 January, 03:41

Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply the heat source and the temperature of the rod increases. The coefficient of linear expansion isα = 2*10-6K-1. a) What is the new length of the rod if the temperature jumps to 80◦C? b) What is the percentage of the volume change of the rod? c) What is the maximal temperature we can allow if the volume should not increase by more thanhalf percent?

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  1. 9 January, 04:01
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    a) 2.00024 m

    b) 0.036%

    c) 436.67°C

    Explanation:

    Given

    Initial length = L₀ = 2 m

    Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

    We can obtain initial volume = V₀ = A₀L₀ = 0.02 * 2 = 0.04 m³

    Initial Temperature = T₀ = 20°C

    Coefficient of linear expansivity = α = (2 * 10⁻⁶) (°C) ⁻¹

    a) New length of the rod after heating to 80°C

    Linear expansion is given as

    ΔL = L₀ * α * ΔT

    ΔL = 2 * 2 * 10⁻⁶ * (80 - 20) = 0.00024 m = 0.24 mm

    New length = old length + expansion = 2 + 0.00024 = 2.00024 m

    b) The percentage of the volume change of the rod.

    Volume expansion is given by

    ΔV = V₀ * (3α) * ΔT

    Volume expansivity ≈ 3 * (linear expansivity)

    ΔV = 0.04 * (3*2*10⁻⁶) * (80 - 20) = 0.0000144 m³

    Percentage change in volume = 100% * (ΔV/V₀) = 100% * (0.0000144/0.04) = 0.036%

    c) The maximal temperature we can allow if the volume should not increase by more than half percent.

    For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

    Percentage change in volume = 100% * (ΔV/V₀)

    0.5 = 100% * (ΔV/0.04)

    (ΔV/0.04) = 0.005

    ΔV = 0.0002 m³

    Then we now investigate the corresponding temperature that causes this.

    ΔV = V₀ * (3α) * ΔT

    0.0002 = 0.04 * (3*2*10⁻⁶) * ΔT

    ΔT = (0.0002) / (0.04 * 3 * 2 * 10⁻⁶) = 416.67°C

    Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C
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