A solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same

Question

A solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube’s temperature in kelvins

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Ben Gia 4 years 2021-08-26T05:55:07+00:00 1 Answers 13 views 0

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  1. thaiduong
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    2021-08-26T05:56:55+00:00
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    Answer:

    Cube temperature = 526.83 K

    Explanation:

    Volume of the cube and sphere will be the same.

    Now, volume of cube = a³

    And ,volume of sphere = (4/3)πr³

    Thus;

    a³ = (4/3)πr³

    a³ = 4.1187r³

    Taking cube root of both sides gives;

    a = 1.6119r

    Formula for surface area of sphere is;

    As = 4πr²

    Also,formula for surface area of cube is; Ac = 6a²

    Thus, since a = 1.6119r,

    Then, Ac = 6(1.6119r)²

    Ac = 15.5893r²

    The formula for radiant power is;

    Q’ = eσT⁴A

    Where;

    e is emissivity

    σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k

    T is temperate in kelvin

    A is Area

    So, for the cube;

    (Qc)’ = eσ(Tc)⁴(Ac)

    For the sphere;

    (Qs)’ = eσ(Ts)⁴(As)

    We are told (Qc)’ = (Qs)’

    Thus;

    eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)

    eσ will cancel out to give;

    (Tc)⁴(Ac) = (Ts)⁴(As)

    Since we want to find the cube’s temperature Tc,

    (Tc)⁴ = [(Ts)⁴(As)]/Ac

    Plugging in relevant figures, we have;

    (Tc)⁴ = [556⁴ × 4πr²]/15.5893r²

    r² will cancel out to give;

    (Tc)⁴ = [556⁴ × 4π]/15.5893

    Tc = ∜([556⁴ × 4π]/15.5893)

    Tc = 526.83 K

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