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29 November, 07:47

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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  1. 29 November, 10:42
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    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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