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March 30, 2017

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A sphere of radius 0.583 m, temperature 39.4°C, and emissivity 0.983 is located in an environment of temperature 85.1°C. At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange?

  • thermal physics - ,

    The area of the sphere is:

    A = 4 pi R^2

    where R is the radius

    The emitted power is given by:

    P_em = emissivity*A*sigma T1^4

    where T1 is the absolute temperature of the sphere

    The absorbed power is given by:

    P_ab = emissivity*A*sigma T2^4

    where T2 is the absolute temperature of the environment.

    sigma is given by:

    sigma = 2 pi^5 k^4/(15 h^3 c^2)

    Here k is Boltzmann's constant, h is Planck's constant and c is the speed of light. sigma is approximately given by:

    sigma = 5.67*10^(-8) Watt/(K^4 m^2)

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