How many days does it take for a perfect blackbody cube (0.0100m on a side, 30.0 degree C)to radiate the same amount of energy that a one-hundred-watt light bulb uses in one hour?

I got no idea to do it. Please give me some hints!!!Thanks!!!

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  1. That is a pretty small cube: 1 cm^3 -- about the size of a lump of sugar. At area of all sides together is
    A = 6 cm^3 = 6*10^-4 m^2. The absolute temperature is T = 303.2 K.
    According to the Blackbody radiation law, the power radiated is
    sigma*T^4*A = 0.5669*10^-7*(303.2)^4* 6.00*10^-4 = 0.2875 W

    A 100W bulb uses 100 Joules in one second, which is 3.6*10^5 J in an hour.

    The cube will need 3.6*10^5 J/(0.2875 J/s)= 1.252*10^s = 14.5 days to radiate that amount of energy.

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  2. The light bulb first
    3600 seconds * 100 Joules /second = 3.6*10^5 Joules

    Now that radiation
    Black body has very good emissivity (e), call it 1

    so Heat current = Area * 1 *5.67*10^-8 W/m^2 deg * T^4
    the constant is the Stefan-Boltzmann constant for radiation
    T is Kelvin = C +273 = 303
    so T^4 = 8.43*10^9
    area of 1 side = .01*.01 = 10^-4 m^2
    6 sides so 6*10^-4 m^2 = A
    Watts = 6*10^-4 * 5.67*10^-8 * 8.43*10^9
    Watts = 287* 10^-3 = .287 watts
    Watts * seconds = Joules = 3.6*10^6 from the light bulb
    so seconds = 3.6*10^6/.287
    = 12.5 * 10^6 seconds
    there are = 86,400 s/day
    12.5 * 10^6 /8.64*10^4 = 1.45*10^2 days
    145 days

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  3. My last line should have read
    1.252*10^6 s = 14.5 days.
    Either I or Damon missed a decimal point. Check us both and find out.

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