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Example · Example 6

Q.A blackened metal sphere of surface area 0.1 m20.1\ \text{m}^2 is maintained at a steady temperature of 727 ∘C727\,^\circ\text{C} (1000 K1000\ \text{K}) inside an enclosure whose walls are at 27 ∘C27\,^\circ\text{C} (300 K300\ \text{K}). Taking the Stefan-Boltzmann constant as σ=5.67×10−8 W m−2K−4\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}, find

(a) the total power the sphere would radiate by Stefan's law alone, and
(b) the net power it actually loses to the enclosure once Boltzmann's correction for the radiation it absorbs back from the surroundings is included.
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Given: A=0.1 m2A = 0.1\ \text{m}^2, T=1000 KT = 1000\ \text{K}, T0=300 KT_0 = 300\ \text{K}, σ=5.67×10−8 W m−2K−4\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}.

  1. Power radiated by Stefan's law alone:

    P=σAT4=5.67×10−8×0.1×(1000)4=5.67×10−9×1012=5670 WP = \sigma A T^4 = 5.67\times10^{-8} \times 0.1 \times (1000)^4 = 5.67\times10^{-9} \times 10^{12} = 5670\ \text{W}

  2. Net power lost, with Boltzmann's correction:

    Pnet=σA(T4−T04)P_{\text{net}} = \sigma A\left(T^4 - T_0^4\right)

    T04=(300)4=8.1×109T_0^4 = (300)^4 = 8.1\times10^{9}, so T4−T04=1012−8.1×109=9.919×1011T^4 - T_0^4 = 10^{12} - 8.1\times10^{9} = 9.919\times10^{11}. …

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