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(g) A black sphere of diameter 4 cm is heated to 400 K when the surrounding temperature is 300 K. What is the rate at which energy is radiated? Given \( \sigma = 6 \times 10^{-8} \, \text{Wm}^{-2} \, \text{K}^{-4} \).

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JU-PHY2nd YearFinalThermal PhysicsRadiationStefan-Boltzmann law; Wien's displacement law. (Topic Practice)JU-PHY - ⚡ অনলাইন প্রশ্নব্যাংক দেখুন 💥
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Another Explanation (5):

The surface area of the sphere is A = 4πr² = 4π(0.02 m)² ≈ 0.005 m².

Using Stefan-Boltzmann's Law, the rate of energy radiated (P) is given by:

P = εσA(T⁴ - T0⁴)

where:

  • σ = 6 × 10⁻⁸ Wm⁻²K⁻⁴
  • ε = 1 (for a black body)
  • A = 0.005 m²
  • T = 400 K
  • T₀ = 300 K

Substituting the values:

P = (1)(6 × 10⁻⁸ Wm⁻²K⁻⁴)(0.005 m²)((400 K)⁴ - (300 K)⁴) ≈ 1.71 W