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JEE Advanced2014PhysicsThermal Properties of MatterActual

Parallel rays of light of intensity I = 912 W m - 2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ = 5.7 × 10 -8 W m - 2 K - 4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to

Options

  1. A330 K
  2. B660 K
  3. C990 K
  4. D1550 K

Correct answer

A. 330 K

Step-by-step solution

Rate of radiation energy lost by the sphere = Rate of radiation energy incident on it ⇒ σ × 4 π r 2 T 4 - 300 4 = 912 × π r 2 ⇒ T = 11 × 10 2 ≈ 330 K

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