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ExamsJEE AdvancedPhysics

A parallel beam of radiation falls on a perfectly conducting (perfectly absorbing) black sphere. The incident radiation intensity on the sphere is 1944 W/m². Stefan's constant sigma = 6*10⁻⁸ W/(m² K⁴). What is the steady-state temperature of the sphere in degrees Celsius? (Use 0 deg C = 273 K)

  1. 27 deg C
  2. 57 deg C
  3. 87 deg C
  4. 117 deg C

Correct answer: 27 deg C

Solution

Equating absorbed and radiated power: I*(pi*r²) = sigma*T⁴*(4*pi*r²). Cancel pi*r²: T⁴ = I/(4*sigma) = 1944/(4*6*10⁻⁸) = 1944/(2.4*10⁻⁷) = 8.1*10⁹. T = (8.1*10⁹)^(1/4). Note 8.1*10⁹ = 81*10⁸, so T = (81)^(1/4)*(10⁸)^(1/4) = 3*100 = 300 K = 27 deg C.

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