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

Stefan's law states that the power radiated by a black body is proportional to the fourth power of its absolute temperature.

The power incident on the black body is given by the intensity multiplied by the area. Let's assume the area of the black body is A. Then, the incident power is 912A W.

In steady state, the power absorbed by the black body is equal to the power radiated by the black body. The power radiated by the black body is given by Stefan-Boltzmann law:

P_radiated = σAT⁴

where:

• P_radiated is the power radiated
• σ is the Stefan-Boltzmann constant (5.7 × 10⁻⁸ Wm⁻²K⁻⁴)
• A is the surface area of the black body
• T is the absolute temperature of the black body

At steady state, the power absorbed equals the power radiated:

912A = σAT⁴

We can cancel A:

912 = σT⁴

Solving for T:

T⁴ = 912 / σ = 912 / (5.7 × 10⁻⁸) = 1.6 × 10¹⁰

T = (1.6 × 10¹⁰)^(1/4) ≈ 660 K

Therefore, the final steady-state temperature of the black body is close to 660K.