Question

For what temperature of a black body, the rate of emitted radiant energy from unit area will be 20 kWm-2 ? (σ = 5.67×10-8 Wm-2 K -4)​

Answer 1

We have to find the temperature of a black body at which the rate of emitted radiant energy from unit area will be 20kW/m² .

solution : From Stefan–Boltzmann law, the rate of emitted radiant energy per unit area is given by, J = σT⁴

where, σ is Stefan–Boltzmann constant.

here, J = 20 kW/m²

and σ = 5.67 × 10⁻⁸ W/m²/K⁴

so, 20 × 10³ W/m² =  5.67 × 10⁻⁸ W/m²/K⁴ × T⁴

⇒ (2/5.67) × 10¹² = T⁴

⇒ $\sqrt[4]{\frac{2}{5.67}}$ × 10³ = T

⇒ T = 0.7054 × 10³ = 705.4K

therefore the temperature of the black body is 705.4K.

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Answer 2

From Stefan–Boltzmann law, the rate of emitted radiant energy per unit area is given by, J = σT⁴

where, σ is Stefan–Boltzmann constant.

here, J = 20 kW/m²

and σ = 5.67 × 10⁻⁸ W/m²/K⁴

so, 20 × 10³ W/m² =  5.67 × 10⁻⁸ W/m²/K⁴ × T⁴

⇒ (2/5.67) × 10¹² = T⁴

⇒  ∜(2/5.67) × 10³ = T

⇒ T = 0.7054 × 10³ = 705.4K