Correct Answer - Option 3 : 1500 K
CONCEPT:
- A perfect black body absorbs all the electromagnetic radiation incident on it.
- The black body radiation spectrum is the spectrum of radiation emitted by a black body when it is at a higher temperature than its surroundings.
-
Stefan-Boltzmann's law states that the total power radiated per unit surface area of a black body is proportional to the fourth power of the temperature of the black body.
⇒ P ∝ T4
⇒ P = kT4
- In the above equation, T is the temperature of the black body in Kelvin, k is the proportionality constant and P is the power radiated per unit surface area of a black body in J/(s.m2).
CALCULATION:
Given - P1 = 1.0 × 105 Joule per sec. per m2, P2 = 81 × 105 Joule/sec/m2, and T1 = 500 K
Case 1:
\(\Rightarrow P_1=kT^4_1\)
\(⇒ k=\frac{P}{T^4} =\frac{1.0 \times 10^5}{500^4}=1.6\times 10^{-6} \frac{J}{s.m^2.K^4 }\)
Case 2:
\(\Rightarrow P_2=kT^4_2\)
\( ⇒ T^4_2=\frac{P_2}{k} =\frac{81.0 \times 10^5}{1.6\times 10^{-6}}=5.0625\times 10^{12} K^4 \\ ⇒ T_2 = 1500 K\)
- Therefore, option 3 is correct.
- For a perfect black body with emissivity ϵ = 1, the value of Stefan-Boltzmann's constant is \(σ = \frac{P}{T^4} = 5.67 \times 10^{-8} \frac{J}{s.m^2.K^4}\).
- For a non-perfect black body with emissivity ϵ < 1, the Stefan-Boltzmann's law modifies as P = ϵσT4 or P = kT4.
