If the radius of star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy protection is Q?
Options
(a) Q / 4 πR²σ
(b) (Q / 4 πR²σ)⁻¹/²
(c) (4 πR²Q / σ)¹/⁴
(d) (Q / 4 πR²σ)¹/⁴
Correct Answer:
(Q / 4 πR²σ)¹/⁴
Explanation:
Stefan’s law for black body radiation Q = σe AT⁴
T = [Q / σ(4πR²) ]¹/⁴ Here e= 1
A = 4πR²