Question

The rate of emission of radiation of a black body at temperature 127°C is Ej . If its temperature is increased to 727°C the rate of emission of
radiation is E2 . The relation between Ej and E, is
(A) E2 = 24 E
(B) 625 E = 16 E2
(C) E2 = 8 E
= E1
(D) 16 Ej = 625 E2 pls give correct answer​

Answer 1

Given:

Rate of emission (power) of radiation of the lack body at temperature 127°C = Ej

Rate of emission at 727°C = E₂

To find:

The relation between Ej and E₂.

Solution:

From the Steffan Bolzmann law:

Power dissipated by black body E = AξσT⁴

Since it is a black body ξ = 1

Here A is the surface area of the block body,

σ is a constant which is 5.67*10⁻⁸ J/m²sK⁴

Therefore E ∝ T⁴

Ej/E₂ = 400⁴/ 1000⁴

400⁴ E₂ = 1000⁴ Ej

256 * 10⁸ E₂ = 10¹² Ej

256 E₂ = 10000 Ej

On reducing it we get:

16 E₂ = 625 Ej

The correct option is option B.