Question

3) When the temperature of a blackbody increases from 900 K to 2300 K, by what factor does the total power of the electromagnetic radiation per-unit- area increase?

Answer 1

Answer:

The electromagnetic radiation per-unit- area increase by 42.65 times.

Explanation:

The radiation per-unit- area (P) is directly proportional to the fourth power of the temperature. Thus,

P ∝ T⁴

For, when the temperature is raised from T₀ to T₁ , the radiation per-unit- area is related to as:

$\frac {P_1}{P_0}=\frac {T_1^4}{T_0^4}$

T₀ =  900 K

T₁ = 2300 K

$\frac {P_1}{P_0}=\frac {2300^4}{900^4}$

So,

P₁ = 42.65 P₀

The electromagnetic radiation per-unit- area increase by 42.65 times.