Answer:
a) The colder body (3000k), b) hearter body c) 12000K body
Explanation:
This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement
Stefan's Law P = σ A e T⁴
Wien displacement law λ T = 2,898 10⁻³ m K
Let's calculate the power emitted for each object.
As they are perfect black bodies e = 1, they also indicate that they have the same area
T = 3000K
P₁ = σ A T₁⁴
T = 12000K
P₂ = σ A T₂⁴
P₂ / P₁ = T₂⁴ / T₁⁴
P₂ / P₁ = (12000/3000)⁴
P₂ / P₁ = 256
This indicates that the hottest body emission is 256 times the coldest body emission.
Let's calculate the maximum emission wavelength
Body 1
T = 3000K
λ T = 2,898 10-3
λ₁ = 2.89810-3 / T
λ₁ = 2,898 10-3 / 3000
λ₁ = 0.966 10-6 m
λ₁ = 966 nm
T = 12000K
λ₂ = 2,898 10-3 / 12000
λ₂ = 0.2415 10-6 m
λ₂ = 214 nm
a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)
b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area
c) The emission of the hottest 12000K body is mainly in UV
d) The hottest body emits more energy in UV and visible
e) No body has greater emission in all zones
