Question

A black body has absolute temperature T and surface area A. The intensity of the radiation emitted by the body is I. Another black body of surface area 2A has absolute temperature 2T. What is the intensity of radiation emitted by this second black body?

I keep getting the answer 16I, but the answer key says it's 32I. How come?

Equations I'm using:

P = εσAT^4

I = P/A

Answer 1

Stefan-Boltzmann Law tells that that power `P` radiated from a black body can be expressed as

`P = εσAT^4`
where `epsilon` is the emissivity, `sigma` is the Stefan–Boltzmann constant, `A` area of radiating surface and `T` is absolute temperature of the radiating body.

As such

`P_1=ε_1σAT^4` ........(1)
and
`P_2=ε_2σA_2T_2^4` ........(2)

Using given values and assuming that emmissivities of both black-bodies is sa emphasized text me we get

`P_2=ε_1σ(2A)(2T)^4`
`=>P_2=32P_1`........(4)

We know that intensity is the power transferred per unit `area`, (this `area` is different from the area of the radiating body. This is the area where intensity is being measured).

Therefore for the same area

`I_2=32I_1`