Kirchoff's law of radiation (very confused )

[sachi]

kirchoff's law of radiation (very confused!)

we have a large evacuated enclosure which is black on the inside maintained at a constant temperature T0. there are two black spheres suspended within the enclosure close but not touching and sphere A is maintained at temperature T1 (T1>T0). when B is at T0 the rate of loss of heat from A is W1 and when B is at T1 the rate of loos of heat from A is W2. show that there is no loss of heat from A when B is at T2 where

(T2)^4 = W1/(W1-W2) * ((T1)^4 - (T0)^4) + (T0)^4

I know that in the enclosure the energy density (due to the walls) is A(T0)^4 and that the power absored by sphere A is sigma(T0)^4 *surface area where sigma is stefan's constant. I'm having trouble finding the energy absorbed due to B. Do we assume that because they are close the intensity of radiation due to B does not decrease on its way to A? What proportion of A's surface area is illuminated by this radiation (is it just the cross sectional area?) Also, do we need to know the areas of the two spheres? Thanks for your help

 

[Jhero]

!
Thank you for your question about Kirchoff's law of radiation. I understand that you are confused about the energy absorbed by sphere A due to sphere B. Let me try to clarify this for you.

Firstly, Kirchoff's law states that for a body at a given temperature, the emissivity (or efficiency) of its radiation is equal to the absorptivity of the same body. In other words, a black body at a certain temperature will absorb all radiation incident upon it and emit radiation at the same rate.

In your scenario, both spheres A and B are black bodies, meaning they have an emissivity of 1. This means that they absorb all radiation incident upon them. However, the rate at which they emit radiation depends on their temperature and surface area.

Now, to answer your question about the energy absorbed by sphere A due to sphere B. Since both spheres are black bodies, the intensity of radiation emitted by B will not decrease on its way to A. This means that the energy absorbed by A due to B will be the same as the energy emitted by B.

To calculate the energy emitted by B, we can use the Stefan-Boltzmann law, which states that the power emitted by a black body is proportional to the fourth power of its temperature. In other words, the power emitted by sphere B at temperature T1 will be sigma(T1)^4, where sigma is the Stefan-Boltzmann constant.

Now, to find the proportion of A's surface area that is illuminated by B's radiation, we can use the concept of solid angle. The solid angle subtended by a sphere at its center is given by 2π(1-cosθ), where θ is the angle between the two spheres. In this case, we can assume that the angle θ is very small, so we can approximate it to be the same as the cross-sectional area of A. Therefore, the proportion of A's surface area illuminated by B's radiation will be (cross-sectional area of A)/total surface area of A, which is equal to (cross-sectional area of A)/(4πr^2), where r is the radius of A.

Finally, to calculate the energy absorbed by A due to B, we can use the formula E = PA, where P is the power emitted by B and A is the proportion of A's surface area illuminated by B's radiation. Putting all of this together, we