# Proof for net heat flow law for radiation

So this has been bothering me for quite a time:

$$\frac{dq}{dt}=e \sigma A(T_1^4-T_0^4)$$

Net radiation is $$emission-radiation$$. According to that if you observe:

Consider a situation where a body is at temperature $$T_1$$ and the surrounding is at temperature $$T_0$$. The rate of emission is

$$\frac{dq}{dt} =e \sigma A T_1^4.$$

So my question is how is the absorption of radiation equal to

$$\frac{dq}{dt} =e \sigma A T_0^4?$$

Is there a proof for this?

I don’t understand how the absorption of radiation is dependent on the surrounding temperature.

The black body radiation spectrum was derived for a closed cavity, where the emission and absorption reached detailed balance (absorption = emission). Then you integrate the whole spectrum to obtain the Stefan-Boltzmann law $$e\sigma A T^4$$.
The atmophere is considered as a black-body cavity, and the object is among the boundary of the cavity, therefore a thermal equilibrium absorption rate, absorb radiation from the "cavity" under thermal equlibrium: $$P_{abs} = e \sigma A T_o^4$$
On the other hand, the object is emiting raditions to the atomosphere. But the emission is running into the vast open space, it is not under thermal equilibrium with temperature $$T_1$$, therefore the black-body radition power can only be adopted as an approximation.