The energy absorbed per unit time by the body at an absolute temperature $T_1$ and kept at a surrounding of higher temperature $T_2$ is $$J=\epsilon\sigma A T_2^4.$$

What my question is that aren't there many objects in the surrounding emitting radiations at temperature $T_2$ and applying Stefan-Boltzmann law for all of them wouldn't the radiations absorbed be greater than J(above)?

Kinda like if you have a blackbody at the center of the spherical shell then shouldn't the shell focus all the radiations on the blackbody effectively heating it beyond the temperature of shell,

$$s_1 - s_2 = \tfrac{1}{2}t^2 \mu g \left( -\tfrac{1}{2} + \frac{M+m}{M} - \frac{m}{2M} \right)$$

• Notice it says "surrounding of higher temperature $T2$." And, notice that there is an $A$ in the formula, presumably for area. Also, in your shell case, the shell will be seeing a lot of itself on the other side, not just the point at the center.
– user93146
Commented Jun 25, 2018 at 14:18
• Why does $T_1$ not appear in your equation? In this case, you can take $T_1$ to be the room temperature. @harambe
– SRS
Commented Jun 25, 2018 at 14:20
• @SRS tlI thought that since the surrounding is emmiting energy then the rate will be dependent only on it
– user195235
Commented Jun 26, 2018 at 9:23
• @puppetsock can you elaborate this as I am afraid I am not able to catch up with what you are saying
– user195235
Commented Jun 26, 2018 at 9:24
• The situation is very ill-defined. Please specify the geometry, form of radiation and absorption coefficients. There are many models that treat this situation of bodies at different temperatures. Check Newton's Cooling law Commented Jun 27, 2018 at 17:33

A spherical perfect black body $B_1$ of temperature $T_1$ completely surrounded by another perfect black body $B_2$ of temperature $T_2$ emits $J_e = \sigma A T_1^4$ and absorbs $J_a = 4 \pi \sigma A T_2^4$. For imperfect black bodies, denoting their absorptivities by $0 \lt \alpha_i \lt 1$ and emissivities by $0\lt\epsilon_i\lt 1$ then we get $J_e = \epsilon_1 \sigma A T_1^4$ and $J_a = \alpha_1 \epsilon_2 \sigma A T_2^4$. If $B_2$ incompletely surrounds $B_1$, that is, if it extend a solid angle of $\Omega \lt 4 \pi$ then $J_a = \Omega \sigma A T_2^4$.