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 Jun 25 '18 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 Jun 25 '18 at 14:20
• @SRS tlI thought that since the surrounding is emmiting energy then the rate will be dependent only on it – user195235 Jun 26 '18 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 Jun 26 '18 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 – ohneVal Jun 27 '18 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 $J_a = \Omega \sigma A T_2^4$.