Interaction between blackbodies in relation to the second law

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In the interaction between two blackbodies, B_1 and B_2, with temperatures T_1 and T_2 (where T_1 > T_2), energy transfer occurs solely through radiation. B_2 can indeed receive energy from B_1, but the second law of thermodynamics dictates that B_1 will always emit more energy to B_2 than it receives back. The net energy exchange will consistently favor B_1, resulting in a positive net heat transfer from B_1 to B_2. The equation presented illustrates that the net heat transfer (dQ) is positive, confirming that energy flows from the hotter body to the cooler one. This behavior aligns with the principles of thermodynamics governing blackbody radiation.
Shunyata
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I am trying to understand the behavior of blackbodies in interaction with each other.

Conditions as follows.

1) There are two blackbodies, say B_1 and B_2, with corresponding temperature T_1 and T_2.

2) Initially T_1>T_2.

3) B_1 and B_2 have thermal contact only through radiation exchange.

And finally my question,

Can B_2 give energy to B_1 or is this prevented by the second law?
 
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Sure it can. In fact, it always will be. However, B1 will always be giving more energy to B2 than the reciprocal effect, so the net energy exchange will always involve B1 giving energy to B2.
 
Is it as simple as

dQ\Big( \frac{1}{T_2} - \frac{1}{T_1}\Big)=dQ\Big( \frac{T_1-T_2}{T_1T_2}\Big)>0,

if dQ=dQ_1 - dQ_2 is the net-heat where positive direction is from B_1 to B_2 and dQ_1>dQ_2?
 
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