Here is the problem, you have an insulated box emitting 240Wm-2 through the outer insulation layer. You peak inside the insulation and see that there are actually ten objects under the insulation, 7 or 70% of the objects are emitting 425Wm-2 and 3 or 30% are emitting 308Wm-2 for a total emission felt at the inside of the insulation of 390Wm-2. Insulation is added so that 3.7Wm-2 of added resistance to flow is uniformity distributed around the box. How much additional energy is stored in each of the ten objects in the interior?

Since the objects in question are not at the same energy state and enclosed in the same insulation blanket there is a variety of ways at looking at the problem. You could ignore the individual objects and just look at the total energy. 390Wm-2 inside, 240Wm-2 outside leaves 150Wm-2 of resistance to energy flow. Increasing the resistance by 3.7Wm-2 would increase the resistance by 3.7/150=0.025 so 1.025*390=399.75Wm-2 would be the impulse increase in the energy of the object. The problem doesn't state that there is constant energy in the insulated part or that the energy through the insulation has to remain constant. If you just magically increased the resistance to energy flow, there would be some thermal inertia which would increase the energy in the insulation to 399.75Wm-2 for some brief moment in time if the thermal mass were small. The energy could overshoot this value and then drop or slowly sneak up on it, but there would be an increase toward 399.75Wm-2.

Another way to look at the problem is that some objects in the box are warmer. There would be energy flowing from those objects to the cooler objects. Since there is energy flowing through the insulation, the warmer objects are losing more energy to the environment. With the warmer at 425Wm-2 and the colder at 308Wm-2, there would be 117Wm-2 net energy flowing from the warmer to the colder. The net flow from the warmer to ambient would be 308Wm-2 matching the flow of the cooler object to ambient. With 308Wm-2 average energy trying to escape and only 240Wm-2 escaping the restriction to flow would be 308-240 or 68Wm-2. Adding 3.7 to 68 would increase the resistance to flow by 3.7/68=0.054 resulting in 1.054 *(.7*425+.3*308)=1.054*390=411Wm-2 as the value that would be approached.

Which one makes the most sense?

Think about if the colder objects are surrounded by the warmer objects. The objects all radiate the same energy in all directions. So now the warmer objects would emit a net flux of 117Wm-2 inward, 425Wm-2 outward which would be restricted by the insulation allowing only 240Wm-2 to escape. 425-240=185Wm-2, Adding 3.7Wm-2 of resistance would decrease the rate of flow by 3.7/185=0.02 leaving 1.02*425=433.5Wm-2.

At 433.5Wm-2 the temperature via S-B would be 295.7 C degrees. The initial temperature of the 425Wm-2 objects was 294.25 C degrees, the increase in temperature would be 1.45 C degrees.

Now let's change the state of the objects. Let's make the warmer objects 300K degrees which would radiate 459Wm-2. 459-240=219, 3.7/219=0.0168 resulting in 1.0169*459=466.8 Wm-2 which would be a S-B temperature of 301.2 K degrees. The warmer the warmest object, the less impact increasing the insulation has.

UPDATE: Since a few people are scratching their heads. Here is something else about this simple problem.

By specifying that the objects inside the insulation are in conditional equilibrium, the energy flow from the warmer to colder objects, 117Wm-2 takes on a whole new meaning. If they are in equilibrium with that flow rate, then the inside of the box is in equilibrium with the outside. That would mean that the ambient heat sink would be 117Wm-2. But wait you say, the flux to ambient is 240Wm-2! Nope, it would be 117Wm-2 which would mean that there is 240-117=123Wm-2 difference. That would be the true effective sink or ambient flux which is a temperature of 57.2 degrees C or 215.8K degrees, the temperature of the tropopause. Some of you may find that interesting.

Now if you were into abstract logic, you would realize that 123Wm-2 is an odd number in more ways than one. Let's see if you can figure out why? If you go back through the calculations above, you may find a hidden surprise.

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