The calorimeter exact operation is complex.
The air is heated by convection.
If you want someone to diligently consider your calculation for
the ht btw the reactor and the box..
Mizuno reports increased excess heat
do due diligence.
Please use Stefans Law with areas and emissivities
to calculate the heat emitted from both surfaces.
Please use the standard convective ht equation
The overall convective HT , "Qc Reactor" is in two parts
not shown in the diagram
1) btw the reactor and the moving air
2) btw the moving air and the box,
1) and 2) have very different values.
Use appropriate areas for the box and for the reactor.. they are very different
State what convective HTCs , h, you are using ... 4, 40 or 400...
State how you justify the HTCs.
Don't use facile shortcuts like "Thus while convective transfer stays roughly proportional to dT"
Is this the THH equation for convective heat transfer?
Then if you come up with missing energy write it down
with units
on a diagram such as this so its easy to understand,