>>12512090It's most easily solved by figuring out the equilibrium final temperature of the system. For reference:
CpFe=0.444 g/ J K #25 c
Cpair=1.0035 g/ J K # 25 c
rhoFe=7.864 g / cm3 #solid, 25c
rhoAir=1.225 g / cm3 #dry air, 25c
So we can figure out Mass of air and Fe pretty easily, MFe and Mair
Keep things simple by assuming constant density and heat capacity.
The final temperature of the iron, Tf_Fe = T0_Fe - \Delta T_Fe
The final temperature of the air, Tf_air = T0_air + \Delta T_air
We can solve for change in temperature:
\Delta T_Fe = T0_Fe - Tf_Fe = \DeltaQ / (CpFe*MFe)
likewise
\Delta T_air = Tf_air - T0_air = \DeltaQ / (Cpair*Mair)
We know Tf_air = Tf_Fe because we are at equilibrium and we also know that \DeltaQ is the same for both because of the first law of thermodynamics, so we add the two equations together and solve for \Delta Q:
\Delta Q = ((T0_Fe - T0_air) *(CpFe*MFe*Cpair*Mair)) / ( Cpair*Mair + CpFe*MFe)
We can then plug \Delta Q into our equation for \Delta T (either one) to find the equilibrium final temperature.
For the 1 m3 iron, the room temperature increases by 80 C.
For the 1 cm3 iron, the room temperature increases by 6 thousandths of a degree.
little python script here:
https://pastebin.com/mEeDS4Gv 
