Heat Flow from Thermal Resistance

This is the payoff of the whole resistance network: once you have added up the layers, the heat flow is just ΔT over R, the exact analogue of I = V/R. A wall assembly totalling 0.066 K/W with 22 K across it passes 22/0.066 = 333 W, and no further physics is required. The same equation run backwards is how a thermal engineer sizes an enclosure: an electronics box dissipating 40 W that may only rise 25 K above ambient needs the path from junction to air to be under 0.625 K/W, total, including the heatsink and the interface pad.

Because resistances are additive and heat flow is not, the intermediate temperatures come free: the drop across any single layer is Q̇ × R for that layer alone. That is how you check whether a wall's dew point falls inside the insulation or safely outside it, and how a plant engineer proves the fouled side of an exchanger is the tube interior rather than the shell. Trap: ΔT here is a difference, so 40 °F of difference is 22.2 K, not 4.4. The calculator handles the conversion, but the arithmetic in your notebook may not.