Heat Flow from Thermal Resistance

\dot{Q} = \frac{\Delta T}{R}

Worked example: 22 K across 0.066 K/W → 333.3 W — press Try an example to run it live, then adjust anything.

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Heat Flow from Thermal Resistance explained

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.

Heat Flow from Thermal Resistance formula

\dot{Q} = \frac{\Delta T}{R}

Where

$\dot{Q}$ = Heat flow rate (W)
$\Delta T$ = Temperature difference (C°)
$R$ = Total thermal resistance (K/W)

Missing one of these? Work it out first, then come back

Heat flow rate — Conduction Through a Pipe Wall, Heat Conduction Rate
Temperature difference — Heat Flux Through Insulation (q = ΔT/R), Heat Conduction Rate
Total thermal resistance — Overall U from Total Resistance, Thermal Resistances in Series