Heat Loss Through an Assembly (Q = A·ΔT/R)

Q˙=A ΔTRtot\dot{Q} = \frac{A \, \Delta T}{R_{tot}}

Worked example: 30 m2 of RSI 3.5 wall at 22 K → 188.6 W — press Try an example to run it live, then adjust anything.

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Heat Loss Through an Assembly (Q = A·ΔT/R) explained

ΔTAQRtot

This is the line item that appears once per surface on every heat-loss worksheet ever drawn up: area times ΔT, divided by R. A 30 m² wall at RSI 3.5 with 22 K across it loses 30 × 22/3.5 = 189 W. In imperial the same arithmetic stays in imperial — 200 ft² of R-20 wall with 40 °F across it loses 200 × 40/20 = 400 BTU/h, which is 117 W — because BTU, hours, feet and Fahrenheit are a self-consistent set. Sum the surfaces, add the infiltration and ventilation load, and you have the design heat loss the boiler or heat pump has to meet.

Three traps, in the order they bite. Use gross area for each assembly type and take the windows and doors out of it, or you will pay for the same square metres twice. ΔT is a difference, so 40 °F of it is 22.2 K, not 4.4 — this page converts correctly, but the arithmetic in your notebook may not. And the R you divide by should be the effective R of the assembly including framing, not the number on the batt: a wall built with R-19 batts performs at about R-13 once the studs are counted, so a heat loss calculated off the label is roughly 30% optimistic. That single substitution is the most common reason a load calculation comes in under the building's measured fuel use.

Heat Loss Through an Assembly (Q = A·ΔT/R) formula

Q˙=A ΔTRtot\dot{Q} = \frac{A \, \Delta T}{R_{tot}}
Where
  • Q˙\dot{Q}= Heat loss rate (W)
  • AA= Assembly area (m²)
  • ΔT\Delta T= Inside-to-outside ΔT (C°)
  • RtotR_{tot}= Total assembly R-value (RSI (m²·K/W))