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Lesson 44 · Heat on the move
Watts through walls
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Watts through walls

So far we have counted heat in lump sums. But through a window, heat runs as a steady rate — joules every second, which is to say watts, the same watts as any power. The machine is P=kA ΔTdP = \dfrac{kA\,\Delta T}{d} (read aloud: P equals k A delta-T over d): PP is the heat flow in watts, kk the material's thermal conductivity in W/(m·°C) — a data-sheet number, glass leaks, foam doesn't — AA the area of wall that is leaking in m², ΔT\Delta T the temperature difference across it in C°, and dd the thickness in metres, underneath because a thicker wall resists.

That placement of dd is the whole physics of insulation: double the thickness, halve the leak — the double-pane logic. And a field note: thickness arrives in millimetres, but the machine runs on metres. A 4 mm pane entered as 4 leaves your answer a thousand-fold wrong, in the embarrassing direction.

ΔTkAPd

P=kAΔTdP = \tfrac{k A \Delta T}{d}

  • PP= Heat flow rate (power)
  • kk= Thermal conductivity (thermal conductivity)
  • AA= Cross-sectional area (area)
  • ΔT\Delta T= Temperature difference (temperature difference)
  • dd= Thickness (length)
Heat Conduction Rate solver →