A 12.0 m² wall consists of 0.100 m brick with thermal conductivity 0.72 W m⁻¹ K⁻¹ and 0.060 m insulation with conductivity 0.040 W m⁻¹ K⁻¹. Indoor and outdoor surface temperatures are 293 K and 273 K. Find the steady conductive power and interface temperature. If the outer surface has emissivity 0.85 and sees surroundings at 263 K, calculate its net radiative loss. Explain why simply adding that radiation to the conduction result may violate steady-state energy balance.

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Thermal resistance = 0.1366 K W⁻¹; conductive power = 146 W; interface temperature ≈ 291.3 K if brick is inside. Net outer radiation ≈ 445 W. The specified outer-surface temperature cannot remain fixed without a matching energy source. Series thermal resistances are R_brick = L/(kA) = 0.100/[0.72(12.0)] = 0.0116 K W⁻¹ and R_ins = 0.060/[0.040(12.0)] = 0.125 K W⁻¹. Total resistance is 0.1366 K W⁻¹, so conductive power through the composite wall is ΔT/R = 20/0.1366 = 146 W, not 12.2 W. The brick temperature drop is P R_brick = (146)(0.0116) = 1.69 K. If brick faces indoors, the interface is therefore 293−1.69 = 291.3 K. Net radiation is εσA(T_s⁴−T_sur⁴) = 0.85(5.67 × 10⁻⁸)(12.0)(273⁴−263⁴) ≈ 445 W. The radiative loss exceeds the calculated conductive supply at the imposed temperatures, so the outer surface would cool unless convection or another source supplies the difference. A true steady-state solution must determine surface temperatures from simultaneous conduction, convection and radiation balances.

Conduction and radiation through a wall

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