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Rock wool to reduce wall heat loss by 80%

Short answer

Existing wall resistance (brick + gypsum) = 0.2262 m²·K/W. To reduce heat loss by 80%, total resistance must become 1.1310 m²·K/W. Required rock wool resistance is 0.9048 m²·K/W, giving thickness L = kR = 0.065 × 0.9048 ≈ 0.059 m (5.9 cm).

An exterior wall of a house consists of a 0.1m layer of common brick (k = 0.7 W/m°C) followed by a 0.04 m layer of gypsum plaster (k = 0.48 W/m°C). What thickness of loosely packed rock wool insulation (k = 0.065 W/m°C) should be added to reduce the heat loss through the wall by 80 percent?

Assumptions: one-dimensional steady conduction; the same inside and outside temperatures before and after adding insulation; surface film resistances are not given and are neglected. Heat flux per unit area is q = ΔT/R_total.

Resistance of the existing wall (per m²)

R_brick = L/k = 0.1/0.7 = 0.1429 m²·K/W R_gypsum = 0.04/0.48 = 0.0833 m²·K/W R_wall = 0.2262 m²·K/W

Required resistance

The heat loss is reduced by 80 %, so the new heat flow is 20 % of the old: q_new = 0.2 q_old.

Since q = ΔT/R and ΔT is unchanged: R_new = R_wall/0.2 = 1.1310 m²·K/W

Rock wool resistance and thickness

R_wool = R_new − R_wall = 1.1310 − 0.2262 = 0.9048 m²·K/W

L_wool = k R_wool = 0.065 × 0.9048 = 0.0588 m

Answer: rock wool thickness ≈ 5.9 cm (about 59 mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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