Equivalent insulation thickness A furnace wall of red brick (thermal conductivity k₁ = 0.7 W/m·K) has thickness 0.50 m. For the same heat loss and the same temperature drop across the wall, what thickness of diatomite earth (thermal conductivity k₂ = 0.14 W/m·K) would provide an equivalent thermal resistance?

Introduction / Context:
Thermal insulation design often requires replacing one material with another while keeping the same conductive heat loss for a given temperature difference. This is a straightforward conduction resistance equivalence problem.

Given Data / Assumptions:

• Brick wall thickness t₁ = 0.50 m, conductivity k₁ = 0.7 W/m·K.
• Diatomite earth conductivity k₂ = 0.14 W/m·K; required thickness t₂ = ?
• One-dimensional steady conduction, same surface area A, and same temperature drop ΔT.

Concept / Approach:
For a plane wall, conductive heat flow Q = k * A * ΔT / t. For the same Q and ΔT with the same area, the ratio k/t must be equal for the two walls. Therefore, t is directly proportional to k when Q, A, and ΔT are held constant.

Step-by-Step Solution:
Set equivalence: k₁ / t₁ = k₂ / t₂.Solve for t₂: t₂ = t₁ * (k₂ / k₁).Compute ratio: k₂ / k₁ = 0.14 / 0.70 = 0.20.Therefore, t₂ = 0.50 * 0.20 = 0.10 m.

Verification / Alternative check:
Compare total resistances: R₁ = t₁/(k₁ A) = 0.50/(0.7A) ≈ 0.714/A; R₂ = 0.10/(0.14A) ≈ 0.714/A; equal as required.

Why Other Options Are Wrong:
0.50 m (same thickness) would lower Q by a factor of 0.14/0.70 = 0.2 (too much resistance). 0.20 m still over-insulates. 0.75 m and 1.00 m are unrelated distractors that significantly overestimate resistance.

Common Pitfalls:
Inverting the proportionality (using t ∝ 1/k) for a fixed Q; remember here we keep Q the same, thus k/t equality applies, leading to t ∝ k.

Final Answer:
0.10 m