---
title: "Rock wool to reduce wall heat loss by 80%"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brick-gypsum-rockwool-80-percent"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brick-gypsum-rockwool-80-percent
site: BCT Engineering
---

# Rock wool to reduce wall heat loss by 80%

Fundamental of Thermodynamics & Heat Transfer · Chapter 7: Introduction to Heat Transfer

- Asked in: 2073 Magh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brick-gypsum-rockwool-80-percent
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

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?

## 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).

## Answer

**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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. If heat loss falls by 80 %, the total resistance becomes
   A. 5 times the old value
   B. 0.2 times the old value
   C. 1.8 times the old value
   D. 0.8 times the old value
   Answer: A. q_new = 0.2 q_old at the same ΔT, so R_new = R_wall/0.2. (Key point: New resistance)

2. Resistance of the existing wall per m^2 is
   A. 0.1 x 0.7 + 0.04 x 0.48
   B. 0.14 / 1.18
   C. 0.7 + 0.48
   D. 0.1/0.7 + 0.04/0.48 = 0.2262
   Answer: D. Resistances L/k of brick and gypsum add. (Key point: Existing wall)

3. The rock wool resistance is
   A. R_wall/0.2 = 1.131
   B. 0.2 R_wall
   C. R_new = 1.131
   D. R_new - R_wall = 0.9048
   Answer: D. The added layer supplies the extra resistance: 1.1310 - 0.2262. (Key point: Insulation resistance)

4. The rock wool thickness is
   A. 0.2262 x 0.065
   B. 0.9048 cm
   C. R/k = 13.9 m
   D. k x R = 0.065 x 0.9048 = 5.9 cm
   Answer: D. L = k R_wool = 0.0588 m. (Key point: Thickness)

## Close variants set in other exams

- [Rock wool to reduce wall heat transfer 50%](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrick-gypsum-rockwool-50-percent) — Fundamental of Thermodynamics & Heat Transfer, chapter 7, asked 2×

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