---
title: "Rock wool to reduce wall heat transfer 50%"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brick-gypsum-rockwool-50-percent"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brick-gypsum-rockwool-50-percent
site: BCT Engineering
---

# Rock wool to reduce wall heat transfer 50%

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

- Asked in: 2069 Bhadra, 8 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-50-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 10cm of common brick (k = 0.8 W/mK) followed by a 4cm layer of gypsum plaster (k = 0.5 W/mK). What thickness of rock wool insulation (k = 0.065 W/mK) should be added to reduce the heat transfer through the wall by 50%?

## Short answer

The existing wall resistance is 0.205 m²·K/W. Halving the heat loss requires doubling the resistance, so the additional rock wool resistance must be 0.205 m²·K/W. Thickness = 0.065 × 0.205 = 0.0133 m (1.33 cm).

## Answer

**Assumptions:** one-dimensional steady conduction; same temperature difference before and after; surface film resistances not given, so neglected. Per m² of wall.

#### Resistance of the existing wall
Brick: R_b = L/k = 0.10/0.8 = 0.1250 m²·K/W
Gypsum plaster: R_g = 0.04/0.5 = 0.0800 m²·K/W
R_wall = 0.2050 m²·K/W

#### Required resistance
Heat transfer is reduced by 50 %, so q_new = 0.5 q_old. With the same ΔT:

R_new = R_wall/0.5 = 0.4100 m²·K/W

#### Rock wool thickness
R_wool = R_new − R_wall = 0.2050 m²·K/W

L_wool = k R_wool = 0.065 × 0.2050 = 0.01333 m

**Answer: rock wool thickness ≈ 1.33 cm (about 13.3 mm).**
(Halving the heat flow means adding a resistance equal to the existing one.)

## Check yourself

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

1. To halve the heat transfer at the same ΔT, the total resistance must become:
   A. Half of the original
   B. The same as the original
   C. Four times the original
   D. Twice the original
   Answer: D. q is proportional to 1/R, so q_new = 0.5 q_old requires R_new = 2 R_wall = 0.41. (Key point: Resistance for 50 % reduction)

2. What resistance must the rock wool add?
   A. 0.205 m2K/W, equal to the existing wall resistance
   B. 0.41 m2K/W, equal to the new total
   C. 0.1025 m2K/W, half the wall
   D. 0.125 m2K/W, the brick only
   Answer: A. R_wool = R_new - R_wall = 0.41 - 0.205 = 0.205 m2K/W. (Key point: Insulation resistance needed)

3. How is the rock wool thickness found from R_wool = 0.205?
   A. L = R/k = 3.15 m
   B. L = k R = 0.065 x 0.205 = 13.3 mm
   C. L = R x (k_brick) = 0.164 m
   D. L = R_wool/2 = 0.1025 m
   Answer: B. For a plane layer R = L/k, so L = kR = 0.065 x 0.205 = 0.01333 m. (Key point: Thickness from resistance)

## Close variants set in other exams

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