---
title: "Polystyrene insulation to cut wall heat loss 55%"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:brick-plaster-polystyrene-55-percent"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/brick-plaster-polystyrene-55-percent
site: BCT Engineering
---

# Polystyrene insulation to cut wall heat loss 55%

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

- Asked in: 2070 Magh, 6 marks
- Video: [Example 3.6 Heat Loss Through Composite Wall | Thermal Resistance Method | Heat Transfer Problem Sol](https://www.youtube.com/watch?v=QiP9stKdbsA) — Education Shop
- 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-plaster-polystyrene-55-percent
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

An exterior wall of a residential building of 25 cm thick brick [k = 0.7 W/m.°C] followed by layers of 2cm thick cement plaster [k = 0.48 W/m.°C] on both sides. What thickness of extruded polystyrene insulation [k = 0.035 W/m.°C] should be added to reduce the heat loss (or gain) through the wall by 55 percent?

## Short answer

The existing wall resistance is 0.4405 m²·K/W. To reduce the heat loss by 55%, the new total resistance must be 0.9788 m²·K/W, so additional insulation resistance is 0.5384 m²·K/W. Polystyrene thickness = 0.035 × 0.5384 = 0.0188 m ≈ 1.9 cm.

## Answer

**Assumptions:** steady one-dimensional conduction; the same inside and outside temperatures before and after insulation; film resistances are not given and are neglected. Per m² of wall.

#### Resistance of the existing wall
Brick: R_b = L/k = 0.25/0.7 = 0.3571 m²·K/W
Two plaster layers: R_p = 2 × 0.02/0.48 = 0.0833 m²·K/W
R_wall = 0.3571 + 0.0833 = 0.4405 m²·K/W

#### Required resistance
The heat flow is reduced by 55 %, so the new flow is 45 % of the old: q_new = 0.45 q_old.

With the same ΔT: R_new = R_wall/0.45 = 0.9788 m²·K/W

#### Polystyrene thickness
R_ins = R_new − R_wall = 0.9788 − 0.4405 = 0.5384 m²·K/W

L_ins = k R_ins = 0.035 × 0.5384 = 0.01884 m

**Answer: extruded polystyrene thickness ≈ 1.9 cm (about 19 mm).**

## Check yourself

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

1. If the heat loss is to be reduced by 55 %, the new heat flow is:
   A. 0.55 q_old
   B. 1.55 q_old
   C. 0.45 q_old
   D. 0.055 q_old
   Answer: C. Reducing by 55 % leaves 45 % of the original heat flow. (Key point: Meaning of 55 % reduction)

2. With the same temperature difference, how is the required total resistance found?
   A. R_new = 0.45 R_wall
   B. R_new = R_wall/0.45 = 0.9788 m2K/W
   C. R_new = R_wall/0.55
   D. R_new = R_wall + 0.55
   Answer: B. q is proportional to 1/R at fixed ΔT, so q_new = 0.45 q_old requires R_new = R_wall/0.45. (Key point: Required total resistance)

3. After finding R_ins = R_new - R_wall = 0.5384 m2K/W, how is the polystyrene thickness obtained?
   A. L = R_ins/k
   B. L = R_ins x k = 0.035 x 0.5384 = 1.9 cm
   C. L = R_ins x k_brick
   D. L = R_ins x 0.45
   Answer: B. For a plane layer R = L/k, so L = kR = 0.035 x 0.5384, about 0.0188 m. (Key point: Insulation thickness)

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