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Heat loss through a wall

Heat and matter · level 7, lesson 1 of 6

A question to start

A Mars habitat is kept at 20 °C inside, with about −60 °C outside. How much heat leaks through its walls each second, and how thick must the insulation be?

The idea

Heat conducts through a slab at a rate

\[P = \frac{kA\,\Delta T}{d},\]

where \(A\) is the area, \(d\) the thickness, \(\Delta T\) the temperature difference across it, and \(k\) the material's thermal conductivity, in watts per metre per kelvin.

Material \(k\), W/(m K)
Copper 400
Steel 50
Concrete about 1.4
Glass about 1.0
Brick about 0.7
Wood about 0.13
Mineral wool about 0.04
Mars soil at InSight's site 0.039
Still air 0.026
Insulating foam board about 0.022
inside 20 °Coutside −60 °Cdheat
100 m² of wall with 10 cm of foam Mars Earth
Inside 20 °C 20 °C
Outside, on average −59 °C 15 °C
\(\Delta T\) 79 K 5 K
Heat lost, \(kA\Delta T/d\) about 1,740 W about 110 W

Worked examples

1. The Mars wall. \(P = 0.022 \times 100 \times 79/0.10 = 1{,}740\) W.

2. Bare concrete instead. 20 cm of concrete: \(P = 1.4 \times 100 \times 79/0.20 = 55{,}000\) W, over 30 times as much.

3. How thick for 500 W? \(d = kA\Delta T/P = 0.022 \times 100 \times 79/500 = 0.35\) m of foam.

Try it

  1. Find the heat lost through 10 m² of brick 0.2 m thick, with 15 K across it.
  2. If the insulation is made twice as thick, what happens to the heat loss?
  3. Why are thin metal walls a poor choice on their own?
  4. Why is the heat loss on Mars so much bigger than on Earth for the same wall?
Answers
  1. \(0.7 \times 10 \times 15/0.2 = 525\) W.
  2. It halves.
  3. Metal conducts heat very well, so heat pours through.
  4. The temperature difference is about 16 times bigger, so the loss is about 16 times bigger.

Common mistake

Using the inside temperature for \(\Delta T\). It is the difference across the wall: 20 − (−59) = 79 K, not 20.

Where it leads