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Physics · Grades 11–12

Steady conduction through a slab

Use a uniform slab, steady temperatures and one-dimensional conduction.

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Rule and assumptions

Use a uniform slab, steady temperatures and one-dimensional conduction.

P = kAΔT/L

Worked examples

For a uniform slab, k = 0.1 W/(m·K), area = 2 m², thickness = 15 cm and temperature difference = 14 K. Assuming steady one-dimensional conduction, find thermal power. Round to 3 decimal places if needed. Answer in W.

P = kAΔT/L 0.1 × 2 × 14 / (15/100) = 18.667

18.667 W

For a uniform slab, k = 0.2 W/(m·K), area = 16 m², thickness = 16 cm and temperature difference = 35 K. Assuming steady one-dimensional conduction, find thermal power. Round to 3 decimal places if needed. Answer in W.

P = kAΔT/L 0.2 × 16 × 35 / (16/100) = 700

700 W

Try two questions

  1. For a uniform slab, k = 0.6 W/(m·K), area = 9 m², thickness = 10 cm and temperature difference = 26 K. Assuming steady one-dimensional conduction, find thermal power. Round to 3 decimal places if needed. Answer in W.

  2. For a uniform slab, k = 0.5 W/(m·K), area = 19 m², thickness = 7 cm and temperature difference = 20 K. Assuming steady one-dimensional conduction, find thermal power. Round to 3 decimal places if needed. Answer in W.

Answers and reasoning

Answers and marking notes
  1. 1404 W

    P = kAΔT/L 0.6 × 9 × 26 / (10/100) = 1404

  2. 2714.286 W

    P = kAΔT/L 0.5 × 19 × 20 / (7/100) = 2714.286

Further reading and review references

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Learning preview · independent educational and translation review required. Suggested grades are not official curriculum alignment.