Question
A uniform wall conducts 120 W of heat at steady state. Its thickness is doubled while its material, area and face temperatures remain the same. What is the new heat-transfer rate?
- a.30 W
- b.60 W
- c.120 W
- d.240 W
Correct answer
B. 60 W
Full reasoning
- 1For a uniform slab, .
- 2Hold k, A and the temperature difference fixed. Only the denominator changes.
- 3Replacing L by 2L gives half the original rate: 120/2 = 60 W.
- 4This comparison assumes steady conduction and fixed face temperatures. A wall that is still warming up is a different problem.
Why each choice is right or wrong
Choice A
This would follow an inverse-square rule; slab conduction is inversely proportional to thickness.
Choice B
Correct. Thickness is in the denominator, so doubling it halves 120 W to 60 W.
Choice C
The thickness matters even when the material and boundary temperatures are unchanged.
Choice D
A thicker wall reduces the rate under these fixed boundary conditions.
Related formula
Steady slab conduction
P = k A ΔT / L
Assumptions: Uniform material, one-dimensional steady conduction and fixed boundary temperatures.
Related topic and practice
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Sources
Original authored exercise; no database question ID and no protected or official exam-bank content. Specialist review pending.
- University Physics Volume 2: Mechanisms of heat transfer — OpenStax. Accessed 2026-09-13. Textbook used as a reference for cited concepts, without reproducing textbook prose or illustrations. Exercise wording and numbers are original; no official exam questions are reproduced.
Sources and corrections
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