A cubic box is being used as an insulator to store soup and keep it hot on a cold day. The wall thickness of the insulator is halved. The exposed area is decreased to a quarter. What would happen to the rate of heat loss?

Question

BlackTom AI · Accepted Answer

Fourier’s law of heat conduction gives the relevant relation: the steady rate of heat loss Q̇ through a wall is Q̇ = k A ΔT / d, so Q̇ ∝ A / d (k, ΔT assumed constant).

1) Relevant knowledge points and concepts
- Fourier’s law for conduction: Q̇ = (k A ΔT) / d.
- For fixed material (k) and fixed temperature difference (ΔT), heat loss scales with exposed area A and inversely with wall thickness d.
- Therefore any change in A or d changes Q̇ in proportion to A/d.

2) Complete reasoning process step by step
- Let the original area be A and original thickness be d. Original heat loss Q̇0 ∝ A / d.
- New area = A' = A/4 (exposed area decreased to a quarter).
- New thickness = d' = d/2 (thickness halved).

3) Calculations with intermediate steps
- New heat loss Q̇' ∝ A' / d' = (A/4) / (d/2).
- Compute (A/4) / (d/2) = (A/4) × (2/d) = (A × 2) / (4 d) = (A/d) × (2/4) = (A/d) × 1/2.
- Thus Q̇' = (1/2) Q̇0. The combined effect is a factor of 1/2.

4) Why the correct answer is correct
- Halving the thickness alone would double heat loss (factor 2), but reducing area to one quarter reduces heat loss by factor 1/4. Multiplying factors: 2 × (1/4) = 1/2. Net effect is half the original rate.

5) Why the other options are incorrect
- (a) Doubling: would require net factor 2, but actual net factor is 1/2, so incorrect.
- (c) Increase by factor of six: no combination of halving thickness and quartering area gives ×6; incorrect.
- (d) No change: clearly incorrect because A and d both changed and Q̇ depends on A/d.

6) Conclusion
- The rate of heat loss will be halved, which matches the provided correct answer (b).

A cubic box is being used as an insulator to store soup and keep it hot on a cold day. The wall thickness of the insulator is halved. The exposed area is decreased to a quarter. What would happen to the rate of heat loss?Single choice

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