A cylindrical flask is being used as an insulator to store tea and keep it hot on a cold day. How will increasing the thickness by 4 times of all walls and halving the contact surface area (without changing the type of material the walls are made of) of the insulator affect the rate of heat loss?

Question

BlackTom AI · Accepted Answer

Heat conduction through a solid wall follows Fourier’s law: Q̇ = k A (ΔT) / d, where Q̇ is the heat-loss rate, k is the material’s thermal conductivity, A is contact area, ΔT is temperature difference, and d is wall thickness.

1) Relevant concepts
- Thermal conductivity k remains unchanged (same material).
- Thickness d appears in the denominator (heat loss ∝ 1/d).
- Contact area A appears in the numerator (heat loss ∝ A).

2) Step-by-step reasoning and calculation
Let the initial heat-loss rate be Q̇1 = k A (ΔT) / d.

Increase thickness by 4×: new thickness d' = 4d → this multiplies the thermal resistance by 4, so that factor alone reduces heat flow by 1/4.

Halve the contact area: new area A' = A/2 → this reduces heat flow by 1/2.

Combine both changes:
Q̇2 = k A' (ΔT) / d' = k (A/2) (ΔT) / (4 d) = [k A (ΔT) / d] × (1/2) × (1/4) = Q̇1 × (1/8).

Thus the new heat-loss rate is one-eighth of the original; equivalently, the rate decreases by a factor of 8.

3) Why the correct answer is correct
Because thickness increase and area decrease act multiplicatively on the Q̇ = kAΔT/d relation (1/4 from thickness × 1/2 from area = 1/8), the heat-loss rate is reduced to 1/8 of its original value.

4) Why the other options are incorrect
- b (doubled): wrong sign and magnitude; both changes reduce heat flow, not increase it.
- c (increase by factor 8): opposite of actual effect (changes decrease, not increase).
- d (quadrupled): incorrect magnitude and direction; the combined effect is a reduction by 8, not an increase by 4.

Conclusion: The rate of heat loss will decrease by a factor of 8, matching answer a.

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