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

Fourier’s law of heat conduction gives the heat-loss rate through a wall as Q̇ = k A ΔT / d, so thicker walls or smaller contact area reduce heat flow.

1. Relevant knowledge points and concepts
- Steady-state conduction: Q̇ = k A ΔT / d, where k is thermal conductivity (material fixed), A is contact area, d is wall thickness, ΔT is temperature difference.
- Thermal resistance viewpoint: R = d/(kA). Increasing R lowers Q̇ since Q̇ = ΔT / R.

2. Complete reasoning process step by step
- Let initial thickness = d and initial area = A. Initial heat loss: Q̇_initial = k A ΔT / d.
- New thickness = 4d (four times original). New area = A/2 (halved).
- Substitute into Fourier’s law for the new configuration.

3. Calculations with intermediate steps
- Q̇_new = k (A/2) ΔT / (4d)
- Simplify numerator and denominator: Q̇_new = (k A ΔT) / (2 * 4 d) = (k A ΔT) / (8 d)
- Compare with initial: Q̇_new = (1/8) * (k A ΔT / d) = (1/8) Q̇_initial

4. Why the correct answer is correct
- The combined effect of quadrupling thickness (factor 1/4 change in Q̇) and halving area (factor 1/2 change) multiplies to 1/8. Thus the rate of heat loss decreases by a factor of 8.

5. Why the other options are incorrect
- (a) Quadrupled: wrong sign and magnitude — increasing thickness reduces, not increases, heat loss.
- (c) Increase by a factor of 8: opposite effect; actual change is a reduction.
- (d) Doubled: incorrect magnitude and direction; heat loss does not increase.

6. Conclusion
- The rate of heat loss decreases by a factor of 8, which matches the provided correct answer (b).

类似问题

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