Difficulty: Medium
Correct Answer: Xc > Xs
Explanation:
Introduction / Context:For brief extraction times from porous solids into a well-mixed liquid, the controlling factor is typically external mass transfer across the wetted external area, not deep internal diffusion. Thus, the surface-area-to-volume ratio (A/V) largely dictates how quickly a given body loses dissolved species initially. This question compares a long sealed-ended cylinder and a sphere of the same diameter.
Given Data / Assumptions:
Concept / Approach:At short times, the rate of salt removal is approximately proportional to external area A (and the mass transfer coefficient), while total salt content scales with volume V. The relevant metric is A/V. For a sphere of radius r, A/V = 3/r. For a long cylinder with sealed ends (radius r), only the curved surface transfers: A/V = 2/r. With equal r (2.5 cm), the sphere has larger A/V than the long cylinder, so it loses salt faster, leaving a smaller fraction remaining after the same time.
Step-by-Step Solution:
Sphere: A/V = 3/r = 3/2.5 cm^-1 = 1.2 cm^-1.Long sealed cylinder: A/V = 2/r = 2/2.5 cm^-1 = 0.8 cm^-1.Larger A/V → faster initial salt removal.Therefore: Xs < Xc, so Xc > Xs.Verification / Alternative check:Dimensional analysis of early-time leaching (or lumped mass-transfer models) shows m(t)/m0 ≈ exp(−kA t / V) at short times; greater A/V decreases m faster.
Why Other Options Are Wrong:
Common Pitfalls:Including end areas for the cylinder (they are sealed); assuming internal diffusion dominates at very short times; forgetting A/V scaling.
Final Answer:Xc > Xs
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