Question

12.4 What is the most important reason that a unit mass of a smaller log burns more rapidly than a unit mass of a larger log?

Answer 1

We can consider an example:-

Consider a small cylindrical log of (say) wood, having radius (r=5) & height (h=7 units), and a large cylindrical log of radius (R=10) & height (H=14 units) both having equal mass density (D = 10 units).

Then surface area of smaller log = 2×π×r×h = 2×(22/7)×5×7=220 sq.units.

Mass of smaller log = density × volume = D × π×r2×h = 10×(22/7)×52×7 = 5500 units mass.

So, surface area per unit mass = 220/5500 = 0.04.

And, surface area of larger log = 2×π×R×H = 2×22/7×10×14=880 sq.units,

Mass of larger log = density × Volume = D × π×R2×H = 10×(22/7)×102×14 = 44000 unit mass.

So, surface area per unit mass = 880/44000 = 0.02.

Clearly surface area per unit mass of smaller log (0.04) is greater than surface area per unit mass of larger log (0.02). Hence, it is clear that smaller log will burn more rapidly.

Here is the reason for your problem with example. Surface area per unit mass is the main reason for slow or rapid burning .

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