Question

It is found that the energy required to reduce particle from a mean diameter of 10 mm to 5 mm is 1 kJ/kg. Using Rittinger's law, what is the energy requirement to reduce the same from a diameter of 1 mm to 0.5 mm?

• A. 5 kJ/kg
• B. 100 kJ/kg
• C. 10 kJ/kg
• D. 1 kJ/kg

Hint:

Rittinger's law depends on the difference of reciprocals of the diameters, not on the ratio of the diameters, so halving a smaller particle costs more energy than halving a bigger one.

Answer 1

The Correct Option is C

Rittinger's law says the energy spent on grinding goes into creating new surface area, and new surface area per unit mass scales with the difference between the reciprocals of final and initial diameter. In symbols, \(E = K_R(1/D_2 - 1/D_1)\).

Plug in the first situation to pin down the constant \(K_R\) for this material. Going from 10 mm down to 5 mm used 1 kJ/kg, so

\[1 = K_R\left(\frac{1}{5} - \frac{1}{10}\right) = K_R(0.1)\]

which gives \(K_R = 10\).

Now use that same constant for the second grind, from 1 mm down to 0.5 mm.

\[E = 10\left(\frac{1}{0.5} - \frac{1}{1}\right) = 10(2-1) = 10\]

Notice that going from 1 mm to 0.5 mm creates far more new surface per kilogram than going from 10 mm to 5 mm, even though both cases halve the diameter, because reciprocal differences grow sharply as particles get smaller. That is exactly why the energy jumps from 1 kJ/kg to 10 kJ/kg instead of staying the same.

\[\boxed{E = 10 \text{ kJ/kg}}\]