Size Reduction and Milling

Using Rittinger's law $E = K_R (1/L_2 - 1/L_1)$, if $K_R = 12\text{ kJ}\cdot\text{mm/kg}$, $L_1 = 4.0\text{ mm}$, and $L_2 = 1.0\text{ mm}$, the energy required $E = 12 \times (1/1.0 - 1/4.0) = 12 \times (1 - 0.25) = 12 \times 0.75$ in kJ/kg is: