Question

(a) A high explosive material employing chemical reactions has an explosive yield of 10⁶ J/kg, measured in joules of energy released per kilogram of material. By what fraction does its mass change when it explodes? (b) What is the explosive yield of a material that produces energy via nuclear-reactions in which its mass decreases by 1 part in 10,000?

Short answer

In an explosive chemical reaction, 106 J/kg explosive yield is a result of the change in mass by $\mathbf{1}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{-3}\mathbf{kg}$. And decrease in mass by 1 part in 10000 results in an explosive yield of$\mathbf{9}\mathbf{\times }{\mathbf{10}}^5\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kg}$}\right.$

Step-by-step solution

Step 1: Determine the change in mass fraction

An explosive yield means the amount of energy released in the conversion of 1 kg of mass in an explosive chemical reaction.

(a) The relation of change in the mass fraction.

$\begin{array}{c}\frac{\Delta m}{m}=\frac{\Delta E}{m}\left(\frac{1}{c^2}\right)\\ =\frac{{10}^{6}J/kg}{{(3\times {10}^{8}m/s)}^2}\\ =1.11\times {10}^{-3}\end{array}$

Determine the explosive yield for  change in mass fraction

(b) If the mass change is ${10}^{-4}$kg, the explosive yield using the above equationwillbeas follows.

$\begin{array}{c}\frac{\Delta E}{m}=\frac{\Delta m}{m}{c}^2\\ ={10}^{-4}\times {(3\times {10}^8)}^2\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$kg$}\right.\\ =9\times {10}^5\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$kg$}\right.\end{array}$

Hence, mass change of${10}^{-4}$ kg will produce an explosive yield of $\mathbf{9}\mathbf{\times }{\mathbf{10}}^5\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kg}$}\right.$.