Answer
See answers.
Work Step by Step
Find the energy released.
$$(20\;kilotons\;TNT)(\frac{5\times10^{12}J}{kiloton})(\frac{MeV}{1.60\times10^{-13}J})$$
$$=6.25\times10^{26}MeV$$
Change this to an amount of uranium. Assume that the energy produced by a single fission is 200 MeV (stated in problems 18-20).
$$(6.25\times10^{26}MeV) \frac{1fission}{200\;MeV}\frac{235g}{6.02\times10^{23}atoms}= 1219.9g\approx 1\;kg$$
b. Calculate the actual mass transformed into energy.
$$E=mc^2$$
$$m=\frac{E}{c^2}=\frac{(20\;kilotons\;TNT)(\frac{5\times10^{12}J}{kiloton})}{(3.0\times10^8 m/s)^2}=1.11\times10^{-3}kg\approx 1\;g$$
