Assuming that the splitting of a `U^(235)` nucleus liberates of `U^(235)` isotope, and the mass of cal with calorific value of `30 kJ//g` which is equivalent to that the to that for one `kg` of `U^(235)`,
(b) the mass of `U^(235)` isotope split during the explosion of the atomic bomb with `30 kt` trotyl equivalent if the calorific value of troty is `4.1 k J//g`.

Explanation

(a) the energy liberated in the fission of `1kg of U^(235)` is
`(1000)/(235)xx6.023xx10^(23)xx200MeV=8.21xx10^(10)kJ`
The mass of coal with equivalent calorific value is
`(8.21xx10^(10))/(30000) kg= 2.74xx10^(6)kg`
(b) The required mass, is
`(30xx10^(9)xx4.1xx10^(3))/(200xx1.602xx10^(-13)xx6.023xx10^(23))xx(235)/(1000)kg =1.49kg`