According to Stefan's law of radiation, a black body radiates energy `sigmaT^(4)` from its unit surface area every second where T is the surface temperature of the black body and `sigma=5.67xx10^(-8)W//m^(2)K^(4)` is known as Stefan's constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When detonated, it reaches temperature of `10^(6)` K and can be treated as a black body.
If surrounding has water at `30^(@)C`, how much water can `10%` of the energy produced evaporate in 1 s ?
`[S_(w)=4186.0" J/kg K and "L_(v)=22.6xx10^(5)" J/kg"]`.

Explanation

Energy obtained at each second,
`U=1.8xx10^(17)" J/s"`
Heat required to convert water into steam at each second,
`Q=10%` of U
`=1.8xx10^(17)xx(10)/(100)`
`=1.8xx10^(16)" J/s"`
Heat used in heating water of .m. mass from `30^(@)C` to `100^(@)C` and then to convert it into steam in each second,
`Q.="mS"_(w)Deltatheta+mL_(v)`
`=mxx4186xx(100-30)+mxx22.6xx10^(5)`
`Q.=2.93xx10^(5)m+22.6xx10^(5)m`
`Q.=25.53xx10^(5)m`
According to question,
`Q=Q.`
`1.8xx10^(16)=25.53xx10^(5)m`
`:.m=(1.8xx10^(16))/(25.53xx10^(5))`
`:.m=0.07050xx10^(11)`
`:.m~~7.0xx10^(9)` kg