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"].

According to Stefan's law of radiation, a black body radiates energy s

1.	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"].
Answer» Solution :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