Ac cording to Stefan' law of radiation, a black body radiates energy `sigma T^4` from its unit surface area every second where T is the surface temperature of the black body and `sigma = 5.67 xx 10^(-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 detoneted, it reachs temperature of `10^6 K` and can be treated as a black body. (a) Estimate the power it radiates. (b) if surrounding has water at `30^@C` how much water can 10% of the energy produced evaporate in 1s ? ` [s_w = 4186.0 J//Kg K and L_(upsilon) = 22.6 xx 10^5 J//kg]` (c ) If all this energy U is in the form of radiation, corresponding momentum is ` p = U//c.` How much momentum per unit time does it impart on unit area at a distance of 1 km ?

Explanation

(a) Power radiated ` = (sigma T^4) xx A = sigma T^4 xx 4 pi R^2 = (5.67 xx 10^(-8)) xx (10^6)^4 xx 4 xx(3.14) xx (0.5)^2`
`= 1.78 xx 10^(17) Js^(-1) ~~ 1.8 xx 10^(17) Js^(-1)` (b) Energy available area second, `U = 1.8 xx 10^(17) J//s.` Actual energy required to evaporate water ` =10% of 1.8 xx 10^(17) J//s = 1.8 xx 10^(16) J//s.` Energy used per second to raise the temperature of m kg of water from `30^@C` to `100^@C` and then into vapour at `100^@C`
` =ms_w Delta theta + m L_(upsilon) = m xx 4186 xx (100 - 30) + m xx 22.6 xx 10^5`
`=2.93 xx 10^5m + 22.6 xx 10^5 m = 25.53 xx 10^5 m J//s`
As per question, `25.53 xx 10^5 m = 1.8 xx 10^(16)`
`or m = (1.8 xx 10^(16))/(25.33 xx 10^5) =7.0 xx 10^9 kg`
(c ) momentum per unit time, `P = (U)/(c ) = (1.8 xx 10^(17))/(3 xx 10^8) = 6xx10^8 kg ms^(-2)` Momentum per unit time per unit area ` P= (p)/(4 pi R^2) = (6xx10^8)/(4xx3.14 xx(10^3)^2) = 47.7 N//m^2`