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

Given, `sigma = 5.67 xx 10^(-8) W//m^(2) kg`
Radius, `= R = 0.5m, T = 10^(6) K`
(a) Power radiated by Stefan's law
`P = sigma AT^(4) = (4pi R^(2))T^(4)`
`= "("5.67 xx 10^(-4) xx 4 xx (3.14) xx (0.5)^(2) xx (10^(6)) 4`
`= 1.78 xx 10^(17) J//s =1.8 xx 10^(17) J//s`
(b) Energy available per second, `U = 1.8 xx 10^(17) J//s = 18 xx 10^(16) 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 + mL_(v) = m xx 4186 xx (100-30) + m xx 22.6 xx 10^(5)`
`= 2.93 xx 10^(5)m+ 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^(6))/(25.53 xx 10^(5)) = 7.0 xx 10^(9) kg`
(c) Momentum per unit time,
`p = (U)/(c)=(U)/(c) = (1.8 xx 10^(17))/(3 xx 10^(8)) = 6 xx 10^(8) kg-m//s^(2) "" [["P = momentum"],["V = energy"],["C = velocity of Light"]]`
Momentum per unit time per unit
area `p = (p)/(4pi R^(2))= (6 xx 10^(8))/(4 xx 3.14 xx (10^(3))^(2))`
`rArr d = 47.7 N//m^(2) " " [4pi R^(2) = "Surface area"]`