# According to Stefan’s law of radiation,

Question:

According to Stefan’s law of radiation, a black body radiates energy σT4 from its unit surface area every second where T is the surface temperature of the black body and σ = 5.67 × 10-8 W/m2K4 is known as Stefan’s constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When denoted, it reaches temperature of 106K and can be treated as a black body.

(a) estimate the power it radiates

(b) if surrounding has water at 30oC, how much water can 10% of the energy produced evaporate in 1 sec?

(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?

Solution:

(a) E = σT4

Total energy = energy radiated from surface area A per second

P = σAT4

σ = 5.67 × 10-8 W/m2K4

R = 0.5 m

T = 106 K

Substituting the values we get,

P = 1.8 × 1017 J/s

(b) P = 18 × 1016 Watt

10% of this energy is used in evaporation of water

E = (10/100)(18)(1016) Watt = 1.8 × 1016 J/s

Energy required to evaporate water in 1 sec using 10% of energy is

$m S_{w}\left(T_{2}-T_{1}\right)+m L=m\left(S_{w}\left(T_{2}-T_{1}\right)+L\right)$

m = 7 109 kg

(c) Momentum per unit time p’ = U/C = 0.6 × 109

P’ = 6 × 108 kg.m/s2

P per unit time at a distance 1 km is

$\frac{6 \times 10^{8}}{4 \pi R^{2}}$

Substituting the values we get P per sec at 1 km on m2 = 47.8 N/m2