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 detonated, it reaches temperature of 106 K and can be treated as a black body.

1. Estimate the power it radiates.
2. 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_v = 22.6 × 10⁵ J/kg]
3. 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?

Answer 1

Given, σ = 5.67 × 10^–8 W/m² kg

Radius = R = 0.5 m, T = 10⁶ K

a. Power radiated by Stefan's law

P = σAT⁴ = (4πR²)T⁴

= 5.67 × 10^–4 × 4 × (3.14) × (0.5)² × (106)⁴

= 1.78 × 10¹⁷ J/s

= 1.8 × 10¹⁷ J/s

b. Energy available per second, U = 1.8 × 10¹⁷ J/s = 18 × 10¹⁶ J/s 

Actual energy required to evaporate water = 10% of 1.8 × 10¹⁷ J/s

= 1.8 × 10¹⁶ J/s

Energy used per second to raise the temperature of m kg of water 30°C to 100°C and then into vapour at 100°C

= ms_w Δθ + mL_v 

= m × 4186 × (100 – 30) + m × 22.6 × 10⁵

= 2.93 × 10⁵ m + 22.6 × 10⁵ 

m = 25.53 × 10⁵ m J/s

As per the question, 25.53 × 10⁵ m = 1.8 × 10¹⁶

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 = U/c`

= `(1.8 xx 10^17)/(3 xx 10^8)`

= `6 xx 10^8` kg-m/s²  ......`[(P = "momentum"),(V = "energy"),(C = "velocity of light")]`

Momentum per unit time per unit 

Area p = `P/(4piR^2)`

= `(6 xx 10^8)/(4 xx 3.14 xx (10^3)^2`

⇒ d = 47.7 N/m²  .......[4πR² = Surface area]