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प्रश्न
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.
- Estimate the power it radiates.
- If surrounding has water at 30°C, how much water can 10% of the energy produced evaporate in 1s? [Sw = 4186.0 J/kg K and Lv = 22.6 × 105 J/kg]
- 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?
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उत्तर
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
= msw Δθ + mLv
= 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]
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