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NCERT Exemplar · Q27

Q.According to Stefan's law of radiation, a black body radiates energy σT4\sigma T^4 from its unit surface area every second where TT is the surface temperature of the black body and σ=5.67×10−8\sigma = 5.67 \times 10^{-8} W/m2^2K4^4 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 10610^6 K and can be treated as a black body.

(a) Estimate the power it radiates.
(b) If surrounding has water at 30∘30^\circC, how much water can 10% of the energy produced evaporate in 1 s? [Sw=4186.0 J/kg K and Lv=22.6×105 J/kg]\left[ S_w = 4186.0 \text{ J/kg K and } L_v = 22.6 \times 10^{5} \text{ J/kg} \right]
(c) If all this energy UU is in the form of radiation, corresponding momentum is p=U/cp = U/c. How much momentum per unit time does it impart on unit area at a distance of 1 km?
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Stefan's law gives the fireball's radiated power as P=σAT4P=\sigma A T^4.

  1. P≈1.78×1017P\approx1.78\times10^{17} W.
  2. Using 10% of one second's energy to first heat then vaporise water gives a mass evaporated of m≈6.97×109m\approx6.97\times10^{9} kg.
  3. Spreading this power over a sphere of radius 1 km and dividing by cc gives a radiation pressure of about 47.2 Pa.

(a) Power radiated

The fireball (radius r=0.5r=0.5 m) is treated as a black body, so its whole surface radiates at rate σT4\sigma T^4 per unit area. Surface area:

A=4πr2=4π(0.5)2=π m2.A = 4\pi r^2 = 4\pi(0.5)^2 = \pi\ \text{m}^2.

At T=106T=10^6 K:

P=σAT4=(5.67×10−8)(π)(106)4=5.67π×1016 W≈1.78×1017 W.P = \sigma A T^4 = (5.67\times10^{-8})(\pi)(10^6)^4 = 5.67\pi\times10^{16}\ \text{W} \approx1.78\times10^{17}\ \text{W}.

(b) Water evaporated using 10% of the energy in 1 s

Energy radiated in 1 s is U=P×1 s≈1.78×1017U=P\times1\,\text{s}\approx1.78\times10^{17} J, so 10%10\% of it is

Uavail=0.10×1.78×1017=1.78×1016 J.U_{\text{avail}} = 0.10\times1.78\times10^{17} = 1.78\times10^{16}\ \text{J}.

To evaporate water starting at 30∘30^\circC, it must first be heated to 100∘100^\circC (ΔT=70\Delta T=70 K) and then vaporised at 100∘100^\circC. Per kilogram:

Qper kg=SwΔT+Lv=(4186)(70)+22.6×105=2.93×105+22.6×105=25.53×105 J/kg.Q_{\text{per kg}} = S_w\Delta T + L_v = (4186)(70) + 22.6\times10^5 = 2.93\times10^5 + 22.6\times10^5 = 25.53\times10^5\ \text{J/kg}.

Mass evaporated:

m=UavailQper kg=1.78×10162.553×106≈6.97×109 kg.m = \frac{U_{\text{avail}}}{Q_{\text{per kg}}} = \frac{1.78\times10^{16}}{2.553\times10^{6}} \approx6.97\times10^{9}\ \text{kg}.

(c) Momentum per unit time per unit area at 1 km

Radiation carries momentum p=U/cp=U/c, so momentum per unit time (force) is F=P/cF=P/c. Spread over a sphere of radius R=1000R=1000 m, the intensity is

I=P4πR2,I = \frac{P}{4\pi R^2},

and the momentum delivered per unit time per unit area (radiation pressure on an absorbing surface) is …

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