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According to Stefan’s 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 \times 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 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 30^{\circ}C, 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=\frac{U}{c}. How much momentum per unit time does it impart on unit area at a distance of 1 km?

Answers (1)

Given: E = \sigma T^{4} per sec per sq. m

Thus, total E is equal to radiated from all surface area A per second will be power radiated by nuclear weapon

Thus, P = \sigma A T^{4}

Now, \sigma = 5.67 \times 10^{-8} W/m^{2}K^{4} &

P = 5.67 \times 10^{-8} \times (4 \times \pi R^{2}) (10^{6})^{4}

            = 5.67 \times 4 \times 3.14 \times \times 0.5 \times 0.5 \times 10^{-8} \times 10^{24} …. (R = 0.5m, T = 106K)

            = 5.67 \times 3.14 \times 10^{24-8} \times 1

Thus, P \cong 18 \times 10^{16} Watt

              = 1.8 \times 10^{17}J/s                       …… (i)

(b) P = 18 \times 10^{16} Watt   from (i)

Now, for evaporation of water 10% of this power is required.

Thus, E = \frac{10}{100}\times 18 \times 10^{16} Watt

            = 1.8 \times 10^{17}J/s

Now, energy required by m kg of water at 30^{\circ}C to evaporate at 100^{\circ}C = E required to heat up water from 30 ^{\circ} C +100^{\circ}C+ E required to evaporate water into vapour

            = mS_{w} (T_{2} - T_{1}) + mL

           = m[S_{w}(T_{2} - T_{1}) + L]

Thus,

18 \times 10^{16 }= m [ 4180 (100-30) + 22.6 \times 10^{5}]

m = \frac{18 \times 10^{16}}{25.5 \times 10^{5}}       

    \cong 7 \times 10^{9}kg

(c) Now, we know that,

Momentum per unit time p^{'}=\frac{U}{c}

                                              = \frac{18 \times 10^{16}}{3 \times 10^{8}}

                                            

Thus,

p^{'}= 6 \times 10^{8} kg ms^{-1 }     

Required momentum per unit time

 =\frac{6 \times 10^{8}}{4 \times 3.14 \times (10^{3})^{2}}

\frac{6 \times 10^{8}}{4 \times 3.14 \times 10^{6} } = 47.77 \frac{kgms^{-1}}{m^{2} }

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