question_answer 49)

                  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 detonated, it reaches temperature of \[{{10}^{6}}\,K\] and can be treated as a black body.                 (a) Estimate the power it radiates.                 (b) If surrounding has water at \[{{30}^{o}}C\], how much water can 10% of the energy produced evaporate in 1 s?                 [\[{{S}_{\omega }}\,=4186.0J/kg\,\,K\]and \[{{L}_{\upsilon }}\,=22.6\times \,{{10}^{5}}\,J/kg\]]                 (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?

Answer:

                  (a) Power radiated\[=A\,\sigma \,{{T}^{4}}\]                 \[=\,4\pi \,{{R}^{2}}\,\sigma \,{{T}^{4}}\]                 \[=4\,\times \,3.14\,\times {{(0.5)}^{2}}\,\times 5.67\,\times {{10}^{-8}}\,\times {{10}^{24}}\]                 \[=17.8\,\times \,{{10}^{16}}\,J\,{{s}^{-1}}\]                 \[\simeq 1.8\,\times \,{{10}^{17}}\,J\,{{s}^{-1}}\]                 (b) Available energy \[=1.8\,\times \,{{10}^{17}}\,J\]                 Energy used to evaporate water \[=10%\] of \[1.8\times {{10}^{17}}\,J\]\[=1.8\times {{10}^{16}}\,J\]                 Energy used to increase the temperature of water from \[{{30}^{o}}C\] to \[{{100}^{o}}C=m\,{{S}_{\omega }}\,DT\]                 \[=m\times 4186\times 70\]                 \[=\left( 293020\text{ }m \right)\text{ }J=\left( 2.93\times {{10}^{5}}m \right)J\]                 Energy used to convert \[m\,\,kg\]  of water into vapours                 \[=m\,{{L}_{u}}=\,(22.6\times \,{{10}^{5}}m)J\]                 \[\therefore \] \[2.93\,\,\times {{10}^{5}}m\,\,+\,22.6\,\,\times \,{{10}^{5}}m\]                 \[=1.8\times {{10}^{16}}\]                 or \[m=\,\frac{1.8\times {{10}^{16}}}{25.53\times {{10}^{5}}}=\,7.0\,\times \,{{10}^{10}}\,kg\]                 (c) \[p=\,\frac{U}{c}=\,\frac{1.8\times \,{{10}^{17}}}{3\times \,{{10}^{8}}}\,=6\,\times {{10}^{8}}\,kg\,\,m{{s}^{-1}}\]                 \[\therefore \] Momentum per unit time per unit area                 (i.e. P) \[=\,\frac{p}{4\pi {{R}^{2}}}\,=\frac{6\times \,{{10}^{8}}}{4\times \,3.14\times {{10}^{6}}}\]                 \[=47.77\,N{{m}^{-2}}\].