Question

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 0.5m. When detonated, it reaches a temperature of ${{10}^{6}}K$ and can be treated as a black body.
(a) Estimate the power it radiates.
(b) If the surrounding has water at $30{}^\circ C$, how much water can 10% of the energy produced evaporate in 1s? $\left[ {{S}_{w}}=4186.0J/kgK,{{L}_{v}}=22.6\times {{10}^{5}}J/kg \right]$
(c) If all this energy U is in the form of radiation, corresponding momentum is
$P=\dfrac{U}{c}$. How much momentum per unit time does it impact on the unit area at a distance of 1km?

Answer 1

As a very first step, one could read the question well and note down the given values. Though the question is lengthy, it is pretty simple to solve and we could approach each sub question one by one. Recall the formulas accordingly and then carryout substitutions and hence find the answers.
Formula used:
Power,
$P=\sigma A{{T}^{4}}$
Energy,
$E'=m{{s}_{w}}\Delta \theta +m{{L}_{v}}$
Momentum,
$P=\dfrac{E}{c}$

Complete step-by-step solution:
(a) Let us recall the expression for power which is given by,
$P=\sigma A{{T}^{4}}$
$\Rightarrow P=\sigma \left( 4\pi {{R}^{2}} \right){{T}^{4}}=5.67\times {{10}^{-8}}\times 4\times 3.14\times {{\left( 0.5 \right)}^{2}}\times {{\left( {{10}^{6}} \right)}^{4}}$
$\therefore P\approx 1.8\times {{10}^{17}}J/s$
Therefore, we found the power radiated to be $P\approx 1.8\times {{10}^{17}}J/s$.
(b) So, we have the energy available per second from the (a) part as,
$E=1.8\times {{10}^{17}}=18\times {{10}^{16}}J$ per second
Now the energy required to evaporate water is 10% of E which would be,
$\dfrac{10}{100}\times 18\times {{10}^{16}}J=1.8\times {{10}^{16}}J$
Now the energy that is used for raising temperature from $30{}^\circ C\to 100{}^\circ C$ and then to vaporize would be,
$E'=m{{s}_{w}}\Delta \theta +m{{L}_{v}}=m\times 4186\times \left( 100-30 \right)+m\times 22.6\times {{10}^{5}}$
$\Rightarrow E'=m\left( 25.53\times {{10}^{5}}J/s \right)$
Now from question, we have, E’=E,
$\Rightarrow m\left( 25.53\times {{10}^{5}} \right)=1.8\times {{10}^{16}}$
$\therefore m=7\times {{10}^{9}}kg$
Therefore, we found the amount of water required to be $m=7\times {{10}^{9}}kg$.
(c) The momentum per unit time could be given by,
$P=\dfrac{E}{c}=\dfrac{1.8\times {{10}^{17}}}{3\times {{10}^{8}}}=6\times {{10}^{8}}kgm/{{s}^{2}}$
Now the momentum per unit time per unit area would be,
$P'=\dfrac{P}{4\pi {{R}^{2}}}=\dfrac{6\times {{10}^{8}}}{4\times 3.14\times {{\left( {{10}^{3}} \right)}^{2}}}$
$\therefore P'=47.7N/{{m}^{2}}$
Therefore, we found the momentum per unit time imparted per unit area to be$P'=47.7N/{{m}^{2}}$.

Note: All the substituted standard values for specific heat of water and latent heat of vaporization are all given in the question. But, to be on the safe side, one could memorize these values for solving the problems where these values won't be given.