Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

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

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

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question