According to Stefan’s law of radiation, a black body radiates energy σT 4 from its unit surface area every second where T is the surface temperature of the black body and σ = 5.67 × 10^-8 W/m²K⁴ 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 106K and can be treated as a black body.

(a) Estimate the power it radiates.

(b) If surrounding has water at 30 C° , how much water can 10% of the energy produced evaporate in 1 s? S_w 4186.0 J/kg K and L_v 22.6 $\times 10^{5} J / k g$

(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?

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

(a) Power radiated by stefan’s law

P= $σ A T^{4} = ($ 4 $π$ R²)T⁴

= 5.67 $\times 10^{- 4} \times 4 \times 3.14 \times {0.5}^{2} \times 10^{6} \times 4$

= 1.78 $\times 10^{17} J / s$ =1.8 $\times 10^{17} J / s$

(b) Energy available per second U= 1.8 $\times 10^{17} J / s$ = 18 $\times 10$ ¹⁶J/s

Actual energy required to evaporate water = 10%of 18 $\times 10^{17} j / s$

=1.8 $\times 10^{16}$ J/s

Energy used per second to raise the temperature of m kg of water from 30⁰C to 100⁰C and then into vapour at 100⁰C

=ms_{w
$? θ$}+mL= m $\times 4186 \times (100 - 30) + m \times 22.6 \times 10^{5}$

= 2.93 $\times 10^{5} m + 22.6 \times 10^{5} m = 22.53 \times 10^{5} m J / s$

As per question , 25.53 $\times 10^{5} m$ = 1.8 $\times 10^{16}$

m= $\frac{1.8 \times 10^{16}}{25.53 \times 10^{5}} = 7 \times 10^{9} k g$

(c) Momentum per unit time

P= U/c= $\frac{1.8 \times 10^{17}}{3 \times 10^{8}} = 6 \times 10^{8} k g - m / s$ ²

Momentum per unit time

Area p= p/4 $π R^{2} = \frac{6 \times 10^{8}}{4 \times 3.14 \times {(10^{3})}^{2}}$

d=47.7N/m²

This is a long answer type question as classified in NCERT Exemplar(a) Power radiated by stefan’s lawP= <math> <mi>σ</mi> <mi>A</mi> <msup> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> <mo>=</mo> <mo>(</mo> </math> 4 <math> <mi>π</mi> </math> R2)T4= 5.67 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>4</mn> <mo>×</mo> <mn>3.14</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>0.5</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>4</mn> </math> = 1.78 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>17</mn> </mrow> </mrow> </msup> <mi>J</mi> <mo>/</mo> <mi>s</mi> </math> =1.8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>17</mn> </mrow> </mrow> </msup> <mi>J</mi> <mo>/</mo> <mi>s</mi> </math> (b) Energy available per second U= 1.8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>17</mn> </mrow> </mrow> </msup> <mi>J</mi> <mo>/</mo> <mi>s</mi> </math> = 18 <math> <mo>×</mo> <mn>10</mn> </math> 16J/sActual energy required to evaporate water = 10%of 18 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>17</mn> </mrow> </mrow> </msup> <mi>j</mi> <mo>/</mo> <mi>s</mi> </math> =1.8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> </math> J/sEnergy used per second to raise the temperature of m kg of water from 300C to 1000C and then into vapour at 1000C=msw <math> <mo>?</mo> <mi>θ</mi> </math> +mL= m <math> <mo>×</mo> <mn>4186</mn> <mo>×</mo> <mfenced separators="|"> <mrow> <mrow> <mn>100</mn> <mo>-</mo> <mn>30</mn> </mrow> </mrow> </mfenced> <mo>+</mo> <mi>m</mi> <mo>×</mo> <mn>22.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </math> = 2.93 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> <mi>m</mi> <mo>+</mo> <mn>22.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> <mi>m</mi> <mo>=</mo> <mn>22.53</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> <mi>m</mi> <mi>J</mi> <mo>/</mo> <mi>s</mi> </math> As per question , 25.53 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> = 1.8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> </math> m= <math> <mfrac> <mrow> <mrow> <mn>1.8</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>25.53</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>7</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> <mi>k</mi> <mi>g</mi> </math> (c) Momentum per unit timeP= U/c= <math> <mfrac> <mrow> <mrow> <mn>1.8</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>17</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </msup> <mi>k</mi> <mi>g</mi> <mo>-</mo> <mi>m</mi> <mo>/</mo> <mi>s</mi> </math> 2Momentum per unit timeArea p= p/4 <math> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mo>×</mo> <mn>3.14</mn> <mo>×</mo> <msup> <mrow> <mrow> <mo>(</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> d=47.7N/m2

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