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Thermal Properties of Matter physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

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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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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alok kumar singh

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Heat Released by block = Heat gain by large Ice block

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(Given a = 5.0 × 10^-4 ⁰C^-1).

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A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

(a) The rate of cooling is constant till milk attains the temperature of the surrounding

(b) The temperature of milk falls off exponentially with time

(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools

(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c), (d) When the hot milk in the table is transferred to the surroundings by conduction, convection and radiation.

According to newton's law of cooling temperature of the milk falls of exponentially. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to surroundings. So option b, c, d satisfy.

Refer to the plot of temperature versus time (Fig) showing the changes in the state of ice on heating (not to scale) Which of the following is correct?

(a) The region AB represents ice and water in thermal equilibrium

(b) At B water starts boiling

(c) At C all the water gets converted into steam

(d) C to D represents water and steam in equilibrium at boiling point

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) During the process AB temperature of the system is 0⁰C . hence it represents phase change that is transformation of ice into water while temperature remains 0⁰C.

BC represents rise in temperature of water from 0⁰C to 100⁰C. now water starts converting into steam which is represented by CD.

Which is only represents by a, d options

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