Ncert Exercises & Solutions

According to Stefan’s law of radiation, a black body radiates the energy $σ T^{4}$ from its unit surface area every second where T is the surface temperature of the black body and $σ = 5.67 \times 10^{- 8} W / m^{2} K^{4}$ 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 a temperature of $10^{6}$ K and can be treated as a black body.

(1) Estimate the power it radiates.

(2) If the surrounding has water at 30°C, how much water can 10% of the energy produced evaporate in 1 s?

$[s_{w} = 4186.0 J / k g K a n d L_{v} = 22.6 \times 10^{5} J / k g]$

(3) If all this energy U is in the form of radiation, the corresponding momentum is p=U/c. How much momentum per unit time does it impart on a unit area at a distance of 1km?

Hint: Use Stefan's law and the principle of calorimetry.

Step 1: Find the power radiated by the body.

Given,

σ = 5.67 \times 10^{- 8} W / m^{2} K^{4}

Radius,

= R = 0.5 m, T = 10^{6} K

(1) Power radiated,

P = σ A T^{4} = σ (4 π R^{2}) T^{4}

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

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

Step 2: Find the amount of water evaporated.

(2) Energy available per second,

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

Actual energy required to evaporate water =

10 % o f 1.8 \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 30 °C to 100 °C and then into vapour at 100°C

= m s_{w} ∆ θ + m L_{v} = 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 = 25.53 \times 10^{5} m J / s

As per the question,

25.53 \times 10^{5} m J / s = 1.8 \times 10^{16}

\Rightarrow m = \frac{1.8 \times 10^{16}}{25.33 \times 10^{5}} = 7.0 \times 10^{9} k g

Step 3: Find the momentum transferred by the energy per unit area.

(3) Momentum per unit time,

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

[P = m o m e n t u m]

U = e n e r g y

c = v e l o c i t y o f L i g h t

Momentum per unit time per unit area,

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

\Rightarrow p' = 47.7 N / m^{2}

[

4 {πR}^{2}

= Surface area]

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions