Stephan's Boltzmann Law MCQ Quiz in বাংলা - Objective Question with Answer for Stephan's Boltzmann Law - বিনামূল্যে ডাউনলোড করুন [PDF]

CONCEPT:

• The greenhouse effect is the process by which radiations from the sun are absorbed by the greenhouse gases and not reflected back into space. This insulates the surface of the earth and prevents it from freezing.
• Due to the increased levels of greenhouse gases, the temperature of the earth has increased to a very high level due to various factors. This has led to several drastic effects.
• The main cause of this environmental issue is the increased volumes of greenhouse gases such as carbon dioxide and methane released by the burning of fossil fuels, emissions from the vehicles, industries and other human activities.

EXPLANATION:

• As the carbon dioxide is one of the main greenhouse gas. So option 4 is correct.

CONCEPT:

• Stefan's Law: According to it the radiant energy emitted by a perfectly black body per unit area per sec (i.e. emissive power of black body) is directly proportional to the fourth power of its absolute temperature i.e.

\(E \propto {T^4}\)

\( ⇒ E = σ {T^4}\)

Where, is a constant called Stefan’s constant.

• The value of Stefan's constant is 5.67 × 10-8 W/m2K4.

CALCULATION:

Given: σ = 5.67 × 10^-8 Watt/ m²K⁴ and T⁴ = 3.6J^-T

According to Stefan's law,

⇒ E = σ T⁴

⇒ E = 5.67 × 10-8 × 3.6 = 20.412 × 10-8 J/m2s

let us assume initial Temperature \(T_1=T\) then final will be

\(T_2=\dfrac{50}{100}T+T=\dfrac{150}{100}T=\dfrac{15}{10}T\)

energy radiated

\(e_1=\sigma A T_1^4\quad \left(1\right)\)

\(e_2=\sigma A T_2^4\quad \left(2\right)\)

\(\%\) increase in radiation

\(\dfrac{e_2-e_1}{e_1}\times 100\)

\(\dfrac{\sigma A T_2^4-\sigma A T_1^4}{\sigma A T_1^4}\times 100\)

\(\dfrac{\left(\dfrac{15}{10}T\right)^4-T^4}{T^4}\times 100\)

\(\dfrac{15^4-10^4}{10^4}\times 10^2\)

\(\dfrac{50625-10000}{10^2}\)

\(=406.25\)

\(\approx 400\%\)

Hence (d) option is correct.

\(P_1=A\sigma T_1^4=\pi r_1^2\sigma T_1^4\)

\(P_2=A\sigma T_2^4=\pi r_2^2\sigma T_2^4\)

\(P_1=P_2\)

so \(\dfrac{r_2}{r_1}=\dfrac{T_1^2}{T_2^2}\)

let us assume initial Temperature \(T_1=T\) then final will be

\(T_2=\dfrac{50}{100}T+T=\dfrac{150}{100}T=\dfrac{15}{10}T\)

energy radiated

\(e_1=\sigma A T_1^4\quad \left(1\right)\)

\(e_2=\sigma A T_2^4\quad \left(2\right)\)

\(\%\) increase in radiation

\(\dfrac{e_2-e_1}{e_1}\times 100\)

\(\dfrac{\sigma A T_2^4-\sigma A T_1^4}{\sigma A T_1^4}\times 100\)

\(\dfrac{\left(\dfrac{15}{10}T\right)^4-T^4}{T^4}\times 100\)

\(\dfrac{15^4-10^4}{10^4}\times 10^2\)

\(\dfrac{50625-10000}{10^2}\)

\(=406.25\)

\(\approx 400\%\)

Hence (d) option is correct.

Calculation:

In this question, we will equate the heat loss from the sphere to heat loss due to radiation.

\( mC \dfrac{dT}{dt} = \sigma A T^4 \)

\( \left( \dfrac{4}{3} \pi r^3 \right) (\rho) C \dfrac{dT}{dt} = \sigma (4 \pi r^2) T^4 \)

\( \dfrac{r \rho C}{3 \sigma} \int_{100}^{200} \dfrac{dT}{T^4} = \int_{0}^{t} dt \)

\( \dfrac{r \rho C}{9 \sigma \times 10^6} \left( 1 - \dfrac{1}{8} \right) = t \)

\( t = \dfrac{7 r \rho C}{72 \times 10^6 \sigma} \)

CONCEPT:

Stefan-Boltzmann's law:

• According to Stefan-Boltzmann's law, the amount of radiation emitted per unit of time from area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature.

⇒ Q = σbAT4

Where σb = Stefan's constant = 5.67×10-8 W/m2 K4, A = area and T = absolute temperature

EXPLANATION:

• From the above, it is clear that the value of the Stefan-Boltzman constant is 5.67×10-8 W/m2 K4. Hence, option 1 is correct.

The correct answer is option 2) i.e. 112.

CONCEPT:

• A black body is a system that absorbs all the electromagnetic radiation incident on it. To maintain thermal equilibrium, this system emits all the radiation at the same rate as it absorbs it. 
• Stefan Boltzmann Law: The total energy radiated per unit surface area of a blackbody across all wavelengths per unit time, is directly proportional to the fourth power of the black body’s thermodynamic temperature.

​It is given by the equation:

 \(E = σ T^4\)

Where E is the energy radiated per unit area, T is the temperature and σ is the Stefan-Boltzmann constant.

CALCULATION:

Given that:

The heat radiated at 227^∘C, E₁ = 7 cal / cm² / s

Initial temperature of the body, T₁ = 227^∘C ⇒ (227 + 273) K = 500 K

Final temperature of the body, T₂ =  727∘C ⇒ (727 + 273) K = 1000 K

From Stefan Boltzmann Law, \(E = σ T^4\)

\(E_1 = σ (T_1)^4\) and \(E_2 = σ (T_2)^4\)

\(\Rightarrow \frac{E_2}{E_1} = (\frac{T_2}{T_1})^4\)

\(\frac{E_2}{7} = (\frac{1000}{500})^4\)

∴ The rate of heat radiated at 727^∘C = E₂ = \(7 \times 2^4 =\) 112 cal / cm² / s

CONCEPT:

• Stefan's Law: According to it the radiant energy emitted by a perfectly black body per unit area per sec (i.e. emissive power of black body) is directly proportional to the fourth power of its absolute temperature i.e.

\(E \propto {T^4}\)

\( \Rightarrow E = \sigma {T^4}\)

Where, is a constant called Stefan’s constant.

• The value of Stefan's constant is 5.67 × 10-8 W/m2K4.

EXPLANATION:

• From the above, it is clear that the energy radiated by a black body is directly proportional to T4. Therefore option 3 is correct.

CONCEPT:

Stefan-Boltzmann law, states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature.

Mathematically, E = σT⁴

(Here, σ = Stefan's constant)

EXPLANATION:

Emissive power E is directly proportional to temperature T.

So, E∝T⁴

Given, T₁ = - 73°C = (-73 + 273) = 200 K

T₂ = 327℃ = (327 + 273) = 600 K

Thus, 

\( \begin{aligned} & \frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4 \\ & \frac{E_2}{E_1} = \left( \frac{600}{200} \right)^4 =3^4 = 81 \\ \end{aligned} \)

Hence the correct answer is option 4.