Tag: stefan's law

Questions Related to stefan's law

Find the radiation pressure of solar radiation on the surface of earth. Solar constant is $1.4kW{{m}^{-2}}$

stefan's law black body radiation heat transfer thermal properties physics

  1. $4.7\times { 10 }^{ -5 }Pa$

  2. $4.7\times { 10 }^{ -6 }Pa$

  3. $2.37\times { 10 }^{ -6 }Pa$

  4. $9.4\times { 10 }^{ -6 }Pa$

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Correct Option: B
Explanation:
METHOD-1:
${Pressure} _{absorbed}=\dfrac{E _{f}}{c}$
${E} _{f}=$ energy flux, $C=$ Speed of light

${ Pressure } _{ absorbed }=\dfrac { 1.4\times 1000 }{ 3\times { 10 }^{ 8 } } =4.66\times { 10 }^{ -6 }Pa$

METHOD-2: (checking the unit,if formula is not remembered)

We know that:
${Power}={force}\times{velocity}$-----(1)
${Pressure}=\dfrac{force}{area}$------(2)

${Force}=\dfrac{power}{velocity}=\dfrac { 1.4\times 1000 }{ 3\times { 10 }^{ 8 } } =4.66\times { 10 }^{ -6 } newton$

Put in equation (2)
${ Pressure } _{ absorbed }=\dfrac { 4.66\times { 10 }^{ -6 }N }{ { m }^{ 2 } } =4.66\times { 10 }^{ -6 }Pa$

The temperature of a black body corresponding to which it will emit energy at the rate of $1 watt/cm^2$ will be

stefan's law black body radiation heat transfer thermal properties physics

  1. 650K

  2. 450K

  3. 350K

  4. 250K

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Correct Option: A
Explanation:

$E\propto { T }^{ 4 }\quad \Longrightarrow \quad E=\sigma { T }^{ 4 }$
$\sigma =5.67*{ 10 }^{ 8 }W{ m }^{ 2 }{ k }^{ 4 }$
$1*{ 10 }^{ -4 }=5.67*{ 10 }^{ 8 }*{ T }^{ 4 }$
$T=648k\cong 650k$



The solar constant for the earth is $\Sigma$. The surface temperature of the sun is $T$ K. The sun subtends an angle $\theta$ at the earth

stefan's law black body radiation heat transfer thermal properties physics

  1. $\Sigma \space \propto \space T^4$

  2. $\Sigma \space \propto \space T^2$

  3. $\Sigma \space \propto \space \theta^4$

  4. $\Sigma \space \propto \space \theta$

Show answer
Correct Option: A
Explanation:

By Stephan Boltzman law,


$P=\sigma (4\pi { R }^{ 2 }){ T }^{ 4 }$

$\theta =\dfrac { 2r }{ R } $ where R is distance between earth and sun and r is radius of earth.

Hence, $\sum {  } =\dfrac { P }{ 4\pi { r }^{ 2 } } =C{ T }^{ 4 }{ \left( \dfrac { R }{ r }  \right)  }^{ 2 }=K{ T }^{ 4 }{ \theta  }^{ 2 }$

Hence $\sum {  } \alpha { T }^{ 4 }$ and $\sum {  } \alpha { \theta }^{ 2 }$

Answer is option A.

In the Orion stellar system the shining of a star is $17\space \times 10^3$ times that of the sun. If the temperature of the surface of the sun $6 \times 10^3 K$ then the temperature of this star will be

stefan's law black body radiation heat transfer thermal properties physics

  1. 273 K

  2. 652 K

  3. 6520 K

  4. 68520 K

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Correct Option: D
Explanation:

The power radiated is directly proportional to the fourth power of the absolute temperature.
$P \propto T^{4}$
$P = kT^{4}$ 

Let P = Power radiated by the sun
Power radiated by star $ = 17 \times  10^{3}P $
$\dfrac{T _{star}^{4}}{T _{sun}^{4}} = 17 \times  10^{3}$

$\dfrac{T _{star}}{T _{sun}} = 11.418$

$T _{star} = 6000 \times  11.418 = 68,511K$

Hence, the answer is option D.

There are two planets $A$ and $B$ at a large distance Planet $A$ is bigger and hotter than planet $B$. The angular diameter of planet $A$ is $40$ minute of arc as seen from planet $B$. The energy received by planet $B$ is $3cal-cm^{-2}$ per minute. Assuming the radiation to be black body in character. Given that stefan costant is $5.67\times 10^{-8}\ Wm^{-2}\ K^{-4}$. The temperature of planet $A$ is

stefan's law black body radiation heat transfer thermal properties physics

  1. $(10.93\times 10^{14})^{1/4}\ K$

  2. $(53.21\times 10^{14})^{1/4}\ K$

  3. $(63.63\times 10^{14})^{1/4}\ K$

  4. $(63.21\times 10^{14})^{1/4}\ K$

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Correct Option: C

A blackened steel plate is put in a dark room after being heated up to a high temperature. A white spot on the plate appears. 

stefan's law black body radiation heat transfer thermal properties physics

  1. brighter than the plate

  2. as bright as the plate

  3. dull as compared to the plate

  4. appears to be yellow

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Correct Option: A

Boltzmann's constant$ K = 1.38 \times 10^{-23} J/k $ The energy associated with helium atom the surface of sun, where surface temperature is 6000 K is

stefan's law black body radiation heat transfer thermal properties physics

  1. $ 1.242 \times 10^{-19} J $

  2. $ 2.484 \times 10^{-19} J $

  3. $ 207 \times 10^{-19} J $

  4. $ 0.621 \times 10^{-19} J $

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Correct Option: A

If in an ideal gas $r$ is radius of molecule, $P$ is pressure, $T$ is absolute temperature and $k$ is Boltzmann's constant, then mean free path $\overline { \lambda  } $ of gas molecules is given as

stefan's law black body radiation heat transfer thermal properties physics

  1. $\dfrac { 4\pi \sqrt { 2 } PT }{ k{ r }^{ 2 } } $

  2. $\dfrac { 4\pi \sqrt { 2 } kT }{ P{ r }^{ 2 } } $

  3. $\dfrac { kP }{ 4\pi \sqrt { 2 } { r }^{ 2 }T } $

  4. $\dfrac { kT }{ 4\pi \sqrt { 2 } { r }^{ 2 }P } $

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Correct Option: D

Solar constant for earth is $2 \mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 } ,$ if distance ofmerary from sun is 0.4 times than distance of earthfrom sun then solar constant for mercury will be? 

stefan's law black body radiation heat transfer thermal properties physics

  1. 12.5$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

  2. 25$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

  3. 0.32$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

  4. 2$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

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Correct Option: C

The solar energy incident on the roof in 1 hour of dimension $ 8m \times 20m$ will be

stefan's law black body radiation heat transfer thermal properties physics

  1. $5.76\times { 10 }^{ 8 }J$

  2. $5.76\times { 10 }^{ 7 }J$

  3. $5.76\times { 10 }^{ 6 }J$

  4. $5.76\times { 10 }^{ 5 }J$

Show answer
Correct Option: A
Explanation:

Here the power per unit area is given, $I=10^3  W/m^2$

So, the total power $=I\times $ area of roof $=10^3\times (8\times 20)=1.6\times 10^5 W$ 
Since power is the energy divided by time so, energy, $E=Pt=1.6\times 10^5 \times (3600)=5.76\times 10^8  J$