Tag: stefan's law

Questions Related to stefan's law

The dimensions of Stefan's constant are

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

  1. $\left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -3 }{ K }^{ -4 } \right] $

  2. $\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -3 }{ K }^{ -3 } \right] $

  3. $\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -3 }{ K }^{ -4 } \right] $

  4. $\left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -3 }{ K }^{ -4 } \right] $

Show answer
Correct Option: D
Explanation:

Power radiated by a body $P = \sigma AeT^4$

where $\sigma$ is the Stefan's constant, $e$ is the emmissivity of the body, $A$ is the surface area of the body and $T$ is its temperature.
Dimensions of power $[P] = [ML^2T^{-3}]$
Dimensions of area $[A] = [L^2]$
Dimensions of temperature $[t] = [K]$
Emmissivity $e$ is a dimensionless quantity.
$\therefore$ Dimensions of Stefan's constant $[\sigma] = \dfrac{[ML^2T^{-3}]}{[L^2] [K^4]}$
$\implies$ $[\sigma] = [M^1 L^0 T^{-3} K^{-4}]$

A black body is heated from $27^oC  $ to $927^oC  $. The ratio of radiation emitted will be:

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

  1. $1 : 4$

  2. $1 : 8$

  3. $1 : 16$

  4. $1 : 256$

Show answer
Correct Option: D
Explanation:

Energy radiated depends on the temperature of the body.
Stefan's law states that the total amount of energy radiated per second per unit area of a perfect black body is directly proportional to the fourth power of the absolute temperature of the surface of the body,ie,
$E\propto { T }^{ 4 }$
or $E=\sigma { T }^{ 4 }$
where $\sigma $ is Stefan's constant. It's value is $5.67\times { 10 }^{ -8 }W{ m }^{ -2 }{ K }^{ -4 }$
Here, ${ T } _{ 1 }=27+273=300K$
          ${ T } _{ 2 }=927+273=1200K$
$\therefore       \dfrac { { E } _{ 1 } }{ { E } _{ 2 } } ={ \left( \dfrac { 300 }{ 1200 }  \right)  }^{ 4 } = 1 : 256$

Two bodies A and B of equal surface area have thermal emissivities of $0.01$ and $0.81$ respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies A and B at wavelengths $\lambda _A$, and $\lambda _B$ respectively. Difference in these two wavelengths is 1 $\mu$. If the temperature of the body A is $5802\  K$, then value of $\lambda _B$ is :

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

  1. $\dfrac{3}{2}\mu m$

  2. $1\mu m$

  3. $2 \mu m$

  4. $\dfrac{3}{4} \mu m$

Show answer
Correct Option: A
Explanation:

We know that as per stephan's boltzman radiation law, $P\ \alpha\  \sigma A{ T }^{ 4 }$.
Since surface area of two bodies is same,

Therefore, ${ \sigma  } _{ A }{ T } _{ A }^{ 4 }={ \sigma  } _{ B }{ T } _{ B }^{ 4 }$. Hence, ${ T } _{ A }=3{ T } _{ B }$.

Now, as per Wein's displacement law, $\lambda T=constant=k$, $\lambda =\dfrac { k }{ T } $.

${ \lambda  } _{ B }-{ \lambda  } _{ A }=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ { T } _{ A } } )=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ 3{ T } _{ B } } )=\dfrac { 2k }{ 3{ T } _{ B } } =1\mu $


${ T } _{ B }=\dfrac { { T } _{ A } }{ 3 } =1934\ K$

Putting in the above equation to calculate $k$ and the solving for ${ \lambda  } _{ B }=\dfrac { k }{ { T } _{ B } } $, we get ${ \lambda  } _{ B }=1.5\mu m$.


A black body at a high temperature $T$ radiates energy at the rate of $U\left( in\quad W/{ m }^{ 2 } \right) $. When the temperature falls to half (i.e $T/2$), the radiated energy $\left( in\quad W/{ m }^{ 2 } \right) $ will be

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

  1. $U/8$

  2. $U/16$

  3. $U/4$

  4. $U/2$

Show answer
Correct Option: B
Explanation:

According to Stefan's law, rate of energy radiated by a black per unit area (inW/m2)(inW/m2) at temperature TT is given by
U=σT4...(i)U=σT4...(i)
when the temperature falls to half (i.e., T/2T/2) 
radiated energy (inW/m2)...(ii)(inW/m2)...(ii)
From (i) and (ii) we get
U′U=(12)4=116U′U=(12)4=116
or U′=U16

 If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is 0? (a stands for Stefan's constant.)

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

  1. $

    \left(\frac{4 \pi R^{2} Q}{\sigma}\right)^{1 / 4}

    $

  2. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{1 / 4}

    $

  3. $

    \frac{Q}{4 \pi R^{2} \sigma}

    $

  4. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{-1 / 2}

    $

Show answer
Correct Option: C

$\dfrac {watt} {kelvin}$ is the unit of 

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

  1. Stefan's constant

  2. Wien's constant

  3. Cooling's constant

  4. Thermal constant

Show answer
Correct Option: A

Assuming the Sun to be a spherical body of radius $R$ at a temperature of $T\ K$. Evaluate the intensity of radiant power, incident on Earth, at a distance $r$ from the Sun where $r _{0}$ is the radius of the Earth and $\sigma$ is Stefan's constant :

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

  1. $\dfrac{R^{2}\sigma T^{4} }{r^{2}}$

  2. $\dfrac{4\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$

  3. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$

  4. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{4\pi r^{2}}$

Show answer
Correct Option: A
Explanation:
Total power radiated by the sun

 $=\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }$

The intensity of power at earth surface

$=\cfrac{\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }}{4\pi { r }^{ 2 }} \\=\cfrac{\sigma { T }^{ 4 } { R }^{ 2 }}{{ r }^{ 2 }}$

The rectangular surface of area $8 cm \times 4 cm$ of a black body at a temperature of $127^0C$ emits energy at the rate of $E$ per second. If both length and breadth of the surface are reduced to half of its initial value, and the temperature is raised to $327^0C$, then the rate of emission of energy will become :

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

  1. $\dfrac{3}{8}E$

  2. $\dfrac{81}{16}E$

  3. $\dfrac{9}{16}E$

  4. $\dfrac{81}{64}E$

Show answer
Correct Option: D
Explanation:
Let $A _1=32$ as given.
Let $A _2$ be the area when length and breadth are reduced by half. Thus the area will be $\dfrac {1}{4}$th of $A _1$
$ \therefore A _2=\dfrac {1}{4} *32=8$
Given $T _1={127}^0C={400}^0K$
Given $T _2={327}^0C={600}^0K$
From Stefan's law $E=\sigma AT^4$
$ \therefore\dfrac{E _1}{E _2}= \dfrac {A _1{T _1}^4}{A _2{T _2}^4}=\dfrac {32*(400)^4}{8*(600)^4}=\dfrac{64}{81}$
$ \therefore \dfrac{E _2}{E _1}= \dfrac{81}{64}$