Tag: stefan's law

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

The dimensions of Stefan's constant are

  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] $
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $[\sigma] = \dfrac{[ML^2T^{-3}]}{[L^2] [K^4]}$
$\implies$ $[\sigma] = [M^1 L^0 T^{-3} K^{-4}]$

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

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

  1. $1 : 4$

  2. $1 : 8$

  3. $1 : 16$

  4. $1 : 256$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, ${ T } _{ 1 }=27+273=300K$
          ${ T } _{ 2 }=927+273=1200K$
$\therefore       \dfrac { { E } _{ 1 } }{ { E } _{ 2 } } ={ \left( \dfrac { 300 }{ 1200 }  \right)  }^{ 4 } = 1 : 256$

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

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 :

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

  2. $1\mu m$

  3. $2 \mu m$

  4. $\dfrac{3}{4} \mu m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ${ \lambda  } _{ B }=1.5\mu m$.


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

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

  1. $U/8$

  2. $U/16$

  3. $U/4$

  4. $U/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

 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.)

  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}

    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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 :

  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. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{4\pi r^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Total power radiated by the sun

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

The intensity of power at earth surface

$=\cfrac{\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }}{4\pi { r }^{ 2 }} \\=\cfrac{\sigma { T }^{ 4 } { R }^{ 2 }}{{ r }^{ 2 }}$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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 :

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

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

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

  4. $\dfrac{81}{64}E$
Reveal answer Fill a bubble to check yourself
D Correct answer
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}$