Tag: stefan's law

Questions Related to stefan's law

A solid shpere and a hollow sphere of the same material and of equal radii are heated to the same temperature

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

  1. both will emit equal amount of radiation per unit time in the beginning.

  2. both will absorbs equal amount of radiation per second from the surrounding in the beginning.

  3. the initial rate of cooling will be the same for both the spheres

  4. the two spheres will have equal temperature at any instant

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Correct Option: A,B
Explanation:

According to many radiation laws like Stefan Boltzmann we know that radiation emission and absorption are a purely surface phenomenon. Since the two bodies are of same material, same radii, and same temperature they will at that instant radiate and absorb at the same rates.
But however since the hollow sphere has lesser mass, the rate at which it's temperature will rise will be different from that of the solid sphere. Hence their rates of cooling would be varied and they would have different temperatures at different times.

A black body at 127$^{o}$C emits the energy at the rate of 10$^{6}$ J/m$^{2}$ s. The temperature of a black body at which the rate of energy emission is 16x10$^{6}$ J/m$^{2}$ s is :

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

  1. $508^{o}C$

  2. $273^{o}C$

  3. $400^{o}C$

  4. $527^{o}C$

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Correct Option: D
Explanation:

Using Stefan's Law:
$\dfrac { { E } _{ 2 } }{ { E } _{ 1 } } ={ \left( \dfrac { { T } _{ 2 } }{ { T } _{ 1 } }  \right)  }^{ 4 }\ { T } _{ 2 }={ T } _{ 1 }\sqrt [ 4 ]{ \dfrac { { E } _{ 2 } }{ { E } _{ 1 } }  } \quad \ { T } _{ 2 }={ (127+273) }\sqrt [ 4 ]{ \dfrac { 16\times { 10 }^{ 6 } }{ { 10 }^{ 6 } }  } =\quad 800K\ { T } _{ 2 }={ 527 }^{ \circ  }C$

Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures of 2T and 3T respectively. The temperatures of the middle (i.e. second) plate under steady state condition is then

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

  1. $\left ( \dfrac{64}{2} \right )^{\dfrac{1}{4}}T$

  2. $\left ( \dfrac{97}{4} \right )^{\dfrac{1}{4}}T$

  3. $\left ( \dfrac{97}{2} \right )^{\dfrac{1}{4}}T$

  4. $\left ( 97\right )^{\dfrac{1}{4}}T$

Show answer
Correct Option: C

The rectangular surface of area $8cm$ $\times$ $4 cm$ of a black body at temperature $127^{\circ}C$ emits energy $E$ per second. If the length and breadth are reduced to half of the initial value and the temperature is raised to $327^{\circ}C$, the rate of emission of energy becomes

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

  1. $\displaystyle \frac{3}{8}E$

  2. $\displaystyle \frac{81}{16}E$

  3. $\displaystyle \frac{9}{16}E$

  4. $\displaystyle \frac{81}{64}E$

Show answer
Correct Option: D
Explanation:

By $Stefan's$ Law

$Power=\sigma A T^4$
$A=length\times breadth$
$\dfrac{P _2}{E}=\dfrac{l _2b _2T _2^4}{l _1b _1T _1^4}$
Here$\dfrac{l _2}{l _1}=\dfrac{1}{2}$       $\dfrac{b _2}{b _1}=\dfrac{1}{2}$       $\dfrac{T _2}{T _1}=\dfrac{3}{2}$

$\implies P _2=\dfrac{81}{64}E$

If the temperature of a hot body is raised by $0.5\%$, then the heat energy radiated would increase by :

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

  1. 0.5% 

  2. 1.0%

  3. 1.5%

  4. 2.0%

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Correct Option: D
Explanation:
The rate of heat energy radiated by black body is given as:
$Q=\sigma T^4A$ where $\sigma$ is the stefen-boltzmann constant, $A$ is the area of the radiating body.
So $Q\propto T^4=kT^4$, where $k$ is the propotionality constant that we have assumed here.
Taking log of above equation, $logQ=log(kT^4)=logk+4logT$
Taking differential of above equation, $\dfrac{dQ}{Q}=0+\dfrac{4dT}{T}$ (because logk is constant).
Here $dQ$ and $dT$ indicate very small change in the $Q$ and $T$ respectively.
Multiplying by $100$,
$\dfrac{dQ}{Q}\times100=4\dfrac{dT}{T}\times100$
$\Rightarrow$ Percentage change in Heat rate $=4\times $ percentage change in temperature
So here, percentage change in rate of heat energy radiated $=4\times0.5=2\%$

A black body is at a temperature of $500$K. It emits its energy at a rate which is proportional to :

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

  1. $500$

  2. $(500)^{2}$

  3. $(500)^{3}$

  4. $(500)^{4}$

Show answer
Correct Option: D
Explanation:

From stefan's boltzmann relation we know that: $E\propto { T }^{ 4 }$
So, $E\propto { 500}^{ 4 }$

The rate of emission of a black body at temperature $27$$^{o}$C is $E _{1}$. If its temperature is increased to $327$$^{o}$C, the rate of emission of radiation is $E _{2}$. The relation between $E _{1} $ and $  E _{2}$ is :

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

  1. $E _{2}=24E _{1}$

  2. $E _{2}=16E _{1}$

  3. $E _{2}=8E _{1}$

  4. $E _{2}=4E _{1}$

Show answer
Correct Option: B
Explanation:

We know the relation for emissive power of body: $\dfrac { { E } _{ 1 } }{ { E } _{ 2 } } =\dfrac { { T } _{ 1 }^{ 4 } }{ { T } _{ 2 }^{ 4 } } $
${T} _{1}=27+273=300\ K$
${T} _{2}=327+273=600\ K$
So, putting all these data in above formulas, we get
${E} _{2}=16{E} _{1}$

The temperature of a black body is increased by $50\%$ . Then the percentage of increase of radiation is approximately

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

  1. 100%

  2. 25%

  3. 400%

  4. 500%

Show answer
Correct Option: C
Explanation:

By $Stefan's$ $Law$,

Power = ${\sigma A {T}^{4}}$ for a black body
$\sigma$ is known as Stefan's constant
$\dfrac{P _1}{P _2}$ = $\dfrac{ T _1^4}{ T _2^4}$
${T _2}$ = $\dfrac{3 T _1}{2}$
$\dfrac{T _1^4}{T _2^4}$= $\dfrac{16}{81}$
By the above equations we get
$P _2$=$\dfrac{81 P _1}{16}$
Percentage increase in radiation = $\dfrac{P _2 - P _1}{P _1}$ x $100$ = $406.25$%

The wave length corresponding to maximum intensity of radiation emitted by a star is $289.8$nm. The intensity of radiation for the star is :

(Stefans constant $=$ 5.6x10$^{-8}Wm^{-2}K^{-4}$, Wien's displacement constant = $2898 \times 10^{-6} mK$ )

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

  1. 5.67 x 10$^{8}Wm^{-2}$

  2. 5.67 x 10$^{4}Wm^{-2}$

  3. 10.67 x 10$^{7}Wm^{-2}$

  4. 10.67 x 10$^{4}Wm^{-2}$

Show answer
Correct Option: A
Explanation:
Given, Wavelength corresponding to maximum radiation $=289.8nm$
Stefans constant$=5.6\times 10^{-8}Wm^{-2}K^{-4}$
From Wien's law,
$\lambda _{max} T$= constant (b)
$b=2898\times 10^{-6}$
$\lambda _{max}=289.8$
$\therefore$ $T=\frac{b}{\lambda _{max}}=\frac{2898\times 10^{-6}}{289.8\times 10^{-9}}=10^4 K$

Intensity of radiation E
$E=\sigma T^4=5.6\times 10^{-8}\times 10^{16}= 5.6\times 10^8 Wm^{-2}$

All bodies emit heat energy from their surfaces by virtue of their temperature. This heat energy is called radiant energy of thermal radiation. The heat that we receive from the sun is transferred to us by a process which, unlike conduction or convection, does not require the help of a medium in the intervening space which is almost free of particles. Radiant energy travels in space as electromagnetic spectrum. Thermal radiations travel through vacuum with the speed of light. Thermal radiations obey the same laws of reflection and refraction as light does. They exhibit the phenomena of interference, diffraction and polarization as light does.
The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the black body) at the same temperature. A black body absorbs and hence emits radiations of all wavelengths. The total energy E emitted by a unit area of a black body per second is given by $E =\sigma T^{4}$ where T is the absolute temperature of the body and $\sigma $ is a constant known as Stefans constant. If the body is not a perfect black body, then $E =\varepsilon \sigma  T^{4}$where $\varepsilon $ is the emissivity of the body.

From stefan-Boltzmann law, the dimensions of Stefans constant $\sigma $ are :

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

  1. $ML^{-2}T^{-2}K^{-4}$

  2. $ML^{-1}T^{-2}K^{-4}$

  3. $MLT^{-3}K^{-4}$

  4. $ML^{0}T^{-3}K^{-4}$

Show answer
Correct Option: D
Explanation:

Stefan's law:
$E = \sigma T^{4}$ 
$\sigma = E/ T^{4}$
$E = energy/(area*time) = ( M L^{2} T^{-2} )/ ( L^{2} T)$
$\sigma =  M T^{-3} K^{-4}$