Stefan's law - class-XI

The solar constant is defined as the amount of heat energy received per second per unit area by a perfect black body placed at the surface of the Earth with its surface being held perpendicular to the direction of the sun's rays.

The value of solar constant is $1388$($\dfrac{watt}{{meter}^{2}}$) or $2$($\dfrac{cal}{{cm}{\times} {min}}$)

Wien's displacement law states maximum intensity wavength $ \lambda _{m}\propto \dfrac{1}{T}$
Also for any photon,$ \lambda \propto \dfrac{1}{\nu}$
Hence, frequency $\nu _m \propto T$
Doubling of temperature leads to doubling of frequency from $\nu _m$ to $ 2\nu _m$
From Stefan's law, power is directly proportional to $T^4$
Hence $ T \rightarrow 2T \Rightarrow E \rightarrow (\dfrac {2T}{T})^4E=16E$

The radiations emitted by the body only depend on the type of surface emitting and the temperature difference between the body and the surroundings.

The intensity is inversely proportional to  square of distance and intensity is directly proportional to square of amplitude. So, amplitude is inversely proportional to distance. 

In steady state, the heat flux through each gap between the plates must be equal. Using the Stefan-Boltzmann law for radiative heat transfer between parallel plates, the heat flux q = sigma * (T1^4 - T2^4). Setting the flux between plate 1 and 2 equal to the flux between plate 2 and 3, we get (2T)^4 - T2^4 = T2^4 - (3T)^4. Solving for T2 gives T2 = ((2^4 + 3^4)/2)^(1/4) * T = (97/2)^(1/4) * T.

Here we know that energy emitted by any body is given by ${E}=\sigma{T^{4}}$

Stefan- Boltzann law, $j^{ \star  }=\varepsilon \sigma T^{ 4 }$ connects temperature T with thermal radiation or irradiance  $j^{ \star  }$.
Thus measuring the irradiance with pyrometer yields the temperature of the body.

The power radiated is directly proportional to fourth power of absolute temperature.
i.e.
$P \propto T^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{T _{1}}{T _{2}})^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{327+273}{427+273})^{4} = 0.53$
Hence the ratio of rate of heat loss = 0.53
Hence option B is correct.

Stefan-Boltzmann law states that total power radiated by a perfectly black body is
$P=A\sigma { T }^{ 4 }$
so the radiation emitted by a perfectly black body is proportional to fourth power of temperature on ideal gas scale.
option (C) is the correct answer.

Refer Stefan's Law of Radiation: $Q = \eta \sigma A \delta{T}^{4}$
where, $\sigma = conductivity\ A$  = area $T$ = difference of the temperature 

By Stefan's Law, rate of thermal radiation is directly proportional to fourth power of temperature of the body.
$Q = \sigma {T}^{4}$

We know that from stefans-boltzman law: $E\propto { T }^{ 4 }$
if temperature will be reduces half form the initial value, then
${E} _{1}\propto ({ \dfrac { T }{ 2 } ) }^{ 4 }$
${E} _{1}\propto\dfrac{E}{16}$

Thermopile is a very sensitive device which converts thermal energy to electrical energy.
It is used to detect thermal radiations.

As we know that: ${E} \ {\propto} \ {T}^{4}$
So, $\dfrac{E}{{T} _{1}^{4}}=\dfrac{{E} _{2}}{{T} _{2}^{4}}$
${E} _{2}=\dfrac{{T} _{2}^{4}\times{E}}{{T} _{1}^{4}}=\dfrac{{546}^{4}}{{273}^{4}}$
${E} _{2}={16E}$

$E=\varepsilon \sigma A{ T }^{ 4 }$

We know that from Stefan's Boltzmann relation: $E\propto { T }^{ 4 }$
if the temperature of the sun will be doubled, then : $E\propto ({ 2T })^{ 4 }$
Hence, $E\  will\  increase \ by \ the \ factor \ of \  { 16}$.

For black body radiation
$E=\sigma{T}^{4}$ or $E\propto {T}^{4}$
Rate of emission of energy $\propto {(300)}^{4}$

The power with which a body(in this case black body) radiates is directly proportional to the fourth power of absolute temperature:
$P = kT^{4}$
i.e.
$P _{1} = kT _{1}^{4}$
If the absolute temperature is doubled,
$P _{2} = k(2T _{1})^{4} = 16kT _{1}^{4}$
Now $\dfrac{P _{2}}{P _{1}} = \dfrac{16}{1}$ 

From the given question,

According to Stefen's Law, the rate of heat radiation from body is proportional to the fourth power of body's temperature.

It is given that,

Temperature of surrounding

  $ {{T} _{0}}={{227}^{0}}C=500\ K $

 $ {{T} _{1}}={{727}^{0}}C=1000\ K $

 $ {{T} _{2}}={{1227}^{0}}C=1500\ K $

 $ {{E} _{1}}=20\ Watt\,\, $

 $ {{E} _{2}}=? $

According to Stefn boltzmann law:

$ E=\sigma {{T}^{4}} $

Or

 $ {{E} _{1}}=\sigma ({{T} _{1}}-{{T} _{0}})^4 $

 $ {{E} _{2}}=\sigma ({{T} _{2}}-{{T} _{0}})^4 $

Taking ratios of above equations:

For $ {{E} _{1}}=20\ Watt $

 $ \dfrac{20}{{{E} _{2}}}={{\left( \dfrac{500}{1000} \right)}^{4}} $

 $ \dfrac{20}{{{E} _{2}}}=\left( \dfrac{1}{16} \right) $

 $ {{E} _{2}}=320\ Watt $

Emissive power varies according to Stefan-Boltzmann law as :

$E=\sigma {{T}^{4}}$

According to Planck’s distribution law:

${{E} _{\lambda }}(\lambda ,T)=\dfrac{{{C} _{1}}}{{{\lambda }^{5}}\left[ \exp (\dfrac{{{C} _{2}}}{\lambda T})-1 \right]}$

The graph shows that the emitted radiation varies with wavelength and also it shows Emissivity.

When a non black body is placed inside a hollow enclosure the total radiation from the body is the sum of what it would emit in the open ( with e<1 ) and the part (1-a) of the incident radiation from the walls reflected by it.

In black body radiation 

Since the temperature of $A$ is higher than the temperature of the surrounding hence $A$ radiates heat much larger than it absorbs heat. Since the temperature of $B$ is lower than the temperature of the surrounding hence $B$ absorbs heat much larger than it radiates.
This process goes on until both $A$ and $B$ reach the temperature $T$.
Even after reaching thermal equilibrium, both bodies keep radiating and absorbing.
Hence options $B$, $D$. 

$E\propto A{ T }^{ 4 }$

The rate of heat loss is dQ/dt = -sigma * A * (T^4 - T0^4). Since Q = mcT and c = alpha * T^3, dQ = m * alpha * T^3 * dT. Equating these, m * alpha * T^3 * dT/dt = -sigma * (4 * pi * R^2) * (T^4 - T0^4). Integrating from 3T0 to 2T0 gives the time.

From Stefan's Law:
$\sigma A\epsilon _AT _A^4=\sigma A\epsilon _BT _B^4$  ....(1)
where, $T _A=5802:K$ is temp of A and $T _B$ is temp of B,

The rate at which a substance radiates is directly proportional to the fourth power of the absolute temperature.
The temperature increases from $300K$ to $324.2K$ which is an increase by $1.080$
Hence the rate at which the metal radiates would increase by $(1.08)^{4}$ = $1.36$

Energy radiated $P=\sigma AT^4$
$ \displaystyle \frac{P _1}{P _2} = \cfrac{T _1^4}{T _2^4}= { \bigg ( \frac {350}{700} \bigg ) }^4 = \frac{10}{P _2} \space or P _2 = 160$

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

By Stephan Boltzman law,

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

The energy received by planet B is given by the Stefan-Boltzmann law applied to the radiation from A reaching B. Using the angular diameter and the flux, one can determine the temperature of A.

According to Kirchhoff's law of radiation, good absorbers are good emitters. A white spot (which is a poor absorber/emitter compared to the blackened surface) will emit less radiation at the same temperature, making it appear darker or less bright than the surrounding blackened surface.

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

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

According to the Stefan-Boltzmann law, the rate of cooling is dQ/dt = -sigma * A * T^4. Since dQ = mc * dT, we have mc * dT/dt = -sigma * A * T^4. Separating variables, T^-4 * dT = -(sigma * A / mc) * dt. Integrating gives (1/3) * T^-3 = (sigma * A / mc) * t. Thus t is proportional to (1/T2^3 - 1/T1^3).

According to stefan's Boltzmann law, the energy radiated per unit time:
$E=\sigma A{ T }^{ 4 }$
It is given that: ${E}={5.67}\times{10}^{4}$
Therefore, ${5.67}\times{10}^{4}={5.67}\times{10}^{-8}\times1\times{T}^{4}$
So, ${T}={1000}K$
${T}={1000-273}={727} \  ^oC$

According to Newton's law of cooling, dT/dt = -k(T - T_room). Integrating this gives ln(T - T_room) = -kt + C. Thus, the plot of the logarithm of the temperature difference versus time is a straight line.

The rate of heat exchange for a body at temperature T1 in a chamber at T2 is proportional to the difference in the fourth powers of the temperatures (T2^4 - T1^4) due to radiative heat transfer.

By Stefan-Boltzmann law black body radiation of energy $E$ is directly proportional to fourth power of $T$ temperature of the black body.
$E\quad \propto \quad { T }^{ 4 }$
If the temperature of the body is reduced to $\dfrac{T}{2}$, the energy of radiation will be $\dfrac{E}{16}$
option (A) is the correct answer.

From the stefan's law: ${E}={\sigma}{A}{T}^{4}$
So, ${E}={\sigma}{A}{273}^{4}$------(1)
Now, for second one ${E} _{1}={\sigma}{A}{546}^{4}$-----(2)
On dividing equation(2) by (1), we get
$\dfrac{{E} _{1}}{E}=\dfrac{{546}^{4}}{{273}^{4}}={16}$

According to Stefan's law, energy emitted is proportional to fourth power of absolute temperature.
So, the same will hold for the sun.
Also, Sun can absorb/emit radiations of all wavelengths - so it is a black body.
So, assertion and reason are both correct.
But reason is not a correct explanation of assertion.
Because all bodies emit energy which is proportional to fourth power of absolute temperature, not only sun.

All bodies radiate heat irrespective of temperature.
Heat radiated per unit time is given by Stefan's law.

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.

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

This is identical to question 534600. The steady state temperature of the middle plate is ( (T1^4 + T3^4) / 2 )^(1/4). With T1=2T and T3=3T, T2 = ((16T^4 + 81T^4)/2)^(1/4) = (97/2)^(1/4) * T.

By $Stefan's$ Law

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

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

By $Stefan's$ $Law$,

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

The latent heat of fusion is $80cal/g$.

We know that emissive power is given as: ${E}={\sigma}{T}^{4}$
${T}^{4}=\dfrac{E}{\sigma}$
${T}=\dfrac{1500}{5.67\times{10}^{-8}}=403.29 K$
Now, we know that radiation from the surface is given as $E=\sigma \epsilon A{ T }^{ 4 }$
${E}={5.67\times{10}^{-8}\times4\times{\pi}\times{0.05}^{2}\times{403.29}^{4}}=47.15 J $

Thermal resistance $=\dfrac{kelvin}{watt}$
Unit of power is watt so power = force$\times$ velocity
Force = mass$\times$acceleration = M$\times$$\dfrac{m}{{sec}^{2}}$
Force = $ML{T}^{-2}$
Power = $M{L}^{2}{T}^{-3}$
So, Thermal resistance $={M}^{-1}{L}^{-2}{T}^{3}{\theta}$
and we know that relation between conductivity and resistevity is that they are reciprocal to each other,so thermal conductivity =${M}^{1}{L}^{2}{T}^{-3}{\theta}^{-1}$

The rate of radiation, emitted by per unit area of a body at temperature $T$ , is given by Stefan-Boltzmann's law as,

Stefan's law states:
$E = \sigma T^{4}$ 
Normally temperature of a body never exceeds more than 1000K. 
As the temperature is quite less compared to the sun's temperature. Photons of this radiation have less energy and hence greater wavelength and fall into infrared region.
Alternately by Wein's displacement law
$ \lambda _{max} T = constant$
By applying above law on sun and normal body, wavelength of normal thermal radiation falls in infrared region.

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

Thermal conductivity =$\dfrac{Watt}{Metre.Kelvin}$

During clear nights object on surface of earth radiate out heat and temperature falls. Hence option (a) is wrong.
The total energy radiated by a body per unit time per unit area $E \propto {T}^{4}$. Hence option (c) is wrong.
Energy radiated per second is given by
$\dfrac { Q }{ t } =PA\varepsilon \sigma { T }^{ 4 }$
$\Rightarrow \dfrac { { P } _{ 1 } }{ { P } _{ 2 } } =\dfrac { { A } _{ 1 } }{ { A } _{ 2 } } { \left( \dfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 4 }={ \left( \dfrac { { r } _{ 1 } }{ { r } _{ 2 } }  \right)  }^{ 2 }\cdot { \left( \dfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 2 }$
$={ \left( \dfrac { 1 }{ 4 }  \right)  }^{ 2 }\left( \dfrac { 4000 }{ 200 }  \right) =\dfrac { 1 }{ 1 } $
$\because    {P} _{1} = {P} _{2}$ hence option (d) is wrong.
Newton's law is an approximate from of Stefan's law of radiation and works well for natural convection. Hence option (b) is correct.

Both statements are true as the animals curl their body in very cold environment so they can reduce their surface area , and reduce the heat leaving their body in form of radiation as heat emitted is directly proportional to surface area.

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

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, ${ T } _{ 1 }=27+273=300K$
          ${ T } _{ 2 }=927+273=1200K$
$\therefore       \dfrac { { E } _{ 1 } }{ { E } _{ 2 } } ={ \left( \dfrac { 300 }{ 1200 }  \right)  }^{ 4 } = 1 : 256$

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

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