Tag: stefan's law

Questions Related to stefan's law

The value of solar constant is approximately  :

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

  1. $ 1340\ watt/m^{2}$

  2. $ 430\ watt/m^{2}$

  3. $ 340\ watt/m^{2}$

  4. $ 1388\ watt/m^{2}$

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Correct Option: D
Explanation:

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 $2$($\dfrac{cal}{{cm}{\times} {min}}$)

A heated body emits radiation which has maximum intensity at frequency $v _m$. If the temperature of the body is doubled

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

  1. the maximum intensity radiation will be at frequency $2v _m$

  2. the maximum intensity radiation will be at frequency $\displaystyle\dfrac{1}{2}v _m$

  3. the total emitted energy will increase by a factor of $16$

  4. the total emitted energy will increase by a factor of $2$

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Correct Option: A,C
Explanation:

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 amount of radiations emitted by a black body depends on its

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

  1. size

  2. mass

  3. temperature

  4. density

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Correct Option: C
Explanation:

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 amplitudes of radiations from a cylindrical heat source is related to the distance are

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

  1. $ A \propto 1/{d}^2$

  2. $\displaystyle A \propto \frac{1}{ d} $

  3. $ A \propto d$

  4. $ A \propto d^2$

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Correct Option: B
Explanation:

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. 

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 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is

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

  1. $(\cfrac{65}{2})^{\frac{1}{4}}T$

  2. $(\cfrac{97}{4})^{\frac{1}{4}}T$

  3. $(\cfrac{97}{2})^{\frac{1}{4}}T$

  4. $(97)^{\frac{1}{4}}T$

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Correct Option: C

The energy emitted by a black body at $727^oC$ is E. If the temperature of the body is increased by $227^oC$, the emitted energy will become

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

  1. 13 times

  2. 2.27 times

  3. 1.9 times

  4. 3.9 times

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Correct Option: B
Explanation:

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

So, at temperature ${T}={727}^{o}C$ energy emitted will be ${E}$
at temperature ${T} _{1}={727+227}={954}^{o}C$ energy emitted will be ${E} _{1}={\sigma}{T} _{1}^{4}$
$\dfrac { E }{ { E } _{ 1 } } =\dfrac { { T }^{ 4 } }{ { T } _{ 1 }^{ 4 } }$
${ E } _{ 1 }=\dfrac { E\times { T } _{ 1 }^{ 4 } }{ { T }^{ 4 } } =\dfrac { E\times { 954 }^{ 4 } }{ { 727 }^{ 4 } } =2.96E$

The radiation emitted by a star $A$ is $10000$ times that of the sun. If the surface temperature of the sun and star $A$ are $6000:K$ and $2000:K$, respectively, the ratio of the radii of the star $A$ and the sun is

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

  1. $300:1$

  2. $600:1$

  3. $900:1$

  4. $1200:1$

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Correct Option: C
Explanation:
Stars can be approximated as black bodies.
Hence by stefan's law, power emitted=P=$\sigma AT^4$
Thus,$ \dfrac{P _A}{P _{sun}}=\dfrac{r _A^2T _A^4}{r _{sun}^2T _{sun}^4}$
$\Rightarrow (\dfrac{r _A}{r _{sun}})^2=10000\times (\dfrac{6000}{2000})^4 \rightarrow \dfrac{r _A}{r _{sun}}=900$
So required ratio is 900:1

In pyrometer , temperature measured is proportional to $\underline{\hspace{0.5in}}$ energy emitted by the body 

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

  1. light

  2. electric

  3. radiation

  4. All the above

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Correct Option: C
Explanation:

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.

Two bodies of same shape and having emissivities 0.1 and 0.9 respectively radiate same energy per second. The ratio of their temperature is :

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

  1. $\sqrt{3}:1$

  2. $1:\sqrt{3}$

  3. $3:1$

  4. $1:3$

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Correct Option: A
Explanation:
$\dfrac{E}{t}=e \sigma A T^4$
From above equation which is Stefan's Law of radiation, it is clear that:
$\dfrac{E _1}{E _2} = \dfrac{{e} _{1}\sigma{T} _{1}^{4}}{{e} _{2}\sigma{T} _{2}^{4}}$

$1 = \dfrac{{0.1}{T} _{1}^{4}}{{0.9}{T} _{2}^{4}}$

$\dfrac{{T} _{1}}{{T} _{2}} = \dfrac{\sqrt{3}}{1} $

Two bodies A and B are kept in an evacuated chamber at $27^oC$. The temperature of A and B are $327^oC$ and $427^oC$ respectively. The ratio of rate of loss of heat from A and B will be

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

  1. 0.25

  2. 0.52

  3. 1.52

  4. 2.52

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Correct Option: B
Explanation:

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.