Tag: stefan's law

Questions Related to stefan's law

The radiation emitted by a perfectly black body is proportional to 

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

  1. temperature on ideal gas scale

  2. fourth root of temperature on ideal gas scale

  3. fourth power of temperature on ideal gas scale

  4. square of temperature on ideal gas scale

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Correct Option: C
Explanation:

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.

The amount of heat energy radiated per second by a surface depends upon:

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

  1. Area of the surface

  2. Difference of temperature between the surface and its surroundings

  3. Nature of the surface

  4. All the above

Show answer
Correct Option: D
Explanation:

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

The thermal radiation emitted by a body is proportional to $T^{n}$ where $T$ is its absolute temperature. The value of $n$ is exactly $4$ for

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

  1. a blackbody

  2. all bodies

  3. bodies painted balck only

  4. polished bodies only

Show answer
Correct Option: B
Explanation:

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

A black body radiates energy at the rate of $E\ watt/m$$^{2}$ at a high temperature $T^{o}K$ when the temperature is reduced to $\left [ \dfrac{T}{2} \right ]^{o}K$ Then radiant energy is

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

  1. $4E$

  2. $16E$

  3. $\dfrac{E}{4}$

  4. $\dfrac{E}{16}$

Show answer
Correct Option: D
Explanation:

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

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 orconvection, 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 oflight. 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 wavelengts. The total energy E emitted by a unit area of a black bodyper 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.

Which of the following devices is used to detect thermal radiations?

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

  1. Constant volume air thermometer

  2. Platinum resistance thermometer

  3. Thermostat

  4. Thermopile

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Correct Option: D
Explanation:

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

The rate of radiation from a black body at $0$$^{o}$C is $E$. The rate of radiation from this black body at $273$$^{o}$C is :

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

  1. $2E$

  2. $E/2$

  3. $16E$

  4. $E/16$

Show answer
Correct Option: C
Explanation:

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

Two spherical black bodies of radii $r _{1} $ and $  r _{2}$ are with surface temperatures $T _{1} $ and $ T _{2}$ respectively radiate the same power. $r _{1} / r _{2}$ must be equal to

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

  1. $(T _{1}/T _{2})^{2}$

  2. $(T _{2}/T _{1})^{2}$

  3. $(T _{1}/T _{2})^{4}$

  4. $(T _{2}/T _{1})^{4}$

Show answer
Correct Option: B
Explanation:

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

Given that both spherical body radiate with same power.
So equating ${E} _{1}={E} _{2}$
${A}=4{\pi}{r}^{2}$
${ r } _{ 1 }^{ 2 }{ T } _{ 1 }^{ 2 }={ r } _{ 2 }^{ 2 }{ T } _{ 2 }^{ 4 }$

$\dfrac { { r } _{ 1 } }{ { r } _{ 2 } } ={ (\dfrac { { T } _{ 2 } }{ { T } _{ 1 } } ) }^{ 2 }$

The temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of :

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

  1. 2

  2. 4

  3. 8

  4. 16

Show answer
Correct Option: D
Explanation:

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, $E\  will\  increase \ by \ the \ factor \ of \  { 16}$.

A black body is at temperature $300K$. It emits energy at a rate, which is proportional to 

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

  1. ${(300)}^{4}$

  2. ${(300)}^{3}$

  3. ${(300)}^{2}$

  4. $300$

Show answer
Correct Option: A
Explanation:

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

If the absolute temperature of a blackbody is doubled, then the maximum energy density

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

  1. Increases to 16 times

  2. Increases to 32 times

  3. Decreases to 16 times

  4. Decreases to 32 times

Show answer
Correct Option: A
Explanation:

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 $\dfrac{P _{2}}{P _{1}} = \dfrac{16}{1}$ 

Hence energy density is increased 16 times.