Tag: physics

Questions Related to physics

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

Show answer
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

Show answer
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. $1200:1$

Show answer
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

Show answer
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. $1:3$

Show answer
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

Show answer
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.

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

Show answer
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
${E} _{1}\propto ({ \dfrac { T }{ 2 } ) }^{ 4 }$
${E} _{1}\propto\dfrac{E}{16}$