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Multiple choice stefan's law black body radiation heat transfer thermal properties 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

  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. $(97)^{\frac{1}{4}}T$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

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

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

  1. 13 times

  2. 2.27 times

  3. 1.9 times

  4. 3.9 times
Reveal answer Fill a bubble to check yourself
B Correct answer
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 ${E} _{1}={\sigma}{T} _{1}^{4}$
$\dfrac { E }{ { E } _{ 1 } } =\dfrac { { T }^{ 4 } }{ { T } _{ 1 }^{ 4 } }$
${ E } _{ 1 }=\dfrac { E\times { T } _{ 1 }^{ 4 } }{ { T }^{ 4 } } =\dfrac { E\times { 954 }^{ 4 } }{ { 727 }^{ 4 } } =2.96E$

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

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

  1. $300:1$

  2. $600:1$

  3. $900:1$

  4. $1200:1$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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

  1. light

  2. electric

  3. radiation

  4. All the above
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 :

  1. $\sqrt{3}:1$

  2. $1:\sqrt{3}$

  3. $3:1$

  4. $1:3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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

  1. 0.25

  2. 0.52

  3. 1.52

  4. 2.52
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

The radiation emitted by a perfectly black body is proportional to 

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. Area of the surface

  2. Difference of temperature between the surface and its surroundings

  3. Nature of the surface

  4. All the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

  1. a blackbody

  2. all bodies

  3. bodies painted balck only

  4. polished bodies only
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  1. $4E$

  2. $16E$

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

  4. $\dfrac{E}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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