Tag: black body radiation

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Questions Related to black body radiation

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

If a graph is plotted by taking spectral emissive power along $y-$axis and wavelength along x-axis is:

  1. Emissivity

  2. Total intensity of radiation

  3. Diffusivity

  4. Solar constant
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

 

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

A spherical body of area A and emissivity $0.6$ is kept inside a perfectly black body. Total heat radiated by the body at temperature T is?

  1. $0.4\sigma AT^4$
  2. $0.8\sigma AT^4$
  3. $0.6\sigma AT^4$
  4. $1.0\sigma AT^4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

The two add up to a black body radiation . Hence the total radiation emitted by the body is $1.0\sigma AT^4$
1.0σAT4.

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

The rate of emission of radiation of ablack body at temperature $27^oC $ is $ E _1 $ . If its temperature is increased to $ 327^oC $ the rate of emission of radiation is $ E _2 . $ The relation between $ E _1 $ and $ E _2 $ is:

  1. $ E _2 = 24 E _1 $
  2. $ E _2 =16 E _1 $
  3. $ E _2 = 8 E _1 $
  4. $ E _2 = 4 E _1 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In black body radiation 

$\dfrac{d\theta}{dt}=(4\pi r^{2})\sigma T^{4}$
If at $T=27^{o}C=300\ K, \dfrac{d\theta}{dt}=E _{1}$
Then, 
$E _{1}=(4\pi R^{2})\sigma(300)^{4}$
If at $T=327^{o}C=600\ k, \dfrac{d\theta}{dt}=E _{2}$
$E _{2}=(4\pi R^{2})\sigma (2)^{4}(300)^{4}$
So, $\dfrac{E _{2}}{E _{1}}=(2)^{4}\dfrac{(4\pi R^{2}\sigma (300)^{4}}{4\pi R^{2}\sigma(300)^{4}}$
$E _{2}=(2)^{4}E _{1}$
$\Rightarrow E _{2}10 E _{1}$
Option $B$ is correct






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

Two identical objects $A$ and $B$ are at temperatures $T _A$ and $T _B$. respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature $T$ ($T _A$ > $T$> $T _B$). The objects $A$ and $B$ attain the temperature $T$ eventually. Select the correct statements from the following

  1. $A$ only emits radiation, while $B$ only absorbs it until both attain the temperature $T$
  2. $A$ loses more heat by radiation than it absorbs, while $B$ absorbs more radiation than it emits until they attain the temperature $T$
  3. Both $A$ and $B$ only absorb radiation, but do not emit it, until they attain the temperature $T$
  4. Each object continuous to emit and absorb radiation even after attaining the temperature $T$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 $B$, $D$. 

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

A planet is at an average distance $d$ from the sun and its average surface temperature is $T$. Assume that the planet receives energy only from the sun and loses energy only through radiation from the surface. Neglect atmospheric effects. If $T$ $\propto d^{-n}$, the value of $n$ is :

  1. $2$
  2. $1$
  3. $\displaystyle \frac{1}{2}$
  4. $\displaystyle \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let P=power radiated by Sun
R=Radius of planet
E=energy received by planet=$\cfrac{P}{4 \pi d^2}\times \pi R^2$
Energy radiated by planet=$(4 \pi R^2)\sigma T^4 $
For thermal equilibriums:
$\Rightarrow \cfrac { P }{ 4\pi d^{ 2 } } \pi R^{ 2 }=4\pi R^{ 2 }\sigma T^{ 4 }\\ \Rightarrow T^{ 4 }\alpha \cfrac { 1 }{ d^{ 2 } } \\ \Rightarrow T\alpha \cfrac { 1 }{ d^{ { 1 }/{ 2 } } } \\ \Rightarrow T\alpha { d }^{ -\cfrac { 1 }{ 2 }  }$
So n=$\cfrac{1}{2}$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

A planet radiates heat at a rate proportional to the fourth power of its surface temperature $T$. If such a steady temperature of the planet is due to an exactly equal amount of heat received from the sun then which of the following statements is true?

  1. The planet's surface temperature varies inversely as the distance of the sun

  2. The planet's surface temperature varies directly as the square of its distance from the sun

  3. The planet's surface temperature varies inversely as the square root of its distance from the sun

  4. The planet's surface temperature is proportional to the fourth power of distance from the sun
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Planet's surface temperature varies inversely as square root of its distance from the Sun.
${ T }^{ 4 }\alpha \cfrac { 1 }{ { d }^{ 2 } } \Rightarrow T\alpha \cfrac { 1 }{ \sqrt { d }  } $
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The radiation emitted by a star $A$ is $1000$ times that of the sun. If the surface temperatures 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

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


$\displaystyle \frac { { E } _{ 1 } }{ { E } _{ 2 } } =\frac { 1000 }{ 1 } =\frac { \pi { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ \pi { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { 2000 }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times 6000^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 } }{ { { r } _{ 2 } }^{ 2 }\times 81 } $

$\displaystyle \frac { { r } _{ 1 } }{ { r } _{ 2 } } =\sqrt { \frac { 1000 }{ 1 } \times \frac { 81 }{ 1 }  } =284.6:1$

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

The number of oxygen molecules in a cylinder of volume $1 \mathrm { m } ^ { 3 }$ at a temperature of $27 ^ { \circ } C$ and pressure $13.8 Pa$ is
 (Boltzmaan's constant $k = 1.38 \times 10 ^ { - 23 } \mathrm { JK } ^ { - 1 }$)

  1. $6.23 \times 10 ^ { 26 }$
  2. $0.33 \times 10 ^ { 28 }$
  3. $3.3 \times 10 ^ { 21 }$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

A solid sphere of mass m and radius $R$ is painted black and placed inside a vacuum chamber. The walls of the chamber are maintained at temperature $T 0$ the initial temperature of the sphere is $3T _0$. The specific heat capacity of the sphere material varies with its temperature $T$ as $\alpha T^3$ where $\alpha$ is a constant. Then the sphere will cool down to temperature $2T _0$ in time ________ ($\sigma$ = Stefan Boltzmann constant)

  1. $\dfrac{m\alpha}{16\pi R^2\sigma}\ell n\left(\dfrac{16}{3}\right)$
  2. $\dfrac{m\alpha}{8\pi R^2\sigma}\ell n\left(\dfrac{4}{3}\right)$
  3. $\dfrac{m\alpha}{8\pi R^2\sigma}\ell n\left(\dfrac{3}{2}\right)$
  4. $\dfrac{m\alpha}{4\pi R^2\sigma}\ell n\left(\dfrac{8}{3}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

In the nuclear fusion, $ _{1}^{2}{H}+ _{1}^{3}{H}\rightarrow _{2}^{4}{He}+n$ given that the repulsive potential energy between the two nuclie is $7.7\times 10^{-14}J$, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant $k=1.38\times 10^{-23}J/K$]-

  1. $10^{7}K$
  2. $10^{5}K$
  3. $10^{3}K$
  4. $10^{9}K$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy    $E \approx kT$

So,     $7.7\times 10^{-14} \approx 1.38\times 10^{-23}\times T$
$\implies \ T\approx  5.6\times 10^9 \ K$
Correct answer is option D.