Tag: stefan's law

Questions Related to stefan's law

Intensity of heat radiation emitted by body is believed to be proportional to fourth power of absolute temperature of the body. The proportionality constant also known as Boltzmann's constant may have possible value of :

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

  1. $5.67\times 10^{-8} watt/K^4 $

  2. $5.67\times 10^{-8} watt/m^2 K^4 $

  3. $5.67\times 10^{-8} J/K^4 $

  4. $5.67\times 10^{-8} Js/K^4 $

Show answer
Correct Option: B
Explanation:

From the given question,

$I\propto T^4$

$I=k T^4$

where $k=$ Stefan-Boltzmann's constant

$k=5.67\times10^{-8} W/m^2K^4 $

The correct option is B.

A black body at a temperature of $227^oC$ radiates heat energy at the rate 5 cal/cm$^{2}-s$. At a temperature of $727^oC$, the rate of heat radiated per unit area in cal/cm$^2$ will be

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

  1. 80

  2. 160

  3. 250

  4. 500

Show answer
Correct Option: A
Explanation:

According to Stefen's Law, the rate of heat radiation from body is proportional to the fourth power of body's temperature.

Thus $P\propto T^4$
$\implies \dfrac{P _2}{P _1}=\dfrac{T _2^4}{T _1^4}$
$=16$
$\implies P _2=80cal/cm^2-s$

For a block body temperature $727^{o}C,$ its rate of energy loss is $20\ watt$ and temperature of surrounding is $227^{o}C.$ If temperature of black body is changed to $1227^{o}C$ then its rate of energy loss will be:

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

  1. $320\ W$

  2. $\dfrac {304}{3}\ W$

  3. $240 W$

  4. $120 W$

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Correct Option: A
Explanation:

It is given that,

Temperature of surrounding

  $ {{T} _{0}}={{227}^{0}}C=500\ K $

 $ {{T} _{1}}={{727}^{0}}C=1000\ K $

 $ {{T} _{2}}={{1227}^{0}}C=1500\ K $

 $ {{E} _{1}}=20\ Watt\,\, $

 $ {{E} _{2}}=? $

According to Stefn boltzmann law:

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

Or

 $ {{E} _{1}}=\sigma ({{T} _{1}}-{{T} _{0}})^4 $

 $ {{E} _{2}}=\sigma ({{T} _{2}}-{{T} _{0}})^4 $

Taking ratios of above equations:

For $ {{E} _{1}}=20\ Watt $

 $ \dfrac{20}{{{E} _{2}}}={{\left( \dfrac{500}{1000} \right)}^{4}} $

 $ \dfrac{20}{{{E} _{2}}}=\left( \dfrac{1}{16} \right) $

 $ {{E} _{2}}=320\ Watt $

The power received at distance $d$ from a small metallic sphere of radius $r(<<d)$ and at absolute temperature $T$ is $P$. If the temperature is doubled and distance reduced to half of the initial value, then the power received at that point will be:

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

  1. $4p$

  2. $8p$

  3. $32p$

  4. $64p$

Show answer
Correct Option: D
Explanation:
Energy received per second i.e., power $P\alpha \dfrac{T^4}{d^2}=k\dfrac{T^4}{d^2}$
if temperature is double than T become 2T and distance become half than d become $\dfrac{d}{2}$
than power $ p _{1}=k\dfrac{(2T)^4}{(\dfrac{d}{2})^2}=64k\dfrac{T^4}{d^2}=64P$
Hence D option is correct.

What is the value of solar constant if the energy received by $ 12$ m$^2$ area in $2$ minutes is $2016$ kJ?

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

  1. $1.4 \times 10^2 J s^{-1}  m^{-2}$

  2. $1400 J s^{-1}m^{-2}$

  3. $84 kJ s^{-1} m^{-2}$

  4. $84 J s^{-1} m^{-2}$

Show answer
Correct Option: B
Explanation:
Solar constant = Energy received by unit area in unit time from Sun
$=\cfrac{2016\times 10^3}{12\times (2\times 60)} \\=1400Js^{-1}m^2$

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

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

  1. Emissivity

  2. Total intensity of radiation

  3. Diffusivity

  4. Solar constant

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Correct Option: A
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.

 

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?

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

  1. $0.4\sigma AT^4$

  2. $0.8\sigma AT^4$

  3. $0.6\sigma AT^4$

  4. $1.0\sigma AT^4$

Show answer
Correct Option: D
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.

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:

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

  1. $ E _2 = 24 E _1 $

  2. $ E _2 =16 E _1 $

  3. $ E _2 = 8 E _1 $

  4. $ E _2 = 4 E _1 $

Show answer
Correct Option: B
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






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

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

  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 $T$

Show answer
Correct Option: B
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 $B$, $D$. 

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 :

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

  1. $2$

  2. $1$

  3. $\displaystyle \frac{1}{2}$

  4. $\displaystyle \frac{1}{4}$

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Correct Option: C
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=$\cfrac{1}{2}$