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

If the temperature of a hot body is raised by $0.5\%$, then the heat energy radiated would increase by :

  1. 0.5%

  2. 1.0%

  3. 1.5%

  4. 2.0%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The rate of heat energy radiated by black body is given as:
$Q=\sigma T^4A$ where $\sigma$ is the stefen-boltzmann constant, $A$ is the area of the radiating body.
So $Q\propto T^4=kT^4$, where $k$ is the propotionality constant that we have assumed here.
Taking log of above equation, $logQ=log(kT^4)=logk+4logT$
Taking differential of above equation, $\dfrac{dQ}{Q}=0+\dfrac{4dT}{T}$ (because logk is constant).
Here $dQ$ and $dT$ indicate very small change in the $Q$ and $T$ respectively.
Multiplying by $100$,
$\dfrac{dQ}{Q}\times100=4\dfrac{dT}{T}\times100$
$\Rightarrow$ Percentage change in Heat rate $=4\times $ percentage change in temperature
So here, percentage change in rate of heat energy radiated $=4\times0.5=2\%$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rate of emission of a black body at temperature $27$$^{o}$C is $E _{1}$. If its temperature is increased to $327$$^{o}$C, the rate of emission of radiation is $E _{2}$. The relation between $E _{1} $ and $  E _{2}$ is :

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

We know the relation for emissive power of body: $\dfrac { { E } _{ 1 } }{ { E } _{ 2 } } =\dfrac { { T } _{ 1 }^{ 4 } }{ { T } _{ 2 }^{ 4 } } $
${T} _{1}=27+273=300\ K$
${T} _{2}=327+273=600\ K$
So, putting all these data in above formulas, we get
${E} _{2}=16{E} _{1}$

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

The temperature of a black body is increased by $50\%$ . Then the percentage of increase of radiation is approximately

  1. 100%

  2. 25%

  3. 400%

  4. 500%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By $Stefan's$ $Law$,

Power = ${\sigma A {T}^{4}}$ for a black body
$\sigma$ is known as Stefan's constant
$\dfrac{P _1}{P _2}$ = $\dfrac{ T _1^4}{ T _2^4}$
${T _2}$ = $\dfrac{3 T _1}{2}$
$\dfrac{T _1^4}{T _2^4}$= $\dfrac{16}{81}$
By the above equations we get
$P _2$=$\dfrac{81 P _1}{16}$
Percentage increase in radiation = $\dfrac{P _2 - P _1}{P _1}$ x $100$ = $406.25$%

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

The wave length corresponding to maximum intensity of radiation emitted by a star is $289.8$nm. The intensity of radiation for the star is :

(Stefans constant $=$ 5.6x10$^{-8}Wm^{-2}K^{-4}$, Wien's displacement constant = $2898 \times 10^{-6} mK$ )

  1. 5.67 x 10$^{8}Wm^{-2}$
  2. 5.67 x 10$^{4}Wm^{-2}$
  3. 10.67 x 10$^{7}Wm^{-2}$
  4. 10.67 x 10$^{4}Wm^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, Wavelength corresponding to maximum radiation $=289.8nm$
Stefans constant$=5.6\times 10^{-8}Wm^{-2}K^{-4}$
From Wien's law,
$\lambda _{max} T$= constant (b)
$b=2898\times 10^{-6}$
$\lambda _{max}=289.8$
$\therefore$ $T=\frac{b}{\lambda _{max}}=\frac{2898\times 10^{-6}}{289.8\times 10^{-9}}=10^4 K$

Intensity of radiation E
$E=\sigma T^4=5.6\times 10^{-8}\times 10^{16}= 5.6\times 10^8 Wm^{-2}$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

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 or convection, 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 of light. 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 wavelengths. The total energy E emitted by a unit area of a black body per 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.

From stefan-Boltzmann law, the dimensions of Stefans constant $\sigma $ are :

  1. $ML^{-2}T^{-2}K^{-4}$
  2. $ML^{-1}T^{-2}K^{-4}$
  3. $MLT^{-3}K^{-4}$
  4. $ML^{0}T^{-3}K^{-4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Stefan's law:
$E = \sigma T^{4}$ 
$\sigma = E/ T^{4}$
$E = energy/(area*time) = ( M L^{2} T^{-2} )/ ( L^{2} T)$
$\sigma =  M T^{-3} K^{-4}$

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

The power radiated by a black body is $P$ and it radiates maximum energy around the wavelength $\lambda  _{o}$ . If the temperature of the black body is now changed so that it radiates maximum energy around a wavelength $3\lambda  _{o}/4$ , the power radiated by it will increase by a factor of :

  1. $4/3$
  2. $16/9$
  3. $64/27$
  4. $256/81$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
From Wein's displacement law, we know that ${\lambda} _{max}{T}={CONSTANT}$

So, let us assume that initial temperature of the body be ${T} _{1}$
and by changing the temperature from ${T} _{1}   \ to \  {T}$, wavelength will change from $\lambda$ to $\dfrac{3{\lambda}}{4}$,so applying the above relation,

${\lambda} _{0}{T} _{1}=\dfrac{3{\lambda} _{0}}{4}{T}$

Hence, ${T}=\dfrac{4}{3}T _1$

and we know that ${P}={e}{{T}^{4}}$

So, ${P'}=\dfrac{256}{81}P$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rays of sun are focussed on a piece of ice through a lens of diameter $5$ cm, as a result of which $10$ grams of ice melts in $10$ min. The amount of heat received from Sun is (per unit area per min)

  1. 4 $cal/cm^{2} \: min$
  2. 40 $cal/cm^{2} \: min$
  3. 4 $J/cm^{2} \: min$
  4. 400 $J/cm^{2} \: min$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The latent heat of fusion is $80cal/g$.

So heat received per unit times is $10g\times 80cal g^{-1}/ 10min=80cal/min$
The are is $\pi r^2=\pi\times(2.5)^2cm^2\approx 20cm^2$
So, amountof heat per unit area per unit time is $80/20=4cal/cm^2\ min$

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

The emissive power of a sphere of radius $5$cm coated with lamp black is $1500$Wm$^{-2}$. The amount of energy radiated per second is.

  1. 15.7 J

  2. 3.14 J

  3. 47.10 J

  4. 4.71 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that emissive power is given as: ${E}={\sigma}{T}^{4}$
${T}^{4}=\dfrac{E}{\sigma}$
${T}=\dfrac{1500}{5.67\times{10}^{-8}}=403.29 K$
Now, we know that radiation from the surface is given as $E=\sigma \epsilon A{ T }^{ 4 }$
${E}={5.67\times{10}^{-8}\times4\times{\pi}\times{0.05}^{2}\times{403.29}^{4}}=47.15 J $

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

Match the physical quantities given in Column I with their dimensional formula given in ColumnII

Column-I Column-II
(a) Thermal conductivity (p) is a dimensionless quantity
(b) Stefans constant (q) $ML^{o}T^{o}K$
(c) Wiens constant (r) $ML^2T^{-3}K^{-1}$
(d) Emissivity (s) $ML^{o}T^{-3}K^{-4}$

  1. a-s, b-r, c-p, d-q

  2. a-r, b-s, c-q, d-p

  3. a-s, b-q, c-r, d-p

  4. a-p, b-q, c-r, d-s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Thermal resistance $=\dfrac{kelvin}{watt}$
Unit of power is watt so power = force$\times$ velocity
Force = mass$\times$acceleration = M$\times$$\dfrac{m}{{sec}^{2}}$
Force = $ML{T}^{-2}$
Power = $M{L}^{2}{T}^{-3}$
So, Thermal resistance $={M}^{-1}{L}^{-2}{T}^{3}{\theta}$
and we know that relation between conductivity and resistevity is that they are reciprocal to each other,so thermal conductivity =${M}^{1}{L}^{2}{T}^{-3}{\theta}^{-1}$

Stefan's constant $=5.64\times$${10}^{-8}$$\dfrac { w }{ { m }^{ 2 }{ k }^{ 4 } } $
$\sigma$=${ M{ L }^{ 0 }{ T }^{ -3 }{ \theta  }^{ 4 } }$

Wein's constant: ${\lambda}{\theta}$ = constant = [$L{\theta}$]

Emissivity is define as the ration of two same quantity so it is dimensionless.