Tag: black body radiation

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Questions Related to black body radiation

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

If temperature of a black body increases from $-73^oC$ to $327^oC$, then ratio of emissive power at the two temperature is

  1. 27 : 1

  2. 81 : 1

  3. 1 : 27

  4. 1 : 81
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$T _1=-73^0C=273^0C-73^0C=200K$
$T _2=327^0C=273^0C+327^0C=600K$
According to Stephen's law of radiation,
$E\propto T^4$
Ratio, $\dfrac{E _1}{E _2}=(\dfrac{T _1^4}{T _2^4})^4=(\dfrac{200}{600})^4$
$\dfrac{E _1}{E _2}=\dfrac{1}{81}$
$E _1:E _2=1:81$
The correct option is D.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

These is a solid cube of sides $1m$. Its temperature is $127^{\circ}C$ and emissivity is $\dfrac{1}{5.67}$. If surrounding temperature is $27^{\circ}C$ then net rate of radiation loss will be:-

  1. $1.05\ KW$
  2. $5.9\ KW$
  3. $0.175\ KW$
  4. $9.5\ KW$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The net rate of heat loss by radiation is given by Stefan-Boltzmann law, P = e * sigma * A * (T1^4 - T2^4). Here, T1 = 127 degrees Celsius = 400 K, T2 = 27 degrees Celsius = 300 K, A = 6 * (1 m)^2 = 6 m^2, and emissivity e = 1/5.67. Substituting sigma = 5.67 x 10^-8 and the temperatures yields a heat loss rate of approximately 1.05 kW.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

The maximum wavelength of radiations emitted at 900 K is $4 \mu m$. What will be the maximum wavelength of radiation emitted at 1200 K?

  1. 3 $\mu m$
  2. 0.3 $\mu m$
  3. 1 $\mu m$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Wien's displacement law, the product of maximum wavelength and absolute temperature is constant, so lambda_1 * T1 = lambda_2 * T2. Substituting the given values, 4 micrometers * 900 K = lambda_2 * 1200 K, which gives lambda_2 = (4 * 900) / 1200 = 3 micrometers.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

Certain substance emit only the wavelength $\lambda  _{1},\lambda  _{2},\lambda  _{3}  \ and  \ \lambda  _{4}$ when it is at a high temperature. When this substance is at a colder temperature, it will absorb only the following wavelength :

  1. $\lambda _{1}$
  2. $\lambda _{2}$
  3. $\lambda _{1}$ and $\lambda _{2} $
  4. $\lambda _{1},\lambda _{2},\lambda _{3}$ and $\lambda _{4} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

With the help of kirchhoff's law  we can say that $\varepsilon =\alpha $
${\varepsilon}= emissivity $
$\alpha=absorptivity $
so kirchhoff's law state that total emissivity of body is equal to total absorptivity.
so ${\lambda} _{1},{\lambda} _{2}, {\lambda} _{3}, {\lambda} _{4}$ will be absorbed.  

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

In steady state

  1. Heat received is partly conducted and partly radiated

  2. Heat is not absorbed

  3. both A and B

  4. All the heat is conducted
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In steady state, heat received by any small element = heat given by it
So, heat is partially radiated and no net heat is absorbed.
We give an argument for this.
suppose that the heat is being given out by the element.
In this case, the element's temperature will decrease.
But, by definition , at steady state, all the temperatures remain constant.
So, this will violate the steady state condition.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

A polished metal plate with a rough black spot on it is heated to about $1400 K$ and quickly taken into a dark room. Which one of the following statements will be true?

  1. The spot will appear brighter than the plate

  2. The spot will appear darker than the plate

  3. The spot and the plate will appear equally bright

  4. The spot and the plate will not be visible in the dark room
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Kirchhoff law, good absorbers are good emitters. Since black spot is good absorbers so it is also a good emitter & will brighter than plate.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

An ideal black body at room temperature is thrown into a furnace.It is observed that

  1. initially it is the darkest body and at later times the brightest.

  2. it is darkest body at all times.

  3. it cannot be distinguished at all times.

  4. initially it is darkest and at later times it cannot be distinguished.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An ideal black body is a good emitter and good absorber. Initially it will absorb all the energy and will appear dark. As soon as it attains the temperature equal to that of the furnace, it will start emitting better than the rest of furnace and hence appears the brightest.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

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

  1. $0.4eAT^{4}$
  2. $0.8eAT^{4}$
  3. $0.6eAT^{4}$
  4. $1.0eAT^{4}$
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 $\epsilon<1$ ) and the part 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 $1.0\epsilon AT^4$.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

The rates of heat radiation from two patches of skin each of area $S$, on a patient's chest differ by $2$%. If the patch of the lower temperature is at $300K$ and the emissivity of both the patches is assumed to be unity, the temperature of the other patch is closest to:

  1. $301.5K$
  2. $306K$
  3. $308.5K$
  4. $312K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P=\sigma e AT^4$

$\cfrac{P _1}{P _2}=(\cfrac{T _1}{T _2})^4\ \cfrac{P+P\times\cfrac{2}{100}}{P}=(\cfrac{T _1}{300})^4\1+\cfrac{2}{100}=(\cfrac{T _1}{300})^4\ \Rightarrow \cfrac{102}{100}=(\cfrac{T}{300})^4\ \Rightarrow T=301.488K\ \quad\simeq 301.5K$

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

Choose the correct statements from the following

  1. Good reflectors are good emitters of thermal radiation.
  2. Burns caused by water at 100$^{o}$C are more severe than those caused by steam at 100$^{o}$C

  3. If the earth did not have atmosphere, it would become intolerably cold.

  4. It is impossible to construct a heat engine of a 100% efficiency.
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation
Statement (1) is incorrect. According to Kirchhoff's law, a good emitter is good absorber.
Statement (2) is incorrect because 100°C steam has more energy compared to 100°C water. 
Statement (3) is correct. The thermal radiation from the sun warms the earth during the day since air is poor conductor of heat, the atmosphere acts as a blanket for the earth and keeps the earth warm during the night. Moon is very cold because it has no atmosphere. 
Statement (4) is also correct , $\eta=1-T _2/T _1$ where $T _2$ is the temperature of the sink. To have 
an efficiency, $\eta = 1 (or\ 100\%)$, $T _2=0K$. Since absolute zero cannot be achieved, even an ideal heat engine cannot have a $100\%$ efficiency.