Kirchhoff's law of thermal radiations-chemistry learners
Kirchhoff's law of blackbody radiations-theoretical proofs
We can see the light and feel the sensation of heat. But we never expected a crazy combination of both. It is blackbody radiation. Blackbody radiation is heat-assimilated electromagnetic light. Generally, radiation is the method of transmission of heat from one point to another without heating the intervening medium. And when the heated object emits radiation, it is called thermal radiation.
Early in the 15th century, people believed ice released cold radiation, and a hot cup of tea discharged hot radiation. The classification of thermal radiation was absurd based on the Prevost hypothesis of heat exchange. It states that the hotness and coldness of a body indicate a rise or fall in its temperature due to thermal energy exchange with the surrounding. So, the body that absorbs more heat from the surrounding becomes hot. The body that releases more heat to the surrounding is cooled. Moreover, it elucidated continuous heat exchange between the body and the surrounding at all finite temperatures.
The hotness and coldness is the thermal state of a body that determines the body's heat flow with the surrounding. Animals curl to protect themselves from cold environments and cloudy winter nights are warmer than clear sky. These realistic observations hinted at an accurate and deep understanding of thermal radiation transferences. Gustav Kirchhoff explained some of such stuff with the assumption of an ideal body. His imaginary black-colored hard solid object played a crucial role in depicting the thermal radiation exchanges by ordinary bodies. It is Kirchhoff's law, our today's blog topic.
Table of contents:
What are Kirchhoff’s thermal radiation laws?
Kirchhoff’s law and
the second law of thermodynamics
Kirchhoff’s experiment and observations
Derivation of Kirchhoff’s radiation laws
Phrases related to Kirchhoff’s law
Applications of Kirchhoff's law
FAQs of Kirchhoff’s law of thermal radiation
Our e-book:
You can download our e-book on blackbody & radiation from our e-book store, "Jayam chemistry adda."
What are Kirchhoff's thermal radiation laws?
Kirchhoff's first radiation law:
It states that the proportion of emissive and absorbing
powers of any material object is constant at every wavelength in thermal
equilibrium. And the emissive power of a perfect blackbody at that wavelength
and temperature gives its value.
Formula: eλ⁄aλ=Eλ
Where,
Eλ =
Emissive power of a perfect blackbody
eλ = Emissive power of an object
aλ = Absorbing power of an object
Explanation:
Kirchhoff's law interprets that the ratio of emissive and absorptive powers of any terrestrial body at wavelength λ in thermal equilibrium conditions is constant. And it is equal to the emissive power of a perfect blackbody at that identical wavelength and temperature conditions.
As a
result, we can calculate the unknown variable of Kirchhoff's formula by putting
the values of the blackbody's emissive power and either absorbing or emissive
capabilities of a material body.
Besides, at a temperature T, the amount of heat radiation
taken or emitted by the body is wavelength-specific.
An ideal blackbody is a fictitious imagination of physicist Gustav Kirchhoff as a complete absorber and emitter of thermal radiation. Hence, its emissive power gives the terrestrial object's absorbing or emissive capabilities.
Moreover, the emissive power of a perfect blackbody is a
universal constant at fixed wavelength and temperature conditions.
Kirchhoff's second radiation law:
The emissivity of a body is numerically equal to its
absorbing power at thermal equilibrium.
Formula: aλ = e
Where,
aλ = Absorbing power of an object at the wavelength λ
e = Emissivity of the object
Explanation:
As discussed earlier, Kirchhoff's imaginary object can
absorb all wavelength light radiations. And it can emit all absorbed radiations
on heating. So, its absorbing power and emissivity are always one.
All other objects' emissivity and absorbing ability remain
below one, and it varies with the nature and composition of the material.
But Kirchhoff's second law proves that any material object
emits all the absorbed light radiations without loss in thermal equilibrium.
Hence, magnitude of absorptivity and emissivity are equal at constant
temperatures, even for terrestrial bodies.
At a thermal equilibrium state, no net transfer of heat and
matter between a system & its surrounding. An object in a closed enclosure
at fixed temperature conditions establishes a thermal equilibrium with its
chamber. Now, the body absorbs and emits equal amounts of thermal radiation to
maintain the thermal equilibrium state. In this way, it obeys the second law of
thermodynamics. Hence, Kirchhoff's second radiation law interprets that the
earthly object's emissivity and absorptivity magnitudes are equal at thermal equilibrium.
Besides the two thermal radiation laws, Kirchhoff elucidated
that perfect absorbers are good emitters and poor absorbers are poor emitters of
thermal radiation.
Kirchhoff's law and the second law of thermodynamics:
The second law of thermodynamics states that there is no net
heat transfer between two bodies at thermal equilibrium.
Kirchhoff's second radiation law proves it. Due to the
body's absorptivity & emissivity being equal at the thermal equilibrium
state.
An existent object kept in the blackbody chamber behaves as
a blackbody at thermal equilibrium. It means the non-ideal object absorbs and
emits equal quantities of thermal radiation at a fixed temperature (T) in the
blackbody enclosure. Consequently, the net heat change is zero in the blackbody
chamber. In this way, Kirchhoff's law obeys the second thermodynamics law.
Kirchhoff's experiment and observations:
Experiment:
Gustav Kirchhoff experimented with a constant temperature
enclosure with two metallic balls. And the two metallic spheres are identical
in shape, size, and nature. One metal sphere's surface is black coated, and the
other one's top is polished white. The two metal balls take some time to attain
the container temperature. After that, the enclosure and the two metallic
spheres are in a state of thermal equilibrium with a fixed temperature of T.
Assumptions:
Generally, the black-colored ball is a good absorber of
thermal radiation. The black ball will intake a large quantum of thermal energy
and becomes hot than the white ball.
Similarly, the white ball is a good reflector of heat and
light. Consequently, it will be a poor absorber of thermal radiation. And it is
colder than the black ball.
Both these assumptions disturb the thermal equilibrium state
of the enclosure and the metallic spheres.
And surprisingly, Kirchhoff did not observe the mentioned
assumptions during his experiment.
Experimental observations:
Both black and white balls expel all absorbed heat radiation
to the enclosure regardless of their absorbing abilities. It maintains a
thermal equilibrium condition in the chamber along with the two metallic
spheres.
Based on these observations, Kirchhoff framed the two
radiation laws.
Discovery of blackbody:
Kirchhoff, during his experiments, found that the ratio of
the absorbing and emissive capacities of material objects are constant at fixed
temperature conditions. To denote their value, he introduced a hypothetical
body in 1860 as a perfect absorber and emitter of thermal radiation and named
it a blackbody.
A blackbody is a solid closed enclosure with an opaque
surface and a pinhole on one of its walls to expel thermal radiation. Besides,
the blackbody can absorb and emit radiations of all wavelengths at thermal
equilibrium conditions. So, the blackbody's absorbing and emissive powers are
higher than other existent objects on earth.
The blackbody detains a fraction of incident light while
absorbing and emitting heat energy at fixed temperature conditions. Heating
disturbs the thermodynamic equilibrium of the blackbody enclosure, where the
blackbody temperature is higher than its surrounding. So, the blackbody expels
all the detained electromagnetic radiation as blackbody radiation. It brings
back the thermal equilibrium state of the blackbody chamber. It is the reason
behind the blackbody radiation emission only on heating.
Derivation of Kirchhoff's radiation laws:
Derivation of Kirchhoff’s first radiation law:
dQ is the amount of heat incident per second per unit area
of both bodies.
dλ is
the incident radiation wavelength that varies between (λ-½)
to (λ+½)
In the case of white metallic sphere:
Energy absorbed per second per unit area of white sphere = aλdQ
aλ is the absorptivity of white sphere at the
wavelength λ.
Energy reflected/transmitted per second per unit area = Total
incident heat – Energy absorbed per second per unit area of the body
Reflecting/transmitting energies of the white sphere = dQ- aλdQ
= dQ (1-aλ)
Energy radiated per second per unit area by the white body = eλdλ
eλ is the emissive power of the white metallic sphere at
the wavelength λ
In thermal equilibrium conditions by the law of energy
conservation,
The amount of heat incident on the body = total amount of heat
radiated from the body
dQ = Energy reflected + energy radiated per second per unit area
of the metallic sphere
dQ = dQ (1-aλ) + eλdλ
On solving the above equation, we get;
dQ = dQ - aλdQ + eλdλ
aλdQ = eλdλ ------------------------------- (1)
In the case of black metallic sphere:
Absorptivity of a perfect blackbody (aλ) = 1
Reflecting/transmitting powers of the blackbody =0
Energy radiated per second per unit area of the blackbody = Eλ dλ
By law of conservation of energy, we have;
dQ = Energy reflected + energy radiated per second per unit area
of the metallic sphere
dQ = 0 + Eλ dλ
dQ = Eλ dλ
By substituting the value of dQ in the above equation (1), we get;
aλ Eλ dλ = eλdλ
Derivation of Kirchhoff’s second radiation law:
Eλ is the emissivity of the blackbody at the wavelength
λ
Eλ is the amount of thermal energy available to the
white sphere for absorption. Being a poor absorber, the white sphere absorbs
less energy than Eλ
The absorptivity of white sphere = aλEλ
Emissivity of white sphere = eEλ
In thermal equilibrium state, the magnitudes of absorbed and
emitted energies remain the same.
aλEλ = eEλ
aλ = e
Phrases related to Kirchhoff’s law:
Absorptivity:
Generally, substances inhale and exhale photons under favorable
conditions. The ratio between the quantities of light absorbed to the total
incident light is called absorptivity. The body's surface that absorbs 100% of
the incident light is a good absorber. And its absorptivity is one.
When a body reflects off a portion of incident light falling on
its surface, its absorptivity lies below one. And these are gray bodies whose
absorptivities vary between 0 and 1.
Likewise, thermal radiations obey the laws of reflection. So, they
get reflected on polished surfaces. It makes polished surfaces poor absorbers
of thermal radiation due to their reflecting abilities. As a result, a shiny
black-colored surface is a poor absorber of thermal radiation than a
black-colored opaque surface. Consequently, a substance is a good absorber only
when its reflecting power is zero.
Emissive power:
At constant temperature T for a particular wavelength λ, the
emissive power is the energy radiated in a vacuum per unit time per unit area
per unit range of wavelength lying between (λ-½) to (λ+½).
A good absorber is always a good emitter. We know a blackbody is a
perfect absorber of all wavelength incident light radiation falling on its
surface. Further, a blackbody is a complete emitter of previously absorbed
light on heating.
Hence, the emissive power of a perfect blackbody is a universal
constant at each wavelength of the electromagnetic spectrum in thermal
equilibrium. It gives the value of the ratio of emissive and absorptive powers
of gray bodies at that wavelength and temperature conditions.
Undoubtedly, we can say poor absorbers are poor emitters of heat
radiation.
Accordingly, when a polished metal with a dark spot on its surface
upon heating to a high temperature in a dark room seems dull. Conversely, the
black speck begins to shine brightly. When heated, the black mark absorbs
radiation, and the shiny metal surface reflects off all radiation.
By Kirchhoff's law, the dark spot emits all absorbed radiations in
the dark room and looks bright. But, the shiny metal became a poor absorber on
heating, hence has low emissive power and seems dull in the dark room.
Thermal equilibrium:
It is the fundamental condition of Kirchhoff's radiation laws. The
magnitude of the heat energy absorbed is equal to that emitted at thermal
equilibrium. And the net heat change is zero. Consequently, both the bodies in
the thermal equilibrium state maintain a fixed temperature. Therefore, we may
feel that the bodies neither intake nor eject heat energy. But they exchange an
equal amount of heat quantum that restricts the heat change.
Additionally, the thermal equilibrium is a dynamic state. So, the
body that is a heat absorber for a while becomes the heat emitter for the next
second. It continues as long as the body is disturbed by an external energy
source. So, an object is said to be in thermal equilibrium despite its heat
energy exchanges until its temperature is constant.
Emissivity:
It is the ratio between the gray body and the perfect blackbody
radiant emittances at temperature T. It defines the gray body's efficacy in
emitting thermal radiation by comparing it with the black body at a particular
wavelength and temperature.
The emissivity of a perfect blackbody is one. So, gray bodies'
emissivity varies between 0 and 1. Gray bodies are poor absorbers and emitters
of thermal radiation compared with black bodies. So, they possess low
emissivities.
As a result, polished silver-coated metal surface emissivity is
less than the lampblack-coated metal surface at the same wavelength and
temperature conditions.
Table-1: Values of absorptivity, emissive power, and emissivity for ideal and non-ideal objects
| Description | Notation | Value | ||
|---|---|---|---|---|
| Phrases relating to Kirchhoff's law | For blackbody | For partial blackbody | For blackbody | For partial blackbody |
| Absorbing power | Aλ | aλ | 1 | 0 to 1 |
| Emissive power | Eλ | eλ | A universal constant at each wavelength in thermal equilibrium state | It depends on Eλ and aλ |
| Emissivity | E | e | 1 | 0 to 1 |
Applications of Kirchhoff's law:
1. Measures thermal electromagnetic energies:
Kirchhoff radiation laws assist in calculating the absorbed and
emitted thermal electromagnetic energies of earthly objects by using the
emissive power of an ideal blackbody. It distinguished the superior and poor
heat energy absorbers and emitters. It aids in heating and lighting purposes.
For example- Cooking utensils design, thermos flasks, china crockery,
decoration materials, etc.
2. In astronomy:
Kirchhoff's law interpreted Fraunhofer's lines of the solar
absorption spectrum by considering the sun as a blackbody. Further, the dark
absorption lines in the outer solar atmosphere form due to thermal energy
absorptions.
3. In spectroscopy:
Exactly the single-positioned spectral lines occurrence in the
emission and absorption spectra of specific element obeys Kirchhoff's law.
According to it, magnitudes of absorptivity and emissivity are equal for
objects at thermal equilibrium. As a result, if we have wavelength data of
spectral lines in either emission or absorption spectrum, we can confirm the
same in the other.
For example- The sodium spectrum shows two intense absorption
lines at 588.99 nm and 589.59 nm. And these spectral lines also occur in the
sodium emission spectrum at the same wavelengths. It confirms that each element
can absorb a specific wavelength of electromagnetic radiation. And it emits the
same wavelength light radiation when stabilized. Consequently, the atomic
spectrum is a fingerprint of the atom.
Examples of Kirchhoff's law:
- Cooking utensils are coated black at the bottom to absorb thermal radiation from the flame to a greater extent. And the shiny upper portion prevents heat loss. It speeds up cooking.
- Sports persons prefer to wear white-colored apparel to reflect the sunlight most effectively. And it keeps them cool.
- Thermos flask's shiny outer walls help to preserve the temperature of fluids inside it as they are poor absorbers and emitters of thermal radiation.
- Decorated china crockery shines brightly in the darkroom on heating to high temperatures. The decorated portions of the utensils are superior heat absorbers. They glow in the dark due to emission of all absorbed thermal radiation.
- Building roofs with white sun proofs reflect sunlight and keep the home cool.
FAQs of Kirchhoff's law of thermal radiation:
1. What are the formulas and units of absorbing power, emissive
power, and emissivity?
Absorbing power:
Absorbing power is the ratio between the amounts of heat absorbed to the total incident heat.
It has no unit. It is a numerical value.
Emissive power:
Emissive power is the quantity of heat radiated in the vacuum per
unit time per unit range of wavelength lying between (λ-½) to (λ+½) at a fixed
temperature of T.
Its SI unit is watts per square meter.
Emissivity:
Emissivity is the ratio of radiant emittances between a body and a
perfect blackbody when they are at thermal equilibrium.
It has no unit. It is a number.
2. Why is daytime hot in desert regions?
Desert sand absorbs heat to a large extent. Hence, it is hot in
the daytime in deserts. At night time, the sand radiates all absorbed heat to
the atmosphere. Consequently, the nighttime is cold in deserts. It is by
Kirchhoff’s law.
3. Why does a hot yellow glass piece appear blue in the darkroom?
A yellow piece of glass, when heated in a dark room, appears blue.
At ordinary temperature, the glass piece looks yellow because it absorbs blue
light and transmits yellow color to a greater extent. While heating, the glass
piece releases absorbed blue light, following Kirchhoff's law. Hence, it is
blue-colored.
4. Two identical metal beakers coated black and white contain the
same amount of hot water of 80 degrees centigrade. Which metal beaker cools
first at room temperature?
Two metallic containers of identical shape and size have alternate
black and white colored surfaces. Additionally, both bowls have an equal
quantity of water heated at 80 degrees centigrade. And they are in room
temperature conditions. Hence, they are not in thermal equilibrium with the
surrounding. Then the water in the black container cools faster. Because we
know the black-coated container is a better heat emitter than the white bowl.
5. How does surface roughness affect emissivity?
Large surface area surfaces are superior absorbers & emitters
of thermal electromagnetic radiation than small surface area surfaces.
Consequently, rough surfaces absorb or emit heat to a greater extent than
polished surfaces due to their large surface area.
6. Does a blackbody emit more energy than it absorbs on heating?
It may sound so, but not correct. A blackbody absorbs
electromagnetic radiation when cold. And it emits thermal electromagnetic
radiation on heating. Following Kirchhoff's second radiation law, the
magnitudes of absorbed and emitted heat quantum are the same in thermal
equilibrium conditions.
7. Does absorptivity and emissivity of a body are equal at a
constant wavelength?
It is false. Following Kirchhoff's law, absorptivity and
emissivity of an object are equal at constant temperature conditions. When the
body is not in thermal equilibrium with the surrounding, the heat flows from
higher temperatures to lower temperature regions. So, their absorptivity and
emissivity magnitudes differ when their temperature is not constant.
8. What is the difference between thermal radiation and blackbody
radiation?
Thermal radiation:
The energy emitted from the body due to its temperature in the
form of radiation is called thermal radiation.
An earthly object can emit thermal radiation.
Thermal radiation follows the law of reflection. And get reflected
from polished surfaces.
Blackbody radiation:
Blackbody radiation is thermal electromagnetic radiation emitted
from a blackbody on heating in thermal equilibrium.
An ideal hot blackbody emits blackbody radiation.
A perfect blackbody's reflecting power is zero.
9. Why do we feel hot in winter when we wear black clothes?
When wearing black clothes, we feel hot in the winter season as
the black color is a good absorber of thermal radiation. And it absorbs all
heat from the surroundings and keeps us warm.
10. When the mercury vapor absorption spectrum shows an intense
blue absorption line at 435.8 nm, Calculate the wavelength of the blue emission
line of mercury using Kirchhoff law of thermal radiation.
Kirchhoff's law states that emissivity and absorbing power of a
body are equal at thermal equilibrium conditions. The wavelength of the
absorption line of mercury is 435.8 nm. Hence, the wavelength of the blue
emission line is 435.8 nm.
11. Does a blackbody surface emit blackbody radiations?
No, the blackbody surface can absorb electromagnetic radiation of
all wavelengths that fall on its opaque surface. The blackbody radiation ejects
from the pinhole on one of the walls of the blackbody enclosure.
12. Why don't the absorbed electromagnetic radiations escape from
the pinhole when the blackbody is cold?
When the blackbody is cold, absorbed electromagnetic radiations
collide with the blackbody's enclosure walls slowly. Consequently, their
chances of escaping from the pinhole are less. On heating, the kinetic energies
of absorbed blackbody radiation increase, and they move fastly. So, their
collisions with the enclosure walls increase enormously. Hence, the fast-moving
thermal photons escape from the blackbody pinhole as blackbody radiation.
13. What is the thermal energy of the system?
Thermal energy is the system's internal energy by virtue of
temperature at thermal equilibrium.
It is the produced heat energy in the system due to the movement
of atoms and molecules.
Examples of thermal energy are;
- Solar energy
- Geothermal energy
- Fuel cell energy
14. Will photons excite when you heat light?
Heat and light are two different energy forms. We cannot heat
light. By the law of conservation of energy, heat converts into light, and
light can convert into heat.
We can heat a material that absorbs or holds light radiations,
such as a blackbody. The supplied thermal energy increases the kinetic energies
of oscillating electrons of the material. Then the electrons excite from a
lower to a higher energy level. The excited state of the electron is unstable.
So, the excited electron jumps back to its original energy state by emission of
thermal electromagnetic radiations or hot photons called blackbody radiation.
Photons are packets of light energy. The physicist Isaac Newton
confirmed their existence through the photoelectric effect and named them.
Further, the energy particles can never excite on the absorption of another
energy form. Only the atomic electrons undergo excitation and de-excitation
processes upon absorption and emission of energy.
15. What are the properties of blackbody radiation similar to that
of thermal radiation?
- Blackbody radiation travels in a straight line.
- It does not require any medium for its propagation.
- Blackbody radiation is isotropic in a homogeneous medium.
- Blackbody radiation does not heat the medium through which it passes.