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Kirchhoff's law formula and its derivation. Numerical problems.pdf
1. Kirchhoff's law of
thermal radiation
A N I N F O R M A T I O N G U I D E
Kirchhoff's law formula and explanation with numerical problems
By,
Jayam chemistry adda
2. 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.
Kirchhoff's law
= emissive power of
a body
= absorbing power
of a body
= emissive power of
a blackbody
At a temperature T, the amount of heat radiation taken
or emitted by the body is wavelength-specific.
Thermal equilibrium is the primary condition of Kirchhoff's
law. In thermal equilibrium conditions, net heat flow is zero.
The magnitude of the absorbed heat energy is equal to
the emitted. Hence, both bodies are at a constant
temperature.
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3. Experimental setup of Kirchhoff's law
Gustav Kirchhoff experimented in a constant
temperature enclosure with two metallic balls. And
two metallic spheres identical in shape, size, and
nature are placed in the closed vessel at a fixed
temperature of T. One metal sphere's surface is
black coated, and the other one's top is polished
white color. 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 at constant
temperature T.
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4. The black ball is a good absorber that takes a high quantum of radiation than the white
ball. So, the black ball should be hotter than the white one. But, it did not happen. We
know the perfect absorbers are good emitters. Consequently, the black sphere emits all
absorbed energy to the box in temperature fluctuations. It maintains thermal equilibrium
in the closed vessel.
In thermodynamic equilibrium, there is no net transfer of matter or energy within or
between the systems. Therefore the blackbody does not emit radiations at the
equilibrium state.
Heating a blackbody disturbs its thermodynamic equilibrium condition. So, the blackbody
emits thermal electromagnetic radiation on heating to attain thermal equilibrium again.
It is the reason behind the blackbody radiation emission only on heating in the
thermodynamic equilibrium state.
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5. 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).
For white metallic sphere:
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Derivation of Kirchhoff's formula
Energy absorbed per second per unit area of white sphere = dQ
is the absorptivity of the 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 - dQ = dQ (1 - )
Energy radiated per second per unit area by the white body = dλ
is the emissive power of the white metallic sphere at the wavelength λ
In thermal equilibrium conditions by the law of energy conservation,
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The amount of heat incident on the body = total amount of heat radiated from the
body
dQ = Reflecting energy + energy radiated per second per unit area of the metal sphere
dQ = (1 - ) dQ + dλ
On solving the above equation, we get;
dQ = dQ - dQ + dλ
dQ = dλ ----------------(1.1)
For black metallic sphere:
Absorptivity of a perfect blackbody ( ) = 1
Reflecting/transmitting powers of the blackbody=0
Energy radiated per second per unit area of the blackbody = dλ
7. dQ = 0 + dλ
dλ = dλ
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By law of conservation of energy, we have;
dQ = Reflecting energy + energy radiated per second per unit area of the metal sphere
dQ = dλ
By substituting dQ value in the equation no.(1.1), we get;
Cancelling dλ on both sides, we get;
It is Kirchhoff's law of thermal radiation that is valid for all non-ideal bodies that
exist.
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Problem-1
By Kirchhoff's law, the formula for emissive power of a blackbody is;
A metal beaker has absorbing power of 500 and emissive power of 250 watts per
square meter at 30 degrees Celsius and 400 nm. What is the emissive power of a
blackbody at the same temperature and wavelength?
=250/500=1/2=0.5 watts per square meter
We have,
= 500
= 250
watts/
square
meter
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Problem-2
By Kirchhoff's law, the formula for emissive power of a blackbody is;
A blackbody has emissive power of 270 watts per square meter at 35 nm in thermal
equilibrium. What is the absorptivity of a copper ball if it has emissive power of 130
watts per square meter?
We have,
= 130/270 = 0.48148
= 270 watts/
square meter
= 130 watts/
square meter
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Problem-3
The absorptivity of sodium bulb= 0.6
A sodium bulb has an absorptivity of 0.6 at 42 degrees Celsius. What is its emissivity
at the same temperature?
The emissivity of the sodium bulb=0.6
As the temperature of the sodium bulb is constant, it is in a thermal equilibrium
state. By using Kirchhoff' s law, we can say that;
The absorptivity and emissivity of a body are numerically equal in thermal
equilibrium conditions.
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Problem-4
e = 2.25/0.75 = 3
What is the emissivity of an iron rod if it has an emissive power of 2.25 watts per
square meter? The emissive power of a blackbody is 0. 75 watts per square meter at
the same temperature and wavelength conditions.
The formula to calculate the emissivity of the iron rod is;
E' = 2.25 watts
per square
meter
E = 0.75 watts
per square
meter
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Problem-5
A gold piece has an emissivity of 0.625 at 1200 degrees centigrade and 254 nm. Then
what is the emissivity of a perfect blackbody at the same temperature and
wavelength conditions?
A perfect blackbody is a good absorber and emitter of electromagnetic light at all
wavelengths in thermal equilibrium conditions.
Hence, its emissivity and absorptivity values are always equal to one.
The emissivity of a perfect blackbody is one. It is the answer.
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law, Don't worry.
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thermal-radiations
