Kirchhoff's law formula and its derivation. Numerical problems

1. A N I N F O R M A T I O N G U I D E Kirchhoff's law of thermal radiation Kirchhoff's law formula and explanation with numerical problems By, Jayam chemistry adda

2. Kirchhoff's 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. = emissive power of a body At a temperature T, the amount of heat radiation taken or emitted by the body is wavelength-specific. = absorbing power of a body 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. = emissive power of a blackbody https://jayamchemistrylearners.blogspot.com/

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. https://jayamchemistrylearners.blogspot.com/

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. https://jayamchemistrylearners.blogspot.com/

5. Derivation of Kirchhoff's formula 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: 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, https://jayamchemistrylearners.blogspot.com/

6. 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 https://jayamchemistrylearners.blogspot.com/

7. By law of conservation of energy, we have; dQ = Reflecting energy + energy radiated per second per unit area of the metal sphere dQ = 0 + dλ dQ = dλ By substituting dQ value in the equation no.(1.1), we get; dλ = dλ 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. https://jayamchemistrylearners.blogspot.com/

8. Problem-1 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? By Kirchhoff's law, the formula for emissive power of a blackbody is; We have, = 500 = 250 watts/ square meter =250/500=1/2=0.5 watts per square meter https://jayamchemistrylearners.blogspot.com/

9. Problem-2 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? By Kirchhoff's law, the formula for emissive power of a blackbody is; We have, = 270 watts/ square meter = 130 watts/ square meter = 130/270 = 0.48148 https://jayamchemistrylearners.blogspot.com/

10. Problem-3 A sodium bulb has an absorptivity of 0.6 at 42 degrees Celsius. What is its emissivity at the same temperature? The absorptivity of 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. The emissivity of the sodium bulb=0.6 https://jayamchemistrylearners.blogspot.com/

11. Problem-4 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 e = 2.25/0.75 = 3 https://jayamchemistrylearners.blogspot.com/

12. 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. https://jayamchemistrylearners.blogspot.com/

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