Physics · Thermal Properties Of Matter · NEET
Emissivity (e) is a pure number from 0 to 1. It compares how much a real surface radiates to how much a perfect black body radiates at the same temperature. A perfect black body has e = 1 (best emitter). A perfect reflector has e = 0. Human skin has e about 0.97, so it emits almost like a black body. Power radiated becomes E = e sigma A T^4, where sigma = 5.67 x 10^-8 W m^-2 K^-4.
Absorptivity (a) is the fraction of radiation falling on a surface that the surface absorbs (rest is reflected or transmitted). It is also a number from 0 to 1. Emissivity is about giving OUT radiation; absorptivity is about taking IN radiation. Kirchhoff's law links them: at the same temperature and wavelength, e = a. So they are two views of the same property of the surface.
Kirchhoff's law says: at a given temperature, the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a perfect black body. A short way to say it: a good absorber is a good emitter (e = a). This is why the inside of a hollow cavity behaves like a black body, and why dark surfaces both heat up fast in sunlight and cool fast at night.
Think of a body in a closed box at fixed temperature (thermal equilibrium). To stay at the same temperature it must emit exactly as much energy as it absorbs. A surface that absorbs a lot must therefore also emit a lot, otherwise it would keep heating up or cooling down. That balance forces emissivity to equal absorptivity. So black (high absorber) = high emitter; shiny silver (low absorber) = low emitter.
Use E = e sigma A T^4 for total power emitted, where T is the absolute (Kelvin) temperature. If you want NET rate of heat exchange with surroundings at temperature T0, use P_net = e sigma A (T^4 - T0^4). Always convert Celsius to Kelvin first by adding 273. Example: for skin at 307 K, area 1.5 m^2, e = 0.97, surroundings 293 K, plug in T = 307 and T0 = 293 in P_net = e sigma A (T^4 - T0^4).
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Emissivity has no unit. It is a pure ratio between 0 and 1, comparing a real surface to a perfect black body at the same temperature.
A perfect black body has emissivity e = 1. It absorbs all radiation falling on it and emits the maximum possible radiation at its temperature.
By Kirchhoff's law they are equal at the same temperature and same wavelength when the body is in thermal equilibrium. This equality e = a is the core result used in NEET problems.
Black paint has high absorptivity, so it absorbs more sunlight and heats up more. By Kirchhoff's law it also emits more, but during the day absorption dominates, so the black car gets hotter.
For a black body power is sigma A T^4. For a real body multiply by emissivity: E = e sigma A T^4. Since e is less than 1, a real body radiates less than a black body at the same temperature.
