An ideal black - body at room temperature is thrown into a furnace. It is observed that? An ideal black - body at room temperature is thrown into a furnace. It is observed that?
(A) initially it is the darkest body and at later times the brightest.
(B) it is the darkest body at all times.
(C) it cannot be distinguished at all times.
(D) initially it is the darkest body and at later times it cannot be distinguished.
Can someone explain me which option is correct and why? I feel that a black body must initially be darkest as it absorbs all energy incident.
 A: The question doesn't say exactly what is meant by darkest, but it seems reasonable to interpret the brightness as the intensity of radiation from the object. In that case the relevant equation is the Stefan-Boltzmann law:
$$ J = \varepsilon\sigma T^4 $$
where $\sigma$ is the Stefan-Boltzmann constant and $\varepsilon$ is the emissivity. The emissivity of a black body is unity by definition, but for objects that are not black bodies the emissivity is less than one and can in principle be arbitrarily small.
If we can treat the furnace as a black body then its emissivity is unity and the brightness of both objects is then simply related to the temperature, which should make the question easy to answer.
If the emissivity of the furnace is less than one then life becomes more complicated, but note that the temperature of the black body and the furnace must end up equal given enough time.
A: The answer may depend on the context, given that it looks like a typical review question, the kind you find at the end of chapters in textbooks. 
As a question about thermodynamic equilibrium and Kirchhoff's law of thermal radiation, the answer would be D:
One can assume that the furnace was at thermodynamic equilibrium at temperature T, just before the ideal black body was thrown in. We know that for an object in the furnace, the emitted power is  $\varepsilon\sigma T^4$, that the absorptivity $\alpha$ is equal to the emissivity $\varepsilon$, and that at thermodynamic equilibrium, the absorbed power is equal to the emitted power. If the absorbed power is $\varepsilon\sigma T^4$ and only a fraction $\alpha = \varepsilon$ of incoming radiation is absorbed, then the reflected radiation must be $(1-\varepsilon)\sigma T^4$ and the total radiation (emitted + reflected) coming from the object is $\sigma T^4$, which is the black-body radiation at temperature T. 
With the radiation from objects in the furnace equal to the black body radiation at the same temperature, it's clear that a cold black body will be darker, and that when it reaches the same temperature, it will be indistinguishable. 
When asked before learning about Kirchoff law or thermodynamic equilibrium, answer A would be reasonable. A black body has higher emissivity than a non-black body, so at the same temperature, it will be brighter than the rest. 
Not necessarily wrong, some furnaces have quite large openings, that could reduce the amount of reflected radiation enough to show a difference. Assuming decent (forced convection, gas burner) heating to compensate the radiation losses...
