Question

A spherical body of area A and emissivity e=0.6 is kept inside a perfectly black body. Total heat radiated by the body at temperature T?
A. 0.4A ${{T}^{4}}$
B. 0.8A ${{T}^{4}}$
C. 0.6A ${{T}^{4}}$
D. A ${{T}^{4}}$

Answer 1

Blackbody is defined as a body which absorbs all the radiations which falls on it. A black body when heated also emits radiation in all the directions. Emissivity is defined as the ratio of the energy radiated from a body to that of radiated from a blackbody. Since, blackbody is a perfect emitter, if the ratio for any body comes out to be one, then that body is also a blackbody.

Complete step by step solution:
To stay in thermal equilibrium, a blackbody must emit radiation at the same rate as it absorbs it. The rate of emission is given by Stefan Boltzmann law, which states that the energy radiated by a body is proportional to the fourth power of its temperature. Writing it mathematically,
$P=\sigma A{{T}^{4}}$, this is for a black body, for any other body it is given by,
$P=\sigma eA{{T}^{4}}$
Since emissivity is given as 0.6, area as A and temperature as T, then in this case it will be given by,
$P=0.6\sigma A{{T}^{4}}$
Assuming the constant to be 1, then it will be given as $P=0.6A{{T}^{4}}$

So, the correct option is C.

Note: If a system is thermally isolated then it is not possible for heat to move from outside into the system and vice versa.While solving such kind of problems we have to keep in mind that all temperature readings have to be put in Kelvin. But if there is change in temperature then there is no issue of converting the given unit into kelvin. Because the difference always remains the same in any unit.