Question 99: Thermodynamics & KTG

Question

In a dark room with ambient temperature $\mathrm{T}_0$, a black body is kept at a temperature T . Keeping the temperature of the black body constant (at T), sunrays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are correct?

Accepted Answer

Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.
The reason given in question is self explanatory. With an increase in temperature, the entire Plank's curve shifts upwards and hence radiation in any spectrum will increase. Reflected energy per unit time will be zero since black body has zero reflectivity and hence it will remain constant.
The bodies which radiate energy in the form of photons can be determined using Kirchoff's
law of radiation. When these photons reach another surface, they may be absorbed, reflected or transmitted. Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.
Therefore, the quantity of radiation absorbed by the black body in unit time will increase, since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increases, Black body radiates more energy in unit time in the visible spectrum, the reflected energy in unit time by the black body remains same.
Hence, the option (A), (B), (C) and (D) are the correct answer.

Answer

The quantity of radiation absorbed by the black body in unit time will increase.

Answer

Since emissivity $=$ absorptivity, hence the quantity of radiation emitted by black body in unit time will increase.

Answer

Black body radiates more energy in unit time in the visible spectrum.

Answer

The reflected energy in unit time by the black body remains the same.

Thermodynamics & KTG

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