I have a question related to a definition (intuitive understanding). It is stated that :

The human body radiates energy as infrared light. The net power radiated is the difference between the power emitted and the power absorbed: $$ P_{net}=P_{emit} - P_{absorb} $$

I can of course understand that we are just subtracting 2 quantities. But why do we call that "net power radiated"? why not simply use the emitted power? What does the subtraction really tell us from the thermodynamic point of view?


1 Answer 1


It tells you the rate that radiant energy is leaving (or entering) the system in total.

At any given time, the body will be both absorbing, and emitting radiation. By just looking at what is emitted or absorbed on their own, you would not be able to determine how much thermal energy is leaving or building up in the system from a net perspective.

The net power will allow you to determine how much the body will heat up or cool down when exposed to that radiation. Neither absorption nor emission on their own would tell you that.

  • $\begingroup$ thanks for your answer. Ok so if I understand correctly, if a body would be emitting more than it is absorbing then the body's temperature is decreasing. Conversely, if it is absorbing more than emitting it's temperature is increasing. right? $\endgroup$
    – John Doe
    Commented Aug 22, 2018 at 15:36
  • $\begingroup$ If we can ignore conduction and convection, yes. $\endgroup$
    – JMac
    Commented Aug 22, 2018 at 16:08
  • $\begingroup$ Ok thanks ! yes let's assume that (just like I remember always neglecting friction in my Physics mechanics classes) $\endgroup$
    – John Doe
    Commented Aug 22, 2018 at 16:16
  • $\begingroup$ Also this makes me think of the idea of computing how much calories are burnt by a "standard" human being during a day (but maybe that's the topic of another question?) Wiki computes it as about 8 MegaJoules /24h based on 100W emission of infrared radiation (which implies that most of our calories are burnt by emitting IR radiation!?) $\endgroup$
    – John Doe
    Commented Aug 22, 2018 at 16:18

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