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enter image description here

What it says is: “The two cans emit different quantities of energy, in the same time interval, even when filled with the same liquid and at the same temperature.”

What I wrote below as an explanation is that the black can absorbs more incident EM radiation and so, in turn, it must also emit more radiant energy in order to be at the same temperature as the silver can. I reckon it simply “just does”. It’s a darker object and so it both emits and absorbs more EM energy than the other. (And its emitted and absorbed wavelengths/frequencies are exactly the same.) Supposing they are both in thermal equilibrium with the surroundings, one being blackish and the other being silverish only means there are more exchanges to and from the dark one than to and from the light one. But the net exchange is equal in both - null.

What about the heat? They are both receiving EM energy and also heat energy from the surroundings. I suppose they both exactly “receive and give” the same heat from and to the surroundings, that is, the total heat “giving + getting” (or just “giving” or just “getting”) is the same in both objects. (And “giving” equals “getting”).

Comparing the two objects: The net exchanges of the two objects are the same, both of heat and EM radiation - zero. The total exchange of heat is the same. The total exchange of EM radiation is “bigger” in the case of the darker object. |If in equilibrium with the environment and between themselves|

Am I right? I know this is a trivial question. I apologize in the case I wasted your time. And also for any mistake I might have made posting the question and/or with the way I referred to things, the terms used and such.

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  • $\begingroup$ The concept of emissivity may be helpful to you here. $\endgroup$ – probably_someone Apr 3 '18 at 1:11
  • $\begingroup$ As you put it it seems correct. But as soon the environment is say, warmer than the liquid, than the black one should warm faster. There is no contradiction between the caption and your explanation. $\endgroup$ – Alchimista Apr 3 '18 at 7:54

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