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Why is it that no heat flowing through the conductor is absorbed by any cross section when the conductor is in a steady state , there was a question on PSE about the same concept , but the answers provided the reason that it is because heat influx in a time $dt$ , in a given cross section is equal to heat efflux in the same time $dt$ , I would want to know why the heat influx and efflux are the same , that wil answer my question effectively.

link to the referred question In heat conduction, what does it actually mean to be in the steady state?

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Energy must be conservered, so thermal energy cannot just disapear. If part of the conductor was absorbing a net amount of heat, its temperature would be increasing, which means you are not in a steady state.

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  • $\begingroup$ I expected that answer , so you say that a steady state is an experimental fact , is there a mathematical proof , that a conductor can reach a so called 'steady state' , if I am wrong? $\endgroup$ Feb 8, 2019 at 15:30
  • $\begingroup$ To determine whether a given system reaches a steady state you need more information about your setup and boundry conditions than you have given in your question. If you want to know more about the conditions underwhich a steady state is reached, I would ask a new question (or better find an existing one) as that discussion is probably too broad for the comment section, and is really a different subject to your current one. $\endgroup$ Feb 8, 2019 at 15:52
  • $\begingroup$ @AdityaPrakash Steady-state is technically never reached. It's actually a limit case as time approaches infinity. However, in practical situations, it is common for a system to approach close enough to this steady state condition that one cannot measure the difference between it and the steady state case. An example might be a piece of metal heated to glowing orange and allowed to cool in a room whose temperature is 21C. Technically the metal does not reach 21C, but eventually it will reach 21.000000000001C, and you likely don't have tools to measure the difference. $\endgroup$
    – Cort Ammon
    Jan 9 at 13:38
  • $\begingroup$ Steady state is nice because many of the equations are much simpler. If it's accurate enough, we can use it. $\endgroup$
    – Cort Ammon
    Jan 9 at 13:38

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