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I am aware that simple DC circuits use a battery which is connected to several elements. In the case that the element is a resistor, I read that the circuit as a whole is an isolated system for energy and that the chemical potential energy in the battery is converted into internal energy of the resistor ( which appears as increased vibrational motion of the atoms of the resistor), essentially: $$∆U_{chem} = -∆E_{int}$$

My question is what exactly constitutes chemical potential energy in a battery? By definition, a potential energy exists between two particles / sub-systems that interact by means of a conservative force. The only conservative interaction I could think of was between free electrons as a product of redox reactions ( like in the Daniel cell ), but this constitutes to electrical potential energy, so I was unsure.

Secondly I was wondering how this energy was transferred over to the resistor. If I consider the resistor to be the system alone, the non-isolated system for energy would lead to an equation that I believe is: $$T_{ET} = ∆E_{int}$$ where $T_{ET}$ represents transfer variable referring to electrical transmission. I am unsure of what this means, but deduced that this would be the most sensible as the other transfer variables were either Work, Heat, Matter Transfer, Electromagnetic Radiation or Mechanical Waves.

Could someone please tell me how the transfer variable for Electrical transmission would work? (if my analysis is right?)

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The chemical reaction in a battery does work to move electrons from the positive to the negative terminal. The electrons repel each other and dissipate their potential energy to collisions with the atoms within the resistor.

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  • $\begingroup$ Can the work done by the battery be quantified by an expression similar to the definition of work? (as the dot product of Force and the displacement of the point of application of force). Furthermore, could you elaborate on the usage of dissipate? Doesn't that suggest that the energy disappears? $\endgroup$
    – Cross
    Commented May 22, 2021 at 17:39
  • $\begingroup$ In an electrical situation the work is qΔV, and the power is iΔV. The dissipated energy increases the internal energy of the resistor and then flows to the surroundings as heat. When the battery goes dead, its energy is gone. $\endgroup$
    – R.W. Bird
    Commented May 22, 2021 at 18:25
  • $\begingroup$ Yes, as I stated, this work is associated with an electrical conservative force, so doesn't it contribute to the "electrical potential energy"? Is it essentially same as the chemical potential energy then?. I understood how the power delivered to the resistor raises it's internal energy $\endgroup$
    – Cross
    Commented May 22, 2021 at 18:43
  • $\begingroup$ The chemical reaction moves electrons to the negative terminal. In this process, chemical potential energy is used to produce the electrical potential energy. $\endgroup$
    – R.W. Bird
    Commented May 23, 2021 at 13:30
  • $\begingroup$ Okay so is my analogy correct? The chemical potential energy associated with the chemical reactions in the cell converts to electrical potential energy associated with the separation of electrons, and this energy is transferred to the resistor by electrical transmission and stored as internal energy in the resistor? $\endgroup$
    – Cross
    Commented May 23, 2021 at 14:32

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