From [1]

First law of thermodynamics: By $\delta Q$, I denote the quantity of energy added to the system by a heating process; by $\delta W$, I denote the quantity of energy lost by the system due to work done by the system on its surroundings; and by $\mathrm{d}U$, I denote the change in the internal energy of the system [final minus initial I assume]. For a closed system in thermodynamics, the first law of thermodynamics may be stated as: $$\delta Q = \mathrm{d}U + \delta W.$$

I've often struggled to know to define the system. So here in for pedagogic purposes I'm intending to work out an example. I'd ask the community to find any kinks in my understanding and help me sort them out.

Example By $m_1$ and $M_1$, I denote two spherical masses, where $M_1 \gg m_1$. By $r_{mM}$, I denote the distance between the two masses. By $t_i$ and $t_f$, I denote an initial and final time, respectively. Let $\left.r_{mM}\right|_{t_f} < \left.r_{mM}\right|_{t_i}$. Allow that within the time interval from $t_i$ to $t_f$ that $\delta Q=0$.


(1) Can you clearly identify what the system is?

(2) Can you determine if $\delta W$ is greater than or less than 0, and explain how come?

(3) Can you determine if $dU$ is greater than or less than 0, and explain how come?

  • $\begingroup$ What are your assessments of the answers so far? $\endgroup$ – Chet Miller Feb 8 '20 at 15:03
  • $\begingroup$ The system is always what you define it to be. Nothing is inherently "a system". $\endgroup$ – Bob D Feb 8 '20 at 15:39

The more general form of the first law of thermodynamics is appropriate for this problem: $$\Delta (U+KE+PE)=Q-W$$where KE is the organized kinetic energy of the system and PE is its potential energy. For the combined system consisting of the two masses interacting gravitationally, $$\Delta U=Q=W=0$$So the first law reduces, for this system, to $$\Delta (KE+PE)=0$$

  • $\begingroup$ I like your answer. However, can you explain the following? There is a force, for example, acting on the smallest mass. The smallest mass displaces considerably towards the larger mass. The direction of force on the smaller mass and the direction of displacement of the smaller mass are parallel. Work should be positive. Do you agree or disagree? Does this inform your answer in any way? Does your answer need to be expanded at all? $\endgroup$ – Michael Levy Feb 8 '20 at 16:43
  • $\begingroup$ If the combination of the two masses is your system, then no external work W is done on this system. If you want to apply the first law to each mass individually, that is OK too, but, recognize that, in the first law, W represents non-gravitational work. So, for each mass, using an inertial frame of reference situated at the center of mass of the two-masses system, $\Delta (KE+PE)=0$. And, of course, the sum of the two equations for the two masses is also equal to zero. $\endgroup$ – Chet Miller Feb 8 '20 at 17:12

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