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I am familiar with isolated systems. They don't interact with the environment and no energy transfer takes place across the boundary of the system. My textbook says,

In an isolated system, if there are only conservative forces acting, its mechanical energy (potential & kinetic) is conserved. While, total energy (potential, kinetic, & internal energy) of an isolated system is conserved if there are non-conservative forces involved.

Is it always the case? Does total energy always remain conserved and no energy transfer takes place across the boundary of an isolated system?

Suppose a book is sliding on a rough horizonal surface. At $t$ = $t_0$, no force acting on it other than friction. It stops after some time. I choose the book and the surface it was sliding on, as my system. It is obviously an isolated system because it is not interacting with the environment (no energy transfer taking place across its boundary).

When the book is sliding, it has some kinetic energy, which reduces to zero when it stops. It's potential energy does not change because the surface is horizonal. The textbook says that its kinetic energy transforms into the internal energy of the book-surface system because both get warm, but total energy of the system (kinetic + potential + internal) remains conserved. The book uses this energy conservation equation

$\Delta{K} + \Delta{E_{int}} + \Delta{U}$ = $0$

I follow it up to this point. What I don't understand is, the book says no energy transfer takes place across the boundary of an isolated system. After the book and the surface get warm, they cool down after some time. Where does the internal energy of the system go? If it transfers to the environment, doesn't it contradict that no energy transfer takes place across the boundary of isolated systems?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    May 27, 2020 at 11:25

3 Answers 3

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After the book and the surface get warm, they cool down after some time. Where does the internal energy of the system go?

It goes further into both bodies. The added internal energy is generated at the surface and then moves by conduction to the farther regions of the bodies. As the energy becomes more dilute, the temperature decreases.

Isolated system is an idealized concept. In practice, the book and the flat body it was sliding on are never isolated system. Some energy is transferred to the outside, across boundary of the system.

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  • $\begingroup$ As the energy become more dilute, the temperature decreases It makes sense, but over time the system cools down completely. It means the energy is eventually transferred to the environment. If this is the case, is there a system that is isolated? Because anything that heats up always cools down, no matter how long it takes $\endgroup$
    – 4d_
    May 27, 2020 at 12:07
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    $\begingroup$ No system is perfectly isolated. It is an idealization. $\endgroup$ May 27, 2020 at 20:11
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What I don't understand is, the book says no energy transfer takes place across the boundary of an isolated system. After the book and the surface get warm, they cool down after some time. Where does the internal energy of the system go? If it transfers to the environment, doesn't it contradict that no energy transfer takes place across the boundary of isolated systems?

The problem is you have not defined the boundary between your "isolated system" and the surroundings (the environment). If heat is able to transfer away from the book and floor to somewhere else, then the book plus the floor can not constitute an isolated system. What about the air in contact with the book and floor? You haven't isolated that from your system. What about what is below the floor? Your system is not isolated from that either.

Let's take your book and floor and locate it in a room that is evacuated of air and contains no furnishings. Let all the walls of the room be complete (no openings), rigid, and be considered perfect thermal insulation (allow no heat transfer). Let the sub-floor (materials below the floor), also be rigid and be perfect thermal insulation. Now the book and the floor can be considered an isolated system. When the book comes to a stop, the kinetic energy lost will result in heat transfer from the contacting surfaces of the book and floor to the interior of the book and floor, increasing their internal energy accordingly, so that total energy is conserved.

Hope this helps.

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  • $\begingroup$ Thanks a lot. It means anything that I see around myself can not constitute an isolated system. Even if I consider the Earth-ball system (when a ball is dropped from a height), potential energy of the system does change into kinetic energy. But when ball hits the surface, all of its kinetic energy transfers away into the surroundings in the form of mechanical waves, heat, sound, etc. This makes the Earth-ball system unisolated. But the set-up you described is an isolated system, but it's practically impossible to find anything like that around us, maybe in a laboratory. Thanks. Very helpful $\endgroup$
    – 4d_
    May 28, 2020 at 3:02
  • $\begingroup$ perhaps I can say that the Earth-ball system is isolated over a certain time interval, but even then I will have to ignore the air drag $\endgroup$
    – 4d_
    May 28, 2020 at 4:14
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    $\begingroup$ @πtimese All valid points. There is no such thing as an isolated system. There's no such thing as perfect thermal insulation, systems of only conservative forces (there is always friction, even in space), reversible processes and thermal reservoirs, etc.. Even the earth is not an isolated system (constantly exchanging radiation and particles). However, the "isolated system" is still a useful construct. If nothing else it establishes boundaries or limits on what is theoretically achievable. Finally your point on time is correct. For limited time we can approach isolation. $\endgroup$
    – Bob D
    May 28, 2020 at 13:25
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Some terms seem to be a bit circular. If energy leaves a system that proves it is not completely isolated. However describing such a system as isolated may be a good enough approximation over a short time.

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