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?