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1

You have not specified how the pressure is controlled in the two systems. If they are each at the triple point pressure of 611.73 Pa there is no reason for heat to exchange and all will stay constant. If the pressures are different from this (and not on the freezing curve) energy can be released if there is heat flow by transferring heat between the ...


0

The original state is not in thermodynamic equilibrium. The liquid side has a lot more heat than the solid side due to the heat of fusion.


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Equilibrium is defined in the original notions of thermodynamics as the asymptotic static state. I.e., by this definition no macroscopical quantity varies in equilibrium. Statistical physics however tells us that the system varies in a certain "random walk" around all the possible states and never stops. We just cannot distinguish most of these states ...


2

I'll assume you want to know the equilibrium points. The Lagrangian tells you everything you need to know about the system. Because variation of generalized momentum is: $$ \frac{dp_k}{dt} = Q_k + \frac{\partial T}{\partial q_k} = -\frac{\partial V}{\partial q_k}+ \frac{\partial T}{\partial q_k} = \frac{\partial L}{\partial q_k} $$ Then: $$ ...


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First of all, let me note that it is misleading to say that we're not even dealing with a particle We are still dealing with a particle, but the state of the particle at a given moment in time is no longer described by a pair $(x(t),p(t))$ of position and momentum in a classical phase space, it is instead a state $|\psi(t)\rangle$ in a Hilbert space. ...


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The final state you propose is admitted by conservation of angular momentum and energy (assuming that $I_B = I_A$), so we're not sunk in the first case. Now lets consider the intermediate cases so see if the transfer is possible all the way through. Simplest case: a passive transfer with no loss to or storage in the transmission itself. We'll take ...



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