What happens to the heat transfer and the internal energy in a system that undergoes reversible adiabatic process back to its original state.
Any internal energy increase due to net heat or work into the system prior to the reversible adiabatic process decreases during the adiabatic process due to work out, so that the internal energy of the system returns to its initial state following the cycle.
The following explanation is offered.
The three reversible processes you described in your comments are shown in the Figure below.
Heat is added in the isochoric process (1-2) and no work is done. Internal energy is therefore increased. Heat is rejected in the isobaric process (2-3), which reduces internal energy. However, work is done on the system, which increases internal energy.
The adiabatic process (3-1), which returns the system to its initial state, is an expansion. Since $Q=0$, the expansion work done by the system comes at the expense of its internal energy, i.e., a causes a decrease in internal energy.
Since the complete cycle 1-2-3-1 returns the system to its initial state, the total change in internal energy is zero. This tells us that there was an increase in internal energy going from 1-2-3 equal in magnitude to the decrease in internal energy in going from 3-1 as a result of the work done in the adiabatic process.
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