Cyclic processes: Over one complete thermodynamic cycle, what is the net change in internal energy (ΔU) of the system?

Introduction / Context:
Cycles are central to engines, refrigerators, and power plants. The first law of thermodynamics applies to cycles with special simplicity because internal energy is a state function. Recognizing the implication for ΔU over a complete cycle avoids conceptual and calculation mistakes.

Given Data / Assumptions:

• System returns to its initial thermodynamic state at the end of the cycle.
• Internal energy U is a state function depending only on state, not path.
• Heat and work over the cycle can be nonzero individually.

Concept / Approach:

The first law for a cycle is ΣQ − ΣW = ΔU_cycle. Since the initial and final states are identical, U_final − U_initial = 0 ⇒ ΔU_cycle = 0. Consequently, the net heat added over a cycle equals the net work done by the system (with sign convention consistency), but internal energy does not accumulate over a cycle.

Step-by-Step Solution:

Verification / Alternative check:

On a P–V diagram, the cyclic integral ∮ δQ − ∮ δW = 0. Internal energy change vanishes even if the enclosed area (net work) is nonzero, consistent with heat–work balance.

Why Other Options Are Wrong:

Positive or negative values would imply a different final state. 'Path dependent' is false for state functions.

Common Pitfalls:

Confusing process path contributions (Q and W, which are path functions) with state functions (U, H, S) when evaluating cycles.

Final Answer:

zero