Meaning of Entropy Increase – What Does It Signify? An increase in the entropy of a system (or universe) represents degradation of energy quality and a reduction in the maximum useful work (availability/exergy) obtainable.

Difficulty: Easy

Correct Answer: degradation of energy

Explanation:


Introduction / Context:
Entropy is a measure of energy dispersion and irreversibility. When entropy increases, less of the system's energy can be converted into useful work. Understanding this connection is vital in power-plant, refrigeration, and process design.


Given Data / Assumptions:

  • Second-law framework with system and surroundings.
  • Availability/exergy defined relative to an environment state.
  • Irreversibilities generate entropy and destroy exergy.


Concept / Approach:

Entropy generation S_gen > 0 accompanies friction, mixing, unrestrained expansion, finite-ΔT heat transfer, and chemical irreversibility. Exergy destruction is T0 * S_gen (per unit time or process), where T0 is ambient temperature. Thus, higher entropy implies lower capacity to do work, i.e., degraded energy quality.


Step-by-Step Solution:

State second law: ΔS_universe = S_gen ≥ 0.Relate exergy loss to entropy generation: X_dest = T0 * S_gen.Interpret physically: more disorder/dispersion → less extractable work.Conclude: entropy increase represents energy degradation (lower availability).


Verification / Alternative check:

Turbine with high irreversibility yields larger S_gen and lower actual work versus isentropic ideal; heat exchanger with large temperature differences generates entropy and reduces potential to do work.


Why Other Options Are Wrong:

Availability does not increase with entropy; temperature/pressure changes are not direct indicators of entropy sign; potential energy is unrelated to entropy definition.


Common Pitfalls:

Equating entropy strictly with 'disorder' without linking to work potential; assuming any temperature rise means entropy rise (not necessarily true).


Final Answer:

degradation of energy

More Questions from Thermodynamics

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion