Why Entropy of Carnot Engine is 0 & Heat Transfer in Refrigerator

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SUMMARY

The entropy of a cyclic Carnot engine is zero because the heat transfer occurs when the system and surroundings are at the same temperature, resulting in no net entropy change. The formula for total entropy change, dS = dS_{sys} + dS_{surr}, confirms that if the temperatures are infinitesimally close, the entropy change is zero. Additionally, in a refrigeration cycle, Q_h represents the heat flow from the hot reservoir, which is negative, while Q_c is the positive heat flow from the cold reservoir.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the Carnot cycle.
  • Familiarity with the concept of entropy in thermodynamics.
  • Knowledge of heat transfer mechanisms in refrigeration cycles.
  • Basic mathematical skills to interpret thermodynamic equations.
NEXT STEPS
  • Study the Carnot cycle in detail, focusing on its efficiency and implications for thermodynamics.
  • Learn about the mathematical derivation of entropy changes in thermodynamic processes.
  • Explore refrigeration cycles, including the roles of Q_h and Q_c in heat transfer.
  • Investigate real-world applications of Carnot engines and refrigeration systems in engineering.
USEFUL FOR

Students of thermodynamics, mechanical engineers, and professionals involved in HVAC systems or energy efficiency research will benefit from this discussion.

AznBoi
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Can someone explain why the entropy of a cyclic carnot engine is equal to 0?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Also in the refigerator section, why is Q_h coming into the heat reservoir when Q_c is the one being transferred??
 
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AznBoi said:
Can someone explain why the entropy of a cyclic carnot engine is equal to 0?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Entropy is 0 because the heat flow occurs when the system and surroundings are at the same temperature.

[tex]dS_{sys} = dQ_{sys}/T_{sys}; dS_{surr} = dQ_{surr}/T_{surr} = - dQ_{sys}/T_{surr}[/tex]

The total entropy change is:

[tex]dS = dS_{sys} + dS_{surr} = dQ_{sys}\left{(}\frac{1}{T_{sys}} - \frac{1}{T_{surr}}\right{)}[/tex]

So if the system and surrounding temperatures are infinitessimally close while heat flows, there is no entropy change. [There is no heat flow during the reversible adiabatic expansion and compression so there is no entropy change during the adiabatic processes (ds=dQ/T = 0/T = 0).]
Also in the refigerator section, why is Q_h coming into the heat reservoir when Q_c is the one being transferred??
A refrigeration cycle takes heat from the cold reservoir and delivers it to the hot reservoir. So Qh (the heat flow from the hot reservoir) is negative and Qc (the heat flow from the cold reservoir) is positive.

AM
 
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