Entropy of a Refrigerator: Is It Constant?

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SUMMARY

The discussion centers on the entropy behavior of a refrigerator during a thermodynamic cycle, specifically addressing a question from an AP Physics exam. The key conclusion is that while heat is removed from the gas, the entropy remains constant because the gas returns to its initial state. This is supported by the principle that entropy is a state variable, meaning it depends only on the state of the system, not the path taken. The equation governing this concept is the change in entropy, defined as ΔS = Q/T, where Q represents heat flow.

PREREQUISITES
  • Understanding of thermodynamic cycles and state variables
  • Familiarity with the concept of entropy in thermodynamics
  • Knowledge of the first law of thermodynamics
  • Ability to interpret and apply the equation ΔS = Q/T
NEXT STEPS
  • Study the principles of thermodynamic cycles in detail
  • Explore the concept of state variables in thermodynamics
  • Learn about the implications of the first law of thermodynamics on energy conservation
  • Investigate the role of heat flow in thermodynamic processes
USEFUL FOR

Students preparing for AP Physics exams, educators teaching thermodynamics, and anyone interested in understanding the principles of entropy in refrigeration systems.

turdferguson
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Homework Statement


Ive got a question regarding a recent AP free response, http://www.collegeboard.com/prod_downloads/ap/students/physics/b_physics_b_frq_03.pdf
Problem 5 is about a simple refrigerator, work is applied and heat is removed keeping temperature constant.

My question is on part (e). I think heat being removed from the gas would cause entropy to decrease. However, the answer key states that because the gas returns to the same state, entropy is the same. Who is right and if it is constant, why?

Homework Equations


change in entropy = Q/T


The Attempt at a Solution


Less heat, less entropy
 
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turdferguson said:

Homework Statement


Ive got a question regarding a recent AP free response, http://www.collegeboard.com/prod_downloads/ap/students/physics/b_physics_b_frq_03.pdf
Problem 5 is about a simple refrigerator, work is applied and heat is removed keeping temperature constant.

My question is on part (e). I think heat being removed from the gas would cause entropy to decrease. However, the answer key states that because the gas returns to the same state, entropy is the same. Who is right and if it is constant, why?

Homework Equations


change in entropy = Q/T

The Attempt at a Solution


Less heat, less entropy
This is one of the conceptual problems that occurs because we talk about heat and not heat flow. We now think of heat as energy but in thermodynamics we use terminology from the mid 1800s as if it were a fluid flowing through a substance. In thermodynamics, Q is heat flow.

In one cycle, heat is added to the gas (AB) then work is done on the gas (BC) and then heat is removed from the gas (CA). The internal energy of the gas is the same when it returns to A. The energy that is removed as heat from C to A consists of the heat added from A to B + the work done in compressing from B to C. Qh = Qc + W, where Qh is the heat delivered to the hot reservoir. Since Qh > Qc, there is heat removed (better to say: a greater flow of heat out of the gas than the heat flow into the gas). However this does not mean that there is energy removed. The energy at A is the same.

Since entropy is a state variable, the entropy of the gas is the same from A back to A. The entropy of the system (consisting of the gas + the hot and cold reservoirs) continually increases, however.

AM
 

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