Second Law of Thermodynamics - Refrigerators

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SUMMARY

The discussion centers on the application of the Second Law of Thermodynamics to a scenario involving heat transfer between thermal reservoirs at temperatures of 300 K and 500 K. The Coefficient of Performance (COP) is calculated as 1.5, leading to a minimum work requirement of 6666.6 J to transfer 10000 J of heat. The net change in entropy is computed but yields an incorrect result, indicating a misunderstanding of the system's nature. The multiplicity factor is incorrectly stated as e^(0/k), which is 1, highlighting a need for clarification on the entropy calculations.

PREREQUISITES
  • Understanding of the Second Law of Thermodynamics
  • Familiarity with the Coefficient of Performance (COP) in thermodynamic systems
  • Knowledge of entropy calculations and their implications
  • Basic principles of heat transfer between thermal reservoirs
NEXT STEPS
  • Study the derivation and implications of the Coefficient of Performance (COP) in refrigeration cycles
  • Learn about entropy changes in irreversible processes and their calculations
  • Explore the concept of multiplicity in statistical mechanics and its relation to thermodynamic systems
  • Investigate the conditions under which the Second Law of Thermodynamics applies to heat transfer scenarios
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Students of thermodynamics, engineers working with refrigeration systems, and anyone interested in the principles governing heat transfer and entropy in thermal systems.

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Homework Statement



Suppose 10000 J is transferred between a large thermal reservoir at a temperature of
300 K to another large thermal reservoir at 500 K. What is the net change in entropy
of the system? Write an expression for the factor by which the multiplicity of the
system changes.

For this system, what is the minimum amount of work required to transfer the energy
between the two reservoirs, and under what conditions will this calculation be valid?

Homework Equations



COP = Qc/W = Qc/(Qh-Qc) = Tc/(Th-Tc)

DeltaS = -(Qc/Tc) + (Qh/Th)

W = Qc/W

Qh = Qc + W


The Attempt at a Solution



COP = k = 300/(500-300) = 1.5
Qc = 10000
W= 10000/1.5 = 6666.6
Qh = 10000 + 6666.6 = 16666.6

Therefore

DeltaS = -(10000/300) + (16666.6/500)
= -33.3333 + 33.3332

This doesn't seem right to me but I don't know why.

Also this would give me a multiplicity increase by a factor of e^(0/k) which is 1. So that surely can't be right.

I apologise if this is really simple and I'm making stupid mistakes.
 
Physics news on Phys.org
This is not a refrigerator problem. This is just a direct transfer of heat from a hot reservoir to a cold reservoir. The OP, for some reason, got the idea that it is a reservoir problem. No.
 

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