Entropy: Heat addition to surrounding.

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SUMMARY

The entropy change of the surroundings during a cooling process where 12007 kJ of heat is lost is calculated using the formula Δs=ΔQ/T. With an ambient temperature of 25 degrees Celsius, the entropy change is determined to be 40.272 kJ/K. The calculation is confirmed as correct, demonstrating a clear understanding of the thermodynamic principles involved.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy.
  • Familiarity with the formula Δs=∫δQ/T for calculating entropy changes.
  • Knowledge of temperature conversion from Celsius to Kelvin.
  • Basic grasp of heat transfer concepts in thermodynamics.
NEXT STEPS
  • Study the implications of entropy changes in various thermodynamic processes.
  • Learn about the second law of thermodynamics and its relation to entropy.
  • Explore real-world applications of entropy in refrigeration and heat engines.
  • Investigate the relationship between heat transfer and entropy in closed systems.
USEFUL FOR

Students of thermodynamics, engineers working with heat transfer systems, and anyone interested in the principles of entropy and its applications in cooling processes.

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If 12007 kJ of heat is lost to the surroundings with an ambient temperature of 25 degrees centigrade during a cooling process, and the ambient temperature of the surroundings is unaffected by the heat addition, what is the entropy change of the surroundings?

If Δs=∫δQ/T, then Δs=ΔQ/T=12007 kJ/(25+273.15)K= 40.272 kJ/K.

Is my thinking process here correct?

Thanks!
 
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afpskierx said:
If 12007 kJ of heat is lost to the surroundings with an ambient temperature of 25 degrees centigrade during a cooling process, and the ambient temperature of the surroundings is unaffected by the heat addition, what is the entropy change of the surroundings?

If Δs=∫δQ/T, then Δs=ΔQ/T=12007 kJ/(25+273.15)K= 40.272 kJ/K.

Is my thinking process here correct?

Thanks!
Yes.
 

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