Finding entropy change when house leaks heat

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SUMMARY

The discussion centers on calculating the change in entropy when heat leaks from a house. Given the internal temperature of 19.2° C and an external temperature of 10.7° C, the heat loss of 5.49 kJ results in a calculated entropy change of 0.01935 kJ/K. However, the correct approach requires considering both the entropy loss inside the house and the entropy gain outside, leading to a total change in entropy of 0.56293 J/K, which was confirmed as correct by the Webassign system.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy.
  • Familiarity with the first law of thermodynamics, including the equations W = QH - QC.
  • Knowledge of temperature conversion between Celsius and Kelvin.
  • Ability to perform calculations involving heat transfer and entropy changes.
NEXT STEPS
  • Study the principles of entropy in thermodynamics.
  • Learn about heat transfer mechanisms and their impact on entropy.
  • Explore the concept of reversible and irreversible processes in thermodynamics.
  • Investigate the application of the second law of thermodynamics in real-world scenarios.
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Students studying thermodynamics, engineers working on HVAC systems, and anyone interested in understanding heat transfer and entropy changes in physical systems.

rubenhero
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Homework Statement


b) The temperature inside a house is Tin = 19.2° C when the temperature outside is Tout = 10.7° C. If Q = 5.49 kJ of heat leak from the house to the outside, find the change in entropy caused by this process.


Homework Equations


W = QH - QC, dS = dQ/T


The Attempt at a Solution


dS = 5.49kJ/283.7K = .0193514276kJ/K

The answer I calculated was wrong, should I include the entropy inside the house?
Any help is appreciated!
 
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Hi rubenhero! :smile:

rubenhero said:

Homework Statement


b) The temperature inside a house is Tin = 19.2° C when the temperature outside is Tout = 10.7° C. If Q = 5.49 kJ of heat leak from the house to the outside, find the change in entropy caused by this process.


Homework Equations


W = QH - QC, dS = dQ/T


The Attempt at a Solution


dS = 5.49kJ/283.7K = .0193514276kJ/K

The answer I calculated was wrong, should I include the entropy inside the house?
Any help is appreciated!

Inside you lose ΔS=-Q/TH.
Outside you gain ΔS=Q/TL.

The question in your problem statement is not entirely clear, but I suspect the total change in entropy is intended, which is the sum of both entropy changes.
 
Thanks you for responding. You were right, I just did the sum of the entropy loss and gain and got an answer of .5629265733 J/K. Webassign marked my answer as right. Thanks again for your help!
 
You're welcome! :)
 

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