How Does Entropy Explain Heat Transfer Between Two Einstein Solids?

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SUMMARY

The discussion centers on the heat transfer between two identical Einstein solids, one in boiling water and the other in ice water, using the concept of entropy. The total change in entropy is expressed as deltaS(total) = deltaS(A) + deltaS(B), where deltaS is calculated using the formula deltaS = deltaQ/T. The key conclusion is that heat will spontaneously transfer from the hotter solid (A) to the colder solid (B), as the total entropy change must be positive, aligning with the second law of thermodynamics. The initial misunderstanding regarding the sign of deltaS(total) was corrected, emphasizing that entropy cannot be negative.

PREREQUISITES
  • Understanding of the second law of thermodynamics
  • Familiarity with the concept of entropy in thermodynamics
  • Knowledge of differential calculus for applying dS and dQ
  • Basic principles of heat transfer between systems
NEXT STEPS
  • Study the second law of thermodynamics in detail
  • Learn about the mathematical formulation of entropy changes
  • Explore the concept of heat transfer in thermodynamic systems
  • Investigate the implications of non-constant temperature on entropy calculations
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Students studying thermodynamics, physics educators, and anyone interested in understanding heat transfer and entropy in thermodynamic systems.

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Homework Statement


Consider two identical Einstein solids. One solid has been placed in boiling water and the other in ice water. Then the two solids are placed in good contact with each other within an insulated box. Using the concept of entropy, prove that heat will spontaneously transfer from the hotter block to the colder block. Explanations that use concepts other than entropy will receive zero credit. Make sure the logic of your explanation is clear. Equations (and perhaps graphs) should make up the bulk of your answer. Use only a minimum number of words.



Homework Equations


deltaS(total)=deltaS(A)+deltaS(B), where A is the hot solid and B is the cold solid
deltaS=deltaQ/T, where Q is energy due to heat transfer, and T is temperature



The Attempt at a Solution


deltaS(total)=deltaS(A)+deltaS(B)=deltaQ(A)/T(A) + deltaQ(B)/T(B)
deltaQ(A)=-deltaQ(B)
T(A) much greater than T(B)
-so deltaS(total) will be negative, which implies that the heat should transfer from the higher entropy object to the lower entropy object
-this is what I have so far, but I turned it in and got zero credit... any suggestions? I'm wondering if I need to use the second law of thermodynamics somewhere...

Thanks!
 
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How does the total change in entropy imply what is happening with the individual bodies, let alone how is it negative? And I do suggest the second law of thermodynamics, especially since you broke it.

Also, you might not want to use delta, since the temperature isn't constant through out the whole process. So it is best to use differentials, such as dS and dQ.
 
Awesome, thanks!

Yeah, I realized yesterday that the deltaS(total) couldn't be negative, since entropy can't be destroyed, so that was a slip on my part. Thanks for pointing it out!
 

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