Classical thermodynamics problem.

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The discussion revolves around a thermodynamics problem involving two identical masses of water at different temperatures, T1 and T2, and the calculation of the global increase in entropy after they reach thermal equilibrium. The initial formula proposed for entropy change was incorrect, leading to confusion. A key point is to consider the equilibrium temperature and the entropy change for each mass of water. After some guidance, the participant realized the correct approach to solve the problem. The conversation highlights the importance of understanding thermal equilibrium and entropy calculations in thermodynamics.
Kalimaa23
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Hm, I'm pretty much stuck on a thermodynamics problem.

If you consider two identical masses of water at temperatures T1 and T2, what is the global increase in entropy after they have reached thermal equilibrium.

The best I could come up with is [del]S = m*C*ln(T2/T1).

The answer listed in the book is 2m*C*ln[(T1+T2)/2Sqrt(T1*T2)]

I am completely and utterly stuck. Any help would be greatly appreciated.

-Dimi
 
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This will probably be moved to HWK. help, but think about what the equilibrium temperature is, and the change in Entropy of each block.
 
Yes, that's it. I've got it. Thanks.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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