Thermodynamic and mechanical work

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Thermodynamic work involves energy transfer due to temperature differences and phase changes, while mechanical work is related to forces acting over distances. In experiments, thermodynamic and mechanical work should ideally yield equal values, but discrepancies can arise from measurement errors, system inefficiencies, or assumptions made in calculations. The discussion emphasizes the importance of understanding the definitions and contexts of each type of work. Participants are encouraged to clarify their definitions and explore the implications of any differences observed in experimental results. Accurate comprehension of these concepts is crucial for effective analysis in thermodynamics and mechanics.
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What is the difference between thermodynamic and mechanical work? Should they show equal values in an experiment? If yes, what possible errors could cause if there is a difference in their values?
 
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WazapMe said:
What is the difference between thermodynamic and mechanical work? Should they show equal values in an experiment? If yes, what possible errors could cause if there is a difference in their values?
[thread moved to the schoolwork forums]

Welcome to PF. Please tell us the definitions of each term that you are familiar with, and tell us what differences you can think of. Per the PF rules, we cannot help you with schoolwork-type questions until you show your best efforts to answer the questions. Thank you. :smile:
 
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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