Gift D.

asked • 11/27/21

A thermodynamic system performs work and at the same time exchanges energy to the outside environment

A thermodynamic system performs work and at the same time exchanges energy to the outside environment. There is no additional help or changes in terms of fuel supply or an increase in the temperature of the system, The work done by the system on the environment is 8,900 J while the change in the internal energy is 3,727 J. How much heat is exchanged between the system and the environment? 

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Grigoriy S. answered • 11/28/21

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Dear Gift! I am just wondering: do you really think that this is a hard question to answer, or you just want to test the expert.?

Let's assume that you do not like to open the textbook and your teacher did not tell you about The First Law of Thermodynamics ( although I am very doubtful about it).

Then let me tell you about " something" that you can find on your own in any decent physics textbook.

The First Law of Thermodynamics is just the energy conservation principle applied to heat, work, and the change in the internal energy.

According to this law: The hit added to the system, Q, is equal to the sum of the work done by the system, W, and the change in its internal energy, ΔU:


Q = W + ΔU


Hope the rest you are capable of handling on your own. Sorry for taking so much of your time for explanation.

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Gift D.

Thank you for your help. Contrary to what you think, I don't think it's a hard question, I know how to use my textbook, and I know the first law of thermodynamics. I just wanted to know if the answer would be 12,627 (W(8900) + delta U (3727)) or 5173 (W(-8900) + delta U (3727)) since the system dies work on the environment and not the other way around. What would be the point of having platforms like this when the "experts" question why students ask their questions (because the expert thinks its easy enough and they should be able to do it on their own). Your response is simply condescending. Considering your explanation didn't include anything I didn't already know, there's no point in hoping I can "handle the rest on my own". Thank you for taking the time to respond.
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11/28/21

Grigoriy S.

tutor
Sorry l just wanted to tease you. The question you asked is much more deeper than everyone thinks. Unfortunately, the confusion began when many textbook rewrote the First law in a bad form. I know personally famous teacher from Texas (she is a reader of AP exam students papers), who told me that she was the one, who insisted in rewriting the law in a form that you now see in official table for AP physics 2 exam. I am very unhappy with this twist. In theoretical physics we write dQ = dU+ dW. For layman person it is better to write delta Q is the sum of change in internal energy and WORK DONE BY THE SYSTEM. I strongly advocated this form. When we create heat engine, we want work done by gas, and want it to be the greatest. Then the efficiency of the engine is the greatest. We want minimum waste of energy in form of internal energy. This is why for us it is important to know how much work would produced, if we give the gas some Q, heating it. This is our main goal! Thank you very much for your question. Free to ask me other questions, and I will be more than happy to answer them. Mr. Grigoriy Shustef
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11/28/21

Grigoriy S.

tutor
Dear Gift! I was not trying to offend you. On the contrary, I wanted to bring everybody's attention to this at first glance "simple" question. As a professional, I am very unhappy that some "smart" people from College Board removed from AP Physics 2 course the core of the thermodynamics curriculum - heat engines. All laws of the thermodynamics make sense only when we have practical use of them. Sadi Carnot invented his cycle for ideal heat engine and showed that it has maximum efficiency. The reason for this, because it is using only isothermal and adiabatic prosses. During those processes we have maximum work DONE BY ENGINE, because any change in gas volume leads to useful work. Conclusion: we need more smart students like you. You understood that there is work DONE BY GAS (and it is positive) and WORK DONE ON GAS (that is negative). Because of the Newton's third law their values are the same, just opposite signs! Thank you again for your digging into the bottom of the problem!!!
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11/28/21

Gift D.

Thank you so much for the details provided. I appreciate it
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11/29/21

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