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For more information, please see full course syllabus of High School Physics
For more information, please see full course syllabus of High School Physics
High School Physics Thermodynamics
Lecture Description
So, we’ve covered the basics of temperature and heat and what happens to gases when temperature rises. Now we’ll end this topic with all the official parts of thermodynamics. There are two laws that we’ll focus on today, neither of which should get too confusing. This is a video that you may have to watch again as the concepts introduced aren’t easily picked up by everyone. Make sure that you understand inside and out what is being said in this video as it’s more conceptual than mathematical. After that, you will have successfully completed one of the final topics in physics!
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1 answer
Thu Mar 3, 2016 2:38 PM
Post by Jayden Luis on March 3, 2016
Why do we put our % oriented backwards?
1 answer
Thu Dec 4, 2014 5:39 PM
Post by Oscar Gil on December 2, 2014
Hello,
I have a question
Five moles of an ideal monatomic gas with an initial temperature of 135∘C expand and, in the process, absorb an amount of heat equal to 1140J and do an amount of work equal to 2040J .
whats the final temp...
I used the first law of thermo. find the delta U is -900
but a solution manual is saying to use deltaU=n(3/2 R) delta T after cause of it being monatomic gas....
I thought the formula was Q=nCdeltaT
and since Q was given I thought maybe to plug it in to that orignal formula
2 answers
Last reply by: javier chichil
Mon Oct 7, 2013 7:59 AM
Post by javier chichil on October 5, 2013
last question: what kind ofscreen you write on? looks so cool. :D
3 answers
Mon Oct 7, 2013 9:34 AM
Post by javier chichil on October 5, 2013
hi agan.
i am working to take a test, and that is why i am active trying to learn as much as i can.
on minute 14:01 there is a formula about efficiency involving temperatures, while on a different reference, they use heat (Q) instead than temperatures.
are they interchangeable concepts ?
thanks
4 answers
Last reply by: javier chichil
Mon Oct 7, 2013 7:50 AM
Post by javier chichil on October 5, 2013
Hi Vincent:
love your way to explain, so clear. just one question, it seems to me that in this example, work is done on the environment ( the system does work towards outside the system) and the sign of the work is positive. however, trying to understand the topic better, saw Dan's explanation and in his, he says that Work is positive if the work is done over the gas, and negative if he gas is the one who is doing work on the environment.
now, i am a little bit confused with the sign conventions. could you please clarify?
thanks
[EDIT BY PROFESSOR: The above brings up a good question about how the sign on work (W) goes. In the answers, you can find a _very_ detailed explanation, but in short, the convention for the sign of work (W) varies from teacher to teacher and textbook to textbook.
In my course, I treat work in Thermodynamics as being positive when the system does work on the outside environment (puts energy into the environment). Similarly, work is negative when the environment does work on the system (puts work into the system).
Notice that the above is the OPPOSITE of what we originally learned when we first studied work. In general, the work of a system is considered positive when it gains energy, but this often gets flipped specifically for Thermodynamics. However, not every teacher/book flips the convention, so you have to pay attention to whomever you're learning from.
If you want more information and discussion on the subject, read the much longer reply I (the professor) have in the answers to this question.]