Q5 on the midterm


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Mary Becerra 2D
Posts: 53
Joined: Fri Sep 29, 2017 7:06 am

Q5 on the midterm

Postby Mary Becerra 2D » Sat Mar 17, 2018 5:45 pm

In step 1 we find that the work done by the system is -158J. How can we make the conclusion that in step 2, our U = +158J? I thought that since there is no heat exchange, w=U.

Jonathan Tangonan 1E
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Re: Q5 on the midterm

Postby Jonathan Tangonan 1E » Sat Mar 17, 2018 7:33 pm

The delta U is +158J because in step 2 it returns to its original internal energy. So, the work the system had done which was -158J returns back to its original state by saying delta U of the second step is +158J.


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