$\Delta U=q+w$

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For question 8.7, how does one differentiate work being done on a system vs. work being done by a system? Also, how does one calculate work once this distinction has been made?

Jaewoo Jo 2L
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Internal energy increased more than the amount of heat added, so the extra energy is from the work done to the system.
Since the equation to calculate work is "w= deltaU-q", plug in the values of change in internal energy and heat absorbed given in question to get "982 J - 492 J = +4.90*10^2 J

Christy Zhao 1H
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On the other hand, if work was done by the system, the internal energy of the system would have decreased.

Nickolas Manipud 1C
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