Question about Question 8.7

[Isa Samad 1L]

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]

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]

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

[Nickolas Manipud 1C]

Work is done ON a system when the system's internal energy increases. This makes sense if you think about a system being compressed (which is work being done ON the system). Since it's being compressed, more pressure is created, leading to more internal energy in that system.