Solution for (a)
The net heat transfer is the heat transfer into the system minus the heat transfer out of the system, or
Q=40.00 J -25.00 J =15.00 J. (15.3)
Similarly, the total work is the work done by the system minus the work done on the system, or
W=10.00 J -4.00 J =6.00 J. (15.4)
Thus the change in internal energy is given by the first law of thermodynamics:
\Delta U=Q-W=15.00 J -6.00 J =9.00 J. (15.5)
We can also find the change in internal energy for each of the two steps. First, consider 40.00 J of heat transfer in and 10.00 J of work out, or
\Delta U_{1}=Q_{1}-W_{1}=40.00 J -10.00 J =30.00 J. (15.6)
Now consider 25.00 J of heat transfer out and 4.00 J of work in, or
\Delta U_{2}=Q_{2}-W_{2}=-25.00 J -(-4.00 J )=-21.00 J. (15.7)
The total change is the sum of these two steps, or
\Delta U=\Delta U_{1}+\Delta U_{2}=30.00 J +(-21.00 J )=9.00 J. (15.8)
Discussion on (a)
No matter whether you look at the overall process or break it into steps, the change in internal energy is the same.
Solution for (b)
Here the net heat transfer and total work are given directly to be Q = – 150.00 J and W = – 159.00 J , so that
\Delta U=Q-W=-150.00 J -(-159.00 J )=9.00 J. (15.9)
Discussion on (b)
A very different process in part (b) produces the same 9.00-J change in internal energy as in part (a). Note that the change in the system in both parts is related to ΔU and not to the individual Q s or W s involved. The system ends up in the same state in both (a) and (b). Parts (a) and
(b) present two different paths for the system to follow between the same starting and ending points, and the change in internal energy for each is the same—it is independent of path.
