A thermodynamic system undergoes a cycle composed of a series of three processes for which Q1 = +10 kJ, Q2 = +30 kJ, Q3 = -5 kJ. For the first process, DE = +20 kJ, and for the third process, DE = -20 kJ. What is the work in the second process, and the net work output of the cycle?

Calculations:

Q + W = DE

Work done in the first process = 20 - 10 = 10 kJ (i.e., work is done on the system)

Work done in the third process = -20 - (-5) = -15 kJ

For a cyclic process, the overall internal energy change is zero. (i.e., DE = 0)

Therefore, DE in the second process = (0 - (20 - 20)) = 0 kJ

Therefore, work done in the second process
= 0 - 30 kJ = **-30 kJ**.

Total work done during the cycle = 10 +
(-15) + (-30) = **-35 kJ** (i.e., 35 kJ of
work is done by the system).

Total work done can also be calculated as follows:

Total Q = Q1 + Q2 + Q3 = 10 + 30 - 5 = 35 kJ

Therefore, net work done during the cycle = 0 - 35 kJ = -35 kJ (the negative sign indicates that work is done by the system).