Question

In a cyclic process, heat transfers are + 14.7 kJ, – 25.2 kJ, – 3.56 kJ and + 31.5 kJ. What is the net work for this cyclic process?

Answer 1

∑Q = (14.7 + 31.5 - 25.2 - 3.56) kJ 

= 17.44 kJ

From first law of thermodynamics (for a cyclic process)

∑Q = ∑W

\(\therefore\) ∑W = 17.44 kJ