From first law of thermodynamics Q=∆U+W or ∆U=Q−W ∴∆U1=Q1−W1=6000−2500=3500J ∆U2=Q2−W2=−5500+1000=−4500J ∆U3=Q3−W3=−3000+1200=−1800J ∆U4=Q4−W4=3500−x For cyclic process ∆U=0 ∴3500−4500−1800+3500−x=0 or x=700J Efficiency, η=