From first law of thermodynamics Q=ΔU+W or ΔU=Q−W ∴ ΔU1=Q1−W1=6000−2500=3500‌J ΔU2=Q2−W2=−5500+1000=−4500‌J ΔU3=Q3−W3=−3000+1200=−1800‌J ΔU4=Q4−W4=3500−x For cyclic process ΔU = 0 ∴ 3500 - 4500 - 1800 + 3500 - x = 0 or x=700‌J Efficiency, η=