From first law of thermodynamics Q=ΔU+W or ΔU=Q−W ∴ ΔU1=Q1−W1=6000−2500=3500JΔU2=Q2−W2=−5500+1000=−4500JΔU3=Q3−W3=−3000+1200=−1800JΔU4=Q4−W4=3500−x For cyclic process ΔU = 0 ∴ 3500 - 4500 - 1800 + 3500 - x = 0 or x=700J Efficiency, η=inputoutput×100=Q1+Q4W1+W2+W3+W4×100=6000+35002500−1000−1200+700×100=95001000×100=10.5%