Solution:
Solution:
From first law of thermodynamics
$$Q =\Delta U +W or \Delta U = Q -W $$
$$ \therefore \Delta U_{1} = Q_{1} -W_{1} = 6000 - 2500 = 3500 J $$
$$ \Delta U_{2}= Q_{2} -W_{2} = - 5500 +1000 = - 4500 J $$
$$ \Delta U_{3} = Q_{3} -W_{3} =-3000 + 1200 = - 1800 J $$
$$ \Delta U_{4} = Q_{4} - W_{4} 3500- x $$
For cyclic process $$ \Delta U = 0 $$
$$ \therefore 3500 - 4500 - 1800 + 3500 - x = 0 $$
or x = 700 J
Efficiency, $$ \eta = \frac{\text{output}}{\text{input}} \times100 $$
$$ = \frac{W_{1} + W_{2} +W_{3} +W_{4}}{Q_{1} +Q_{4}} \times100 $$
$$ = \frac{\left(2500 - 1000 -1200 + 700\right)}{6000+3500} \times100 $$
$$ = \frac{1000}{9500} \times100 = 10.5 \% $$