A system undergoes a cycle of three steps and returns to its initial state. It absorbs $300 J$ of heat in step 1, releases $120 J$ in step 2 and absorbs $40 J$ in step 3. The net work done by the system in the cycle is

Answer: A

For a cycle

Δ U = 0

, so

q_{n e t} = w_{b y}

q_{n e t} = 300 - 120 + 40 = 220 J

, hence the system does

220 J

of net work. Adding all three heats as positive gives

460 J

, and

140 J

results from subtracting step 3 instead of adding it.