# Why is work done by the system in a Carnot engine?

In a Carnot engine, represented above, it is said that work is done by the system, but in reality , work is done by the system only in the isothermal and adiabatic expansions (from 1 to 3 in the picture above) then work is done on the system, as it contracts in an isothermal and adiabatic process (3 to 1). So why do we say that work is done only by the system in this engine, as the above pictures suggests, does it refer only to the first 2 processes (the expansions)? And the same goes to the reverse process, the refrigerator engine, where it is said that work is done on the system, and obviously in that case i still don't understand why it is said that work is done on the system.

• It refers to a full cycle, so, globally, it does work, altough it's only doing work in two periods. – FGSUZ Jan 11 at 21:22
• The net work from all four segments of the cycle is not zero. It is equal to the area inside the "parallelogram." – Chester Miller Jan 11 at 21:29
• If you calculate the individual works in all 4 processes, add them all together, you'll get a total work $> 0$ (i.e., out of the system). – Drew Jan 12 at 0:04