Work and power for a square thermodynamic cycle

[Effect]

Homework Statement

An engine goes through a "square" circuit process (is btw "circuit process" this the correct term in English?) according to the diagram below:

a) How much work is performed by the engine during one revolution in the circuit process?
b) What effect does the engine deliver if it takes 0.2 s for the circuit process to go through one revolution?

Homework Equations

W = p*V
P = W*t

The Attempt at a Solution

For a) we have simply the area within the square, which is 1.6 kJ.

For b) if we have W = 1.6 kJ and t = 0.2 s, then we should get P = 1.600 J * 0.2 s = 320 W. Why is this not correct?

 

[andrevdh]

You should divide, not multiply - power is the rate at which work is done by the engine.

 

[Effect]

Ah yes, of course. Thanks.