About *Carnot's cycle* and *engine*

Question

we know the Carnot's cycle diagram, so A, B, C, D are points regarding the state of the system. Consider a piston with weight on it. -----AB is just making the piston go up Q = W by providing heat no change in internal energy (this is done by lowering some weight on the top of it ) -----BC is stopping the heat supply and lowering some more weight so the pressure decreases as the work is done by the expense of internal energy w = -du -----CD is putting the system on a cooler surface adding the weight we took out in BC so the work was done on the system and the change in internal energy due to that work was converted into heat to the reservoir (cold one) so the heat you gave in AB was given back ----DA: Now you add some more weight the same as you took away in AB this is more work done on the system and the internal energy rises to the same number as it was at point A hence completing the cycle the energy was transferred from the hot reservoir to the cold one and nothing else happened this is was our whacky way to deliver energy and nothing else

What im saying is that total work is zero, also our book gives this equation Q1(heat in) - Q2(heat out) = W, but here Q1 = Q2 so total work is zero.

But in the figure the total work is the area under the curve, please explain what that area under the curve says.

Answer 1

It isn't only that you are adding and removing mass from the top of the piston that matters. What also matters is the piston elevation at which the mass is being added and removed.

If m represents the amount of mass on the (massless) piston at any stage of the process, then the pressure is related to m by: $$P=\frac{mg}{A}$$where A is the area of the piston. Similarly, if h represents the elevation of the piston at any start of the process, then the volume of gas is related to h by: $$V=Ah$$. So the mass on the piston is directly related to the pressure, and the elevation of the piston is directly related to the gas volume.

Next, we're going to focus on the changes in potential energy of the surroundings as we add and remove mass from the piston. The change in potential energy of the surroundings when the amount of mass on the piston decreases by a differential amount dm is just $$d(PE)=-ghdm$$But, from the previous equations, this is equal to $$d(PE)=-g\left(\frac{V}{A}\right)\left(\frac{A}{g}dP\right)=-VdP$$

If we integrate this equation for the potential energy change of the masses in the surroundings over an entire Carnot cycle, we obtain $$\Delta (PE)=nR(T_H-T_C)\ln{\frac{V_B}{V_A}}=W =Q_H-Q_C$$