Net work output of an engine performing Carnot cycle?

My problem gives me a Carnot cycle heat engine with water as its working fluid, with $T_H$, $T_L$, and the fact that it starts from saturated liquid to saturated vapor in the heating process.

I need to find the net work output of this engine. So my solution goes like this:

$$\eta_{HE} = 1 - \frac {T_L}{T_H}$$

So I get $\eta_{HE}$. And then I also know that

$$\eta_{HE} = \frac {W_{HE}}{Q_H}$$

Since my goal is $W_{HE}$, I know I'll need to solve for $Q_H$ first. Since I have the state and temperatures of water at both states, I can get $h_H$ and $h_L$. I know that $h_L - h_H$ is some quantity of energy, and it must be either $Q_H$ or $Q_L$. The problem is, I don't know which it is. And I assume it is $h_L - h_H$ because it is heated first and then cooled. I hope this is correct.

I know that the change in enthalpy is the energy either released or absorbed by the system. In this case, since the temperature moves from higher to lower, it seems to me that the system is absorbing energy and hence this must be $Q_H$.

But it also makes sense to me to say that $h_L - h_H$ is the energy released by the system (as in an exothermic reaction), so it can be $Q_L$ too, because that is the heat rejected by the system.

So my question is, for the general case, which is it? Or is it situational? Does its being $Q_H$ or $Q_L$ depend on the resulting sign when I do my calculations? For this problem, the sign of $h_L - h_H$ that I computed is positive. Then this looks like heat entering the system and so it is $Q_H$. Am I correct?

Or, am I completely on the wrong track and should be chasing down another solution?

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Also, I know I tag this as homework (as is the practice on this site for homework-y questions), but it's not actually graded homework. I'm just doing problems. –  markovchain Mar 27 '13 at 4:35

Since the question was posted 10 months ago, I guess that the OP has already forgotten about it. Anyway, somebody might still be interested in it. So here's the explanation:

$$\eta_{HE} = \frac {W_{HE}}{Q_H}$$
is correct. It is worth noting that $W_{HE}$ is Your net work output and $Q_H$ is the value of heat acquired from heat source (not the difference between the heat acquired and heat released - this one is exactly equal to $W_{HE}$).
Usually $h$ means specific enthalpy, so if $h_L$ and $h_H$ mean the values of specific enthalpy before and after heating respectively, $m(h_H-h_L)$ is Your value of heat acquired from the heat source $Q_H$, where $m$ is the mass of the working fluid that performs the Carnot cycle. Specific enthalpy at the end of heating process will always be higher than at the beginning, so $m(h_H-h_L)$ is always going to be positive.
If, as You said, You are able to obtain the values of $h_L$ and $h_H$ this solution will work regardless of the starting and end points parameters - specific enthalpy is a definite description of thermodynamic state.