# Where heat goes in and out for a thermodynamic cycle? (ideal cycle and cycles in general)

I'm trying to work out the efficiency for this cycle. I've found out the work and heat for all legs of the cycle because they tell me how to do it in the lecture notes. I get that the efficiency of the thermodynamic cycle is $\frac{Work}{Q_{in}}$ which is $1-\frac{Q_{out}}{Q_{in}}$.

But I am unable to figure out where the heat goes in or out for any cycle. Can someone please tell me how to figure this out, and how I can tell just by looking at the cycle, because all the questions ask me where the heat goes in and out before any of the calculations. Right now all my Work and Heat calculations are in terms of volume and pressure.

Are there any formulas I should know to help me get the efficiency in terms of temperature?

I'll assume that the working substance is an ideal gas.

You need to make use of the First Law of Thermodynamics:

$$Increase\ in\ internal\ energy\ of\ gas = Heat\ into\ gas\ – Work\ done\ by\ gas$$ That is $$\Delta U = Q - W$$

I'll explain what to do in A and B and leave C and D to you.

In A, $W$ is clearly positive ($p$ and $\Delta V$ both positive). But $\Delta U$ is also positive because there must be a temperature rise in order for the volume to increase at constant pressure ($pV=nRT$). Therefore, from the First Law, $Q$ is positive, that is heat must flow in. This should make sense: the heat input allows the gas both to gain internal energy and to do work.

In B no work is done, but the internal energy falls ($\Delta U$ negative). This is the only way the gas pressure could fall at constant volume ($pV=nRT$). Therefore, from the First Law, $Q$ is negative – heat must flow out.

Hope you get the idea. Don't forget that you can also look at the whole cycle. Overall, $\Delta U = 0$, as the gas is back in the state it started, so $Q=W$ overall.