# Carnot engine efficiency and entropy

I think I am not understanding a key concept about thermodynamics. I have worked out the heat and work changes for a particular Carnot cycle. In fact, there is no heat change as we have 2 isothermal processes $$\Delta T =0$$ and 2 adiabatic processes (no heat added) but only changes in work. Thus I do not understand the equation for the efficiency of the Carnot Engine: $$\eta = 1 - \frac{Q_c}{Q_h}$$ Also, to calculate the change in entropy for the Carnot cycle (which should be $$0$$), I need to use: $$ds = \oint \frac{dQ_{rev}}{T}$$ This doesn't make sense, since to my understanding there is no change in heat anyways.

The changed heat in a isothermal process is not 0.

$$dU = dQ - dW$$

$$dT=0 \implies dU = 0$$ , so for an ideal gas we have:

$$Q_{A} = W_{A} = \int_{V_0}^{V} PdV = \int_{V_0}^{V} nrT_a \cdot \frac{1}{V}dV = nrT_a \cdot ln(\frac{V}{V_0})$$

$$Q_{B} = W_{B} = \int_{V}^{V_0} PdV = \int_{V}^{V_0} nrT_b \cdot\frac{1}{V}dV =- nrT_b \cdot ln(\frac{V}{V_0})$$

So the total changed heat $$Q_A +Q_B$$ is:

$$Q_t = nr (T_a -T_b) \cdot ln(\frac{V}{V_0})$$

$$\eta = \frac{Q_t}{Q_a} =1 - \frac{|Qb|}{|Qa|} = 1 - \frac{Tb}{Ta}$$

The change of entropy is given by:

$$dS = \frac{dQ}{T} \implies \Delta S = \int \frac{dQ}{T}$$

Since T is constant:

$$\Delta S_a = \frac{Q_A}{T_a} = nr \cdot ln(\frac{V}{V_0})$$

$$\Delta S_b = \frac{Q_B}{T_b} = - nr \cdot ln(\frac{V}{V_0})$$

$$\Delta S = \Delta S_b + \Delta S_a = 0$$

• Hi thank you so much. So you are rearranging the first law to $dQ = dU + dW$ and it is not the heat change that is nought, but the change in internal energy: $\Delta U = \frac 32 nR \Delta T$ and since $\Delta T =0, dU= 0$ so $dQ = dW$ ? Feb 15, 2020 at 14:15
• yes that is right Feb 15, 2020 at 14:30