Why does $\oint\frac{\delta Q}{T} = 0$ when evaluated over a Carnot cycle?

Question

In my thermodynamics schoolbook, it is asked to apply the Clausius inequality to a Carnot cycle. According to the book the answer is to evaluate $\oint\frac{\delta Q}{T}$ over the cycle which should yield $$ \oint\frac{\delta Q}{T} = \frac{Q_H}{T_H}+\frac{Q_C}{T_C}=0 $$

1. $Q_H$ is the heat transferred from the hot reservoir (positive)
2. $T_H$ is the temperature of the hot reservoir
3. $Q_C$ is the heat transferred to the cold reservoir (negative)
4. $T_C$ is the temperature of the cold reservoir

While it is easy enough to show this is true using $\delta Q =TdS$, my schoolbook doesn't introduce entropy until later. Therefore, my question is how to prove this equality without using entropy ?

Answer 1

I only know a long ride for this, sorry :/

You have to imagine two processes iabf and if that have the same work $\Delta W_{iabf}=\Delta W_{if}$: (Dashed lines are adiabatic curves and solide lines are isothermes)

The internal energy variation is the same for both processes and $\Delta Q_{ab}=\Delta Q_{if}$.

Now imagine a system S' that provides energy $\Delta Q$ to the cycle C(system S). The only way to enter energy is through an isotherm, with temperature $T_0=T_H$. This works as the hot reservoir. The ab isotherm has $T<T_0$, which is your $T_C$. The system S' transfers heat($\Delta Q' $) to S: $$\Delta Q' = T_H \frac{\Delta Q}{T_C}$$

The cd path has the opposite direction. Try drawing one Carnot Cycle above the other and you will see they cancel each other.

Now imagine infine mini carnot cycles inside C, the sum of all cycles cancel each other:

$$ \Sigma \Delta Q' = \oint_C T_C \frac{dQ}{T} = 0 $$ *T varies in each mini cycle.

Moving $T_C$ $$\oint_C \frac{dQ}{T} = 0$$

I believe your book assume reversible processes from the begining. Otherwise, according to the second law:

$$\oint_C \frac{dQ}{T} \leq 0$$