# How can I understand the two things coming from the thermodynamic second law?

The second law of thermodynamics has been stated in several different ways. For example:

$$\textbf{Clausius Statement:}$$ No cyclic process is possible, whose only effect is the transfer of heat from a colder body to a hotter body.

then one argues that one can derive the following two things just from this "simple" law:

$$\textbf{1.an absolute thermodynamic scale of temperature;}$$

$$\textbf{2.a function of state called entropy : S}$$

But how that's possible? Can someone give me some advice to understand two expressions coming from the second law?

3. From (C) and (K) one can prove the Carnot Theorem which asserts that no heat engine can have a better efficiency than a reversible engine and all reversible engine have the same efficiency. Then compute the efficiency of a particular reversible engine (e.g. ideal gas in the Carnot cycle) and get $$\eta_R=1-\frac{T_c}{T_h},$$ where $T_h$ and $T_c$ are the temperatures of the hot and cold sources, respectively. As an intermediate step you shall get the relation $$\frac{Q_h}{T_h}=\frac{Q_c}{T_c}.\tag{1}$$ Given the universality of reversible engine efficiency, one can arbitrarily define the temperature of the cold source $T_c$, measure - mechanically - the efficiency of the engine and then the temperature $T_h$ is determined by $$T_1=\frac{T_2}{1-\eta_R}. \tag{2}$$ This is an absolute temperature since it can be checked by any reversible engine with no ambiguities. Note that this not define temperature, since this is done only by the zeroth law. However the temperature defined by the zeroth law is highly dependent on substances, thermometric properties and scales. The second law (through the efficiency of a reversible heat engine and Carnot Theorem) gives a temperature which is independent of any substance, property or scale. There is more about the difference between them in this answer.
Suppose each infinitesimal (isotherm) part of this cycle is exchanging heat $\Delta Q_i$ with a (external) source at a temperature $T_i$. Then you sum all contributions $\Delta Q_i/T_i$, use Eq. (1) and in the limiting case it gives $$\oint \frac{dQ}{T}=0.$$
5. If The above closed integral vanishes for any cycle, then it is easy to show that the value of $\int_a^b \frac{dQ}{T}$ is independent of the (reversible) processes linking the states $a$ and $b$. This allows us to define a state function $S$ such that $$\int_a^b \frac{dQ}{T}=S(b)-S(a).$$ This state function is called entropy.