Second law of thermodynamics

All answers correspond to different laws of thermodynamics (including the zeroth law). The correct answer and other phrases of the 2nd law are

(A) There is no machine that does nothing but transfer heat from the colder to a warmer system.

(B) There is no machine (perpetual motion machine of the 2nd kind) that does nothing but extract heat from a system and turn it completely into work.

(C) A heat engine that removes the heat $Q$ from a hot system, converts the fraction into the work $W$ and transfers the rest to a cold system, never has a higher efficiency $\eta = W / Q$ than a reversible heat engine with the Carnot efficiency $\eta_\text{rev} = 1 - T_\text{cold} / T_\text{hot}$ .

(D) In a closed system, entropy can only increase and become maximal in equilibrium.

The equivalence of all formulations of the 2nd law can be proved:

• $(A) \Rightarrow (B) \equiv \neg (B) \Rightarrow \neg (A)$ : If there is a perpetuum mobile of the 2nd kind (PM), it could extract heat from the cold system and, with the work done, operate a reverse heat-power machine (heat pump). Therefore we get a perfect heat pump (PHP), that transfers heat from cold to hot without additional work.
• $(B) \Rightarrow (C) \equiv \neg (C) \Rightarrow \neg (B)$ : If there were a heat engine with $\eta > \eta_\text{rev}$ , it could be paired with a reversible heat engine that works as a heat pump and transfers the heat $Q'$ from the cold system back to the hot system, resulting in a perpetuum mobile of the 2nd kind.
• $(C) \Rightarrow (D)$ : A heat engine removes heat from the hot system ( $d Q_\text{hot} <0$ ) and adds some of the heat to the cold system ( $d Q_\text{cold}> 0$ ). The entropy changes accordingly $d S = \frac{d Q_\text{hot}}{T_\text{hot}} + \frac{d Q_\text{cold}}{T_\text{cold}} \geq |d Q_\text{hot}| \left( - \frac{1}{T_\text{hot}} + \frac{1 - \eta_\text{rev}}{T_\text{cold}}\right) = 0$ In equilibrium ( $T_\text{hot} = T_\text{cold}$ ) the entropy becames maximum since there is no heat flow ( $dQ = 0$ ).
• $(D) \Rightarrow (A) \equiv \neg (A) \Rightarrow \neg (D)$ :If there is a perfect heat pump which lets the heat $dQ$ flow from cold to hot, it follows $dS = \frac{dQ_\text{hot}}{T_\text{hot}} + \frac{dQ_\text{cold}}{T_\text{cold}} = dQ \frac{T_\text{cold} - T_\text{hot}}{T_\text{hot} T_\text{cold}} < 0$