Past exam of the physics B course of the University of Cambridge / 2015 / Paper 2 / Section C / C13

The Second law of thermodynamics restricts which energy-conserving processes can occur. In the Kelvin-Planck statement of the second law, no cyclic engine can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work. In the Clausius statement of the second law, no process can have as its sole effect transferring heat from a colder reservoir to a hotter one. An engine or refrigerator combined with the reverse of a reversible device shows that a violation of either statement would violate the other.

A heat engine takes heat $Q_h>0$ from a reservoir at $T_h$, rejects $Q_c>0$ to one at $T_c<T_h$, and returns its working substance to the initial state. Energy conservation gives $W=Q_h-Q_c$ and efficiency $\eta =W/Q_h=1-Q_c/Q_h$. Carnot's theorem states that no engine between these reservoirs is more efficient than a reversible one, and that all reversible engines between them have the same efficiency. Otherwise, drive a reversible engine backwards with the work of a more efficient engine; the composite cycle transfers heat from cold to hot without external work, contradicting the Clausius statement.

A reversible Carnot cycle consists of two isotherms joined by two reversible adiabats. For an ideal gas, $Q_h=N{k}_B{T}_h\log (V_2/V_1)$ and $Q_c=N{k}_B{T}_c\log (V_3/V_4)$. The adiabats give $V_2/V_1=V_3/V_4$, soCarnot's theorem makes this independent of the working substance and defines the thermodynamic absolute-temperature scale consistently with ideal-gas temperature.^[ref]

For any reversible cycle, $\oint \delta Q_{\mathrm{rev}}/T=0$. This makes the integral between two equilibrium states independent of the reversible path, defining the state function entropy:Heat itself is path dependent; entropy is not. An irreversible change can be evaluated using any imagined reversible path between its equilibrium endpoints. For example, a fixed amount of ideal gas with constant $C_V$ has $\Delta S=C_V\log (T_B/T_A)+N{k}_B\log (V_B/V_A)$.

For a general cycle the Clausius inequality is $\oint \delta Q/T_{\mathrm{boundary}}\le 0$. The entropy balance for a process is $\Delta S\ge \int \delta Q/T_{\mathrm{boundary}}$, with equality for reversible heat exchange and no entropy production. The surroundings' entropy must be included, giving $\Delta S_{\mathrm{universe}}\ge 0$. A system's own entropy can decrease by exporting entropy; an isolated system's cannot.

The engine's reservoirs have net entropy change $Q_c/T_c-Q_h/T_h\ge 0$, sowith equality for a reversible engine. Friction, heat flow through a finite temperature difference and unrestrained expansion produce entropy and reduce usable work. Thus energy conservation fixes the balance of heat and work, while the second law fixes their direction and the attainable efficiency.^[ref]