Past exam of the physics B course of the University of Cambridge / 2015 / Paper 2 (Physics B)

For entropy gain from impact and thermal equilibration, let the large lake be a reservoir at an initially unspecified temperature $T_L$, in kelvin, and take the final meteorite temperature to be $T_L$. Its initial temperature is $T_i=1273.15\mathrm{K}$. The meteorite loses thermal energy $mc(T_i-T_L)$, and its kinetic energy $m{v}^2/2$ is also dissipated into the lake. ThusandBoth heat equilibration and dissipation contribute positively. TODO: the lake temperature is not supplied, so a unique numerical entropy change needs that input. As an explicitly assumed room-temperature estimate, $T_L=300\mathrm{K}$ gives $\Delta S_{\mathrm{universe}}\simeq 2.93\times 10^3\mathrm{J}{\mathrm{K}}^{-1}$. Taking $273$–$300\mathrm{K}$ instead gives about $3.3$–$2.9\mathrm{kJ}{\mathrm{K}}^{-1}$; this shows the estimate's sensitivity without asserting an unprovided lake temperature. The model uses the stated constant heat capacity, no phase change of the meteorite, and a lake large enough that its temperature change is negligible.^[ref]

For the image force on a line charge above a conducting plane, with no imposed external electric field, use the method of images: replace the plane by a parallel image line of charge $-\lambda$ at height $-h$. The image is $2h$ from the real line, and its field at the real line has magnitude $\lambda /(4\pi {ϵ}_{0}h)$. ThereforeHere $\hat{\mathbf{z}}$ points upward from the plane. The force is attractive and normal to the conductor, for either sign of $\lambda$. The image reproduces the induced surface charge field; it is not an extra physical charge, and no factor $1/2$ is inserted in the force on the actual line.

Interpret the two radial turning values as $r_{\min }=r_0$ and $r_{\max }=4r_0$; the printed maximum/minimum order is reversed. The potential is $V(r)=A{r}^2/2$. At either turning point,Equating these two energies gives $L^2=mA{r}_{\min }^2{r}_{\max }^2$. A radial impulse changes the radial velocity and energy but exerts no torque about the centre, so it preserves $L$. For radial-impulse circularization of a harmonic central orbit, the circular orbit satisfies $L^2/(m{r}_c^3)=A{r}_c$. HenceThe original orbit crosses this radius. Cancelling its radial velocity there leaves the tangential speed required for the circular orbit; the impulse is not applied at an original radial turning point.

For a thick good conductor in the normal skin effect regime, the skin-effect surface resistance is determined byIdentical guide geometry then giveswhere the numerical ratio assumes both coatings are nonmagnetic, with $\mu \simeq {\mu }_0$. A magnetic stainless-steel grade would require its magnetic permeability; conductivity alone does not fix that alternative.

Using ${\mu }_0\simeq 4\pi \times 10^{-7}\mathrm{H}{\mathrm{m}}^{-1}$, the gold skin depth at the stated frequency isThe metal coating thickness relative to skin depth sets the useful scale. Thus a layer of order $1\mu \mathrm{m}$ begins to be useful, and a few skin depths, roughly $2$–$3\mu \mathrm{m}$, substantially screen the poorer substrate. This is a thickness scale, not a specified attenuation threshold. It assumes a continuous coating thicker than the relevant microscopic surface scale.^[ref][ref]

Take upward as positive and interpret the stated $97\mathrm{kg}$ as the total mass immediately before launch, including the harpoon. The remaining probe mass is $M=95.6\mathrm{kg}$. Conservation of momentum during the short firing interval givesThe harpoon speed used here is relative to the comet, as printed. From $v_{\mathrm{esc}}^2=2G{M}_c/R$, the near-surface gravitational acceleration isFor the near-surface recoil height with constant downward thrust, the upward motion has downward acceleration $a=g_c+F/M$. ThereforeSince $H/R\simeq 1.8\times 10^{-4}$, treating gravity as constant is consistent. This uses only the stated gravity and thrust after firing; a taut tether could reduce this free-flight height, but no tether length or elasticity is specified. If instead $97\mathrm{kg}$ meant the probe mass after firing, the pre-firing mass would be $98.4\mathrm{kg}$ and its momentum would leave the probe still moving downward, with no upward excursion. This illustrates nearly cancelling recoil momenta in payload ejection. The mass convention matters because the launch momenta nearly cancel.

At an interface with unit normal $\hat{\mathbf{n}}$ pointing from medium 1 to medium 2, the magnetic boundary conditions areThe first follows from a pillbox applied to $\nabla \cdot \mathbf{B}=0$. The tangential magnetic field jump from a free surface current follows from a shrinking rectangular Ampère loop; ${\mathbf{K}}_f$ is the free surface current. With no free sheet current, normal $B$ and tangential $H$ are continuous. In linear isotropic media $\mathbf{B}=\mu \mathbf{H}$, so neither all components of $B$ nor all components of $H$ generally remain continuous.

For magnetic field-line refraction at a permeability interface, with angles measured from the normal, these conditions give $\tan {\theta }_2/\tan {\theta }_1={\mu }_2/{\mu }_1$. They determine the bending of magnetic flux at a material interface. In a slender core electromagnet, flux continuity and Ampère's law lead to the magnetic-circuit estimateAn air gap can dominate the reluctance even when the iron core has high magnetic permeability. Boundary conditions also determine demagnetizing fields; fringing, saturation and hysteresis limit the simple linear circuit model. A high-magnetic permeability material redirects flux, whereas a perfect-conductor boundary is a different physical assumption.^[ref]

An inductor stores magnetic energy and opposes changes of current through electromagnetic induction. For a fixed linear device, the flux linkage is $\Psi =N\Phi =LI$, the self-induced electromotive force is $-LdI/dt$, and the passive terminal voltage is $V=LdI/dt$. HenceThe voltage leads a sinusoidal current by a quarter cycle. A long solenoid has $B\simeq \mu NI/l$ and $L\simeq \mu N^{2}A/l$; its energy agrees with the volume integral of $B^2/(2\mu )$.

For two coupled coils, ${\Psi }_1=L_1{I}_1+\mathcal{M}{I}_2$, ${\Psi }_2=\mathcal{M}{I}_1+L_2{I}_2$, with signs set by coil orientations. Reciprocity gives the same mutual inductance in both equations. For the energy stored in a nonlinear inductor, $U=\int Id\Psi$ replaces the simple quadratic formula, and hysteresis can dissipate energy. Real coils also have resistance, core loss and parasitic capacitance, so an ideal inductance model has a finite useful frequency range.^[ref]

In a linear, nondispersive, lossless medium, the field energy density and energy flux areThe Poynting theorem follows by dotting the two curl Maxwell equations with the other field and subtracting:The right-hand side is power transferred from the field to free charges. In vacuum the energy density is ${ϵ}_0{E}^2/2+B^2/(2{\mu }_0)$.

Static electric or magnetic fields store energy even though their time derivatives vanish. Static Poynting flow occurs because static crossed fields can also have a nonzero Poynting flux: for example, a steady current-carrying coaxial line has radial electric and azimuthal magnetic fields, carrying energy along the line toward the load. Time independence therefore does not imply zero energy flow.

For phasors whose real parts represent the physical fields,For a travelling plane wave in vacuum the electric and magnetic contributions are equal and $⟨S⟩=c⟨u_{\mathrm{em}}⟩$. In dispersive matter the simple macroscopic energy formula needs modification to include the material response; it is not a universal formula for plasma energy.^[ref]

Reflected and transmitted waves arise because the fields must satisfy the electromagnetic boundary conditions. With no free surface charge or current, tangential $E,H$ and normal $D,B$ are continuous. At normal incidence between lossless media, define the real wave impedance $Z_j=\sqrt{{\mu }_j/{ϵ}_j}$. Writing $r=E_r/E_i$ and $t=E_t/E_i$ in a common transverse electric-field direction givesFor the power normalization of Fresnel amplitudes, power fractions are $R=|r{|}^2$ and $T=(Z_1/Z_2)|t{|}^2$, so $R+T=1$. An electric-field amplitude coefficient alone is not a transmitted power fraction.

At oblique incidence, phase matching along the interface gives Snell's law, $n_1\sin {\theta }_i=n_2\sin {\theta }_t$. Applying the same boundary conditions separately to the two linear polarizations gives the Fresnel equations. For nonmagnetic dielectrics the parallel-polarized reflection vanishes at $\tan {\theta }_B=n_2/n_1$. Above the critical angle into a lower-index lossless medium, the transmitted field is evanescent and reflection is total. A perfect conductor instead has zero tangential electric field and unit reflected power; finite conductivity introduces penetration and absorption.^[ref]

Displacement current makes Ampère's law compatible with charge conservation. The macroscopic Maxwell equations areTaking the divergence of the last equation and using Gauss's law gives $\nabla \cdot {\mathbf{J}}_f+{\partial }_t{\rho }_f=0$. Omitting ${\partial }_t\mathbf{D}$ would wrongly require the conduction current to be divergence free during charging. The displacement-current density ${\partial }_t\mathbf{D}$ has current-density units, but need not represent free charges crossing a region.

The surface independence of Ampère circulation during capacitor charging illustrates the need directly. A surface spanning the circuit can cut the wire or pass between the plates. For ideal parallel plates, $Q={ϵ}_{0}AE$ andThe corrected circulation of $\mathbf{H}$ is independent of that surface choice. Free conduction current stops at a plate; displacement current supplies the gap contribution.

In vacuum the two time-dependent curl equations support waves. Taking a curl of Faraday's law in a source-free region yieldsThe corresponding magnetic equation is identical. For a plane wave, $\mathbf{k}\cdot \mathbf{E}=\mathbf{k}\cdot \mathbf{B}=0$ and $\mathbf{B}=\mathbf{k}\times \mathbf{E}/\omega$. The fields carry energy through the Poynting flux. Dropping displacement current everywhere would remove these freely propagating solutions.^[ref]

In a metal, conduction and displacement compete. For a uniform ohmic material with ${\mathbf{J}}_f=\sigma \mathbf{E}$, constant $ϵ,\mu$, and transverse waves,Use $e^{i\mathbf{k}\cdot \mathbf{r}-i\omega t}$. The dispersion relation and conduction-to-displacement current ratio areWhen $\sigma \gg \omega ϵ$, $k\simeq (1+i)/\delta$ with $\delta =\sqrt{2/(\mu \sigma \omega )}$: fields are screened within a skin depth and energy is dissipated by Joule heating. This quasistatic limit neglects displacement current locally compared with conduction; it does not invalidate the full Maxwell law. At sufficiently high frequency, conductivity and permittivity are frequency dependent, so a constant-$\sigma$ approximation must be replaced by a material model.

In an unmagnetized cold plasma, electron inertia gives a cutoff. With immobile ions and negligible collisions, $m_e\dot{\mathbf{v}}=-e\mathbf{E}$ and $\mathbf{J}=-n_{e}e\mathbf{v}$ give $\sigma (\omega )=i{n}_e{e}^2/(m_e\omega )$. For the transverse wave cutoff in a cold unmagnetized plasma, combining this carrier current with vacuum displacement current givesTransverse waves propagate above ${\omega }_p$ and are evanescent below it. The group velocity is $c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$ even though the phase velocity exceeds $c$. The cold model also has longitudinal charge-density oscillations at ${\omega }_p$, distinct from transverse light waves. Collisions add damping, and pressure or magnetic fields modify these simple dispersions. To avoid carrier-current double counting in effective permittivity, once the carrier response is included in ${ϵ}_{\mathrm{eff}}$ it must not be counted a second time as a separate conductivity. Displacement current therefore connects circuit continuity, vacuum radiation, conductor screening and plasma propagation within one consistent field theory.^[ref]

A torque-free rigid body has constant angular momentum $\mathbf{L}$ in inertial space. Its angular velocity need not be parallel to $\mathbf{L}$, because $\mathbf{L}=\mathbf{I}\mathbf{\omega }$. In body principal axes, Euler equations for a torque-free rigid body areFor a symmetric top $I_1=I_2=I_{\perp }$, $I_3=I_{\parallel }$, the component ${\omega }_3$ is constant andThis is the signed rotation frequency of the transverse angular-velocity components in the body frame. In space the symmetry axis instead obeysIts angle to $\mathbf{L}$ is constant, with $\cos \theta =I_{\parallel }{\omega }_3/L$. The body-frame and space-frame frequencies of free symmetric-top precession describe different frames. A spherical body has $\mathbf{\omega }\parallel \mathbf{L}$ and no corresponding wobble. For a triaxial body, rotations about the largest and smallest principal-inertia axes are stable, whereas rotation about the intermediate axis is unstable.^[ref]

For generalized coordinates $q_i$ with Lagrangian $L(q,\dot{q},t)$, the conjugate momentum isA coordinate absent from $L$ is cyclic, and its conjugate momentum is conserved. More generally, Noether's theorem connects continuous symmetries of the action to conserved quantities: spatial translations give linear momentum and rotations give angular momentum. If $L$ has no explicit time dependence, the conserved energy function isThe momentum conjugate to an angle is an angular momentum, not a linear velocity. Canonical momentum can also differ from mechanical momentum: for $L=\frac{1}{2}m{v}^2+q\mathbf{A}\cdot \mathbf{v}-q\varphi$, $\mathbf{p}=m\mathbf{v}+q\mathbf{A}$. Thus a conserved conjugate momentum must be identified from the actual Lagrangian and its symmetry, rather than assumed to equal $m\mathbf{v}$. Gauge-dependent canonical momenta require the corresponding gauge choice or gauge-covariant symmetry interpretation.^[ref]

For positions measured from the chosen origin, the inertia tensor isIt relates angular momentum and rotational kinetic energy to angular velocity:The real symmetric tensor has orthogonal principal axes, in which $\mathbf{I}=\mathrm{diag}(I_1,I_2,I_3)$. Its eigenvalues are nonnegative moments of inertia. Rotation about a principal axis has $\mathbf{L}\parallel \mathbf{\omega }$; a general rotation does not. The moment about a unit axis $\mathbf{n}$ is $I_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{In}$.

Changing from the centre of mass to a parallel origin displaced by $\mathbf{a}$ gives the tensor parallel-axis theorem,For example, a uniform solid sphere has $\mathbf{I}=(2M{R}^2/5)1$, while a thin rod along $x$ has principal moments $0,M{l}^2/12,M{l}^2/12$ in the ideal zero-radius limit. The tensor is needed to describe off-axis rotation and torque-free precession; a single scalar inertia only suffices after an axis is specified.^[ref]

For steady, inviscid, incompressible flow, Euler's equation integrated along a streamline givesThis is Bernoulli's equation. The constant is common to different streamlines if the flow is also irrotational; incompressibility alone does not imply that. Viscous loss or work by a pump requires additional terms.

A Pitot-static tube compares stagnation pressure $p_0$ at a forward-facing opening with nearby static pressure $p$. For equal height, the stagnation point has $v=0$, soThis measures flow speed when compressibility corrections are negligible.

In a flowmeter using the Venturi effect, continuity gives $A_1{v}_1=A_2{v}_2$. For a horizontal constriction with $A_2<A_1$, Bernoulli gives $p_1-p_2=\rho (v_2^2-v_1^2)/2$, henceThe fluid accelerates and static pressure falls in the throat. This is a second practical application of the conversion between pressure and kinetic energy; real instruments need calibration for flow losses.

A normal mode is a motion in which all coordinates have a fixed relative pattern and oscillate at one frequency. Near a stable equilibrium, let the quadratic kinetic and potential energies be $T={\dot{\mathbf{q}}}^{\mathsf{T}}\mathbf{M}\dot{\mathbf{q}}/2$ and $V={\mathbf{q}}^{\mathsf{T}}\mathbf{Kq}/2$. With positive-definite mass matrix $\mathbf{M}$, the equations and harmonic ansatz giveThe stiffness and mass matrices are symmetric. For distinct frequencies, their symmetry gives $({\omega }_s^2-{\omega }_r^2){\mathbf{a}}_r^{\mathsf{T}}\mathbf{M}{\mathbf{a}}_s=0$, so the mode vectors are orthogonal in the mass inner product. A basis can also be chosen in a degenerate eigenspace. Expanding $\mathbf{q}={\sum }_s{Q}_s{\mathbf{a}}_s$ decouples the equations into ${\ddot{Q}}_s+{\omega }_s^2{Q}_s=0$; initial displacements and velocities determine the coefficients. Positive stiffness gives real frequencies, a zero eigenvalue can describe a free symmetry motion, and a negative one signals instability rather than an oscillatory mode.^[ref]

For two equal masses between fixed walls with three equal springs, the stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and the mass matrix is $m1$. The patterns $(1,1)$ and $(1,-1)$ have frequencies $\sqrt{k/m}$ and $\sqrt{3k/m}$. An initial displacement of one mass excites both modes and can produce beats.

For a uniform taut string fixed at both ends, the wave equation ${\partial }_t^{2}y=c_s^2{\partial }_x^{2}y$, with $c_s=\sqrt{\mathcal{T}/{\mu }_{\ell }}$, has patterns and angular frequenciesThe boundary conditions select the spectrum. Fourier expansion of arbitrary admissible initial data provides the mode amplitudes, illustrating the extension from a finite set of coordinates to a field.

For the normal modes of inductively coupled LC circuits, take two identical circuits with self-inductance $L$, capacitance $C$ and mutual inductance $\mathcal{M}$, where $|\mathcal{M}|<L$. The capacitor charges obey $L{\ddot{q}}_1+\mathcal{M}{\ddot{q}}_2+q_1/C=0$ and the exchanged equation. In-phase and opposite-phase charge patterns haveThe positive inductance matrix for two identical coils is the physical counterpart of positive mass. Here inductance plays the role of the mass matrix. Mechanical, wave and electrical examples therefore share the same eigenvalue structure. Weak forcing excites modes according to its spatial overlap with each pattern; damping broadens the resonances, while strong nonlinearity can invalidate a fixed linear-mode decomposition.

A liquid-gas thermodynamic critical point is the endpoint of the coexistence curve where distinct liquid and gas densities merge. In the van der Waals illustration, write the original constants as $a_v,b_v$ to distinguish them from the later polynomial coefficients. The critical isotherm has a horizontal inflection in $p(V)$; setting its first two volume derivatives to zero givesBelow this temperature the homogeneous van der Waals loop contains an unstable region; physical equilibrium instead separates into liquid and gas.^[ref]

For the critical point of a cubic density equation of state, assume $b,c>0$. Its density derivatives areBecause $n=N/V$ is a smooth monotone function of $V$ at finite positive density, a stationary inflection obeys $p_n=p_{nn}=0$ as well. Solving these equations and substituting back givesThe symbols $b,c$ here have different units from the van der Waals excluded-volume parameter. With $x=n/n_c$, $t=T/T_c$, the isotherms becomeThe spinodal densities of a cubic gas delimit the two-extrema regime. The curves are monotone for $t>1$, have a stationary inflection at $(x,p/p_c)=(1,1)$ for $t=1$, and have two extrema $x=1\pm \sqrt{1-t}$ for $0<t<1$.

For the critical-density compressibility of a cubic gas, since ${\kappa }_T=(1/n)(\partial n/\partial p{)}_T$, evaluation at the critical density givesIt diverges as $T\to T_c^{+}$ in this mean-field model. For $T<T_c$ it is negative on the homogeneous branch at $n_c$, violating mechanical stability: a pressure increase would expand rather than compress that branch.

The actual equilibrium below criticality is coexistence of two phases, with equal pressure and chemical potential and different densities. A sample's mean density can change by changing phase proportions. The unstable homogeneous branch is not a realizable equilibrium isotherm. The Maxwell construction in density coordinates requires care: the equal-area construction is performed in $p$ versus volume, not by using unweighted area in $p$ versus density. The truncated polynomial is an approximate mean-field equation, so its predictions need not give quantitative critical behaviour for a real gas.

The Boltzmann distribution assigns probability $P_i=e^{-E_i/(k_{B}T)}/Z$ to each microstate. For the equilibrium height distribution in a uniform gravitational field, take a dilute gas above a floor in a constant gravitational field. Uniform cross-section and no relevant upper boundary give the normalized height densityThusThe gravitational potential is linear in height, so its mean is not a quadratic-mode contribution of $k_{B}T/2$.

The equipartition states that each independent quadratic term in a classical equilibrium Hamiltonian contributes mean energy $k_{B}T/2$, provided the relevant degrees of freedom are thermally accessible. Equivalently, $⟨x_i\partial H/\partial x_i⟩=k_{B}T$ when the phase-space integration by parts has vanishing boundary terms. Three translational velocity components therefore contribute $3k_{B}T/2$.

For rotational equipartition in a rough-ball gas, collisions exchange both translational and rotational energy. In the intended three-dimensional classical equilibrium model there are three translational and three rotational quadratic terms, giving $3k_{B}T$ in total kinetic energy per ball. Adding $⟨U_g⟩=k_{B}T$, and conserving the initial energy $mgh$ per ball, givesRoughness is relevant because it allows rotation to thermalize. The effective mechanical temperature of released rough balls describes the balls' mechanical degrees of freedom, not their internal material temperature; finite ball size and excluded-volume effects are neglected.^[ref]

The height tail isThis can be nonzero because collisions redistribute energy between balls; an individual ball is not restricted to its original energy. The exponential distribution is the large-number statistical approximation under the stated effective-equilibrium assumption.

For the angularly resolved molecular arrival flux, let $f(v)$ be normalized by ${\int }_0^{\infty }f(v)dv=1$. Isotropic velocity directions have probability $d\Omega /(4\pi )$. Integrating over azimuth gives $(1/2)\sin \theta d\theta$, while the speed normal to a stationary piston is $v\cos \theta$. For incident directions $0\le \theta \le \pi /2$, the arrival flux is thereforeAn elastic collision transfers normal momentum $2mv\cos \theta$ to the piston. Integrating the momentum flux givesNo specific Maxwellian speed distribution was needed for this isotropic kinetic-pressure identity.

Use the figure's $+x$ direction toward the piston and let its inward speed $u>0$ mean piston velocity $-u\hat{\mathbf{x}}$. For reflection from a translating planar piston, in the piston frame the incoming normal velocity is $v_x+u$. Elastic reflection reverses it, and transforming back givesThe exact kinetic-energy change in this ideal massive-piston model is $\Delta K=2mu{v}_x+2m{u}^2$. Thus, to first order in the small piston speed,The first-order collision-flux approximation for a slow piston applies: the incident collision flux changes at order $u$, but its product with the order-$u$ energy increment changes the energy rate only at order $u^2$. Hence one may use the stationary flux at leading order. For piston area $S$,Here $dV/dt=-Su$. This compression work from molecular impacts, with no heat flow, is the first law for work done on the gas by compression. The derivation assumes $u$ small compared with thermal speeds and an approximately isotropic distribution at the piston; rapid compression can produce nonequilibrium pressure and higher-order corrections.

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]