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Use absolute temperatures $T_c=278.15\mathrm{K}$ and $T_h=298.15\mathrm{K}$. Let ${\dot{Q}}_c$ be the heat extracted from the ground, ${\dot{Q}}_h$ the heat delivered indoors, and $P$ the electrical input power. For a reversible heat pump, the entropy transferred per unit time balances:Consequently the coefficient of performance isThe requested heat delivery is a rate, since the duration is unspecified:Of this, $13.9\mathrm{kW}$ comes from the ground and $1.0\mathrm{kW}$ from the electricity. A coefficient of performance greater than one is consistent with conservation of energy because the heat pump transports heat as well as converting electrical energy.

An electrical transformer transfers alternating electrical power between windings through a shared changing magnetic flux. A core with large magnetic permeability confines most of the flux, so both windings link approximately the same flux $\Phi (t)$. Faraday's law gives induced electromotive forcesFor an ideal electrical transformer, negligible winding resistance and leakage flux imply the terminal-voltage magnitude ratioThe relative phase depends on the winding sense and terminal convention. At a loaded secondary, Lenz's law makes the secondary current oppose the flux change. The primary draws an additional current which approximately cancels the secondary magnetomotive force: $N_p{I}_p\simeq N_s{I}_s$ in magnitude when the magnetizing current is negligible. Consequentlyfor a resistive load, using root mean square values. More generally the ideal device conserves instantaneous power, with the appropriate current signs. A load impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_p/N_s{)}^2{Z}_L$, enabling impedance matching and electrical isolation.

For sinusoidal flux, $\Phi ={\Phi }_{\max }\sin \omega t$ gives $V_{\mathrm{rms}}=\omega N{\Phi }_{\max }/\sqrt{2}$. At a given supply voltage, reducing frequency increases the required peak flux and can drive the core into magnetic saturation. Real losses include winding Joule heating, magnetic hysteresis, eddy currents and leakage flux; a laminated core suppresses eddy-current loops. A steady direct current produces no continuing secondary electromotive force, while its small winding resistance can allow a damaging primary current after switching transients.

Distinguish electrical current density ${\mathbf{J}}_e$ from heat flux ${\mathbf{J}}_Q$. Uniform current across the circular section givesby Ohm's law. The local Joule heating rate is electrical work per unit volume and time:For a segment of length $\ell$, integration gives $H\pi a^2\ell =I^2\ell /(\sigma \pi a^2)=I^{2}R$, confirming the usual resistor power. The model takes electrical conductivity constant and ignores temperature-induced current redistribution.

In a long wire away from its ends, the steady temperature is independent of axial position and azimuth. The heat equation reduces toMultiplying by $\rho$ and integrating givesA finite temperature and nonsingular flux at the axis require $C_1=0$. Applying $T(a)=T_0$ determines $C_2=T_0+H{a}^2/(4\kappa )$. ThereforeThis is the steady temperature of a uniformly heated cylindrical wire. Its hottest point is at the centre, with $T(0)-T_0=I^2/(4{\pi }^2{a}^2\kappa \sigma )$. The outward heat flux is $J_Q(\rho )=-\kappa T^{\prime }(\rho )=H\rho /2$, so the power crossing a cylindrical surface per unit length, $2\pi \rho J_Q(\rho )=H\pi {\rho }^2$, equals the heating generated inside it.

Replacing the fixed surface temperature by Newton's law of cooling changes the boundary condition but not the radial derivative in steady state. At the surface,HenceHere $\gamma$ is a surface heat transfer coefficient in $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, unrelated to the surface tension symbol used in C10. An independent energy-balance check uses a length $\ell$: the electrical input $I^2\ell /(\sigma \pi a^2)$ must equal the surface loss $2\pi a\ell \gamma [T(a)-T_{\mathrm{air}}]$, which gives the same answer.

The complete temperature profile isThermal conductivity controls the internal gradient, but cancels from the surface-temperature rise because all generated power must eventually cross the prescribed surface. This assumes positive constant $\gamma$, uniform electrical conductivity and negligible end or additional radiative losses; temperature-dependent material properties could make the heating and boundary problem nonlinear.

For an ideal electrical transformer, let $n=N_p/N_s$. The reflected secondary load is $R_{\mathrm{in}}=n^2{R}_L$. Matching a $4\Omega$ load to an $8\Omega$ source therefore requiresWith the generator replaced by its internal resistance, its resistance referred to the secondary isThis is the source output resistance relevant to matching and maximum power transfer.

To find the windings, use the inductance of a toroidal winding. The mean magnetic path is $\ell =2\pi (0.10)\mathrm{m}$, the core area is $A=30\times 10^{-4}=0.003{\mathrm{m}}^2$ and $\mu =500{\mu }_0$. ThusThe ideal secondary turn count is $N_s=500/\sqrt{2}=353.55$, giving the practical nominal designWith integer turns the irrational ratio cannot be exact: $500:354$ reflects the $4\Omega$ load as $7.98\Omega$, an approximately $0.25\%$ resistance mismatch. These turn counts use the stated $0.75\mathrm{H}$ self-inductance; no operating frequency was specified from which to set a separate minimum inductance.

The $200\Omega$ condition must refer to the unloaded magnetizing reactance, not to the matched loaded input resistance. With $L_p=0.75\mathrm{H}$,The secondary self-inductance is $L_s=L_p/n^2=0.375\mathrm{H}$, so its corresponding unloaded coil impedance satisfiesIf output impedance means this transformed coil reactance, $100\Omega$ is its minimum magnitude. It is distinct from the $4\Omega$ transformed source resistance. This is the magnetizing reactance and reflected resistance in a transformer distinction. A finite magnetizing inductance adds a parallel branch, $Z_{\mathrm{in}}=(8\Omega )\parallel (i\omega L_p)$, so exact purely resistive matching is an ideal large-reactance approximation; interpreting the loaded input itself as both $8\Omega$ and at least $200\Omega$ would be impossible.

In a homogeneous Ohmic conductor, $\mathbf{J}=\sigma \mathbf{E}$. Combining this with Maxwell's equations givesFor a good conductor, $\sigma \gg \omega ϵ$, so conduction dominates displacement current. With time dependence $e^{-i\omega t}$ and depth $z>0$, take $E\propto e^{ikz}$. ThenThe field is consequently proportional to $e^{-z/\delta }{e}^{iz/\delta }$: it both attenuates and changes phase with depth. The skin depth is the distance for amplitude to fall by $e^{-1}$; power density falls as $e^{-2z/\delta }$, so its attenuation length is $\delta /2$.

Physically, changing fields induce currents whose magnetic field opposes penetration. Equivalently, after neglecting displacement current, magnetic diffusion obeys ${\partial }_t\mathbf{B}=(\mu \sigma {)}^{-1}{\nabla }^2\mathbf{B}$. Faster changes have less time to diffuse into the metal. Increasing frequency, conductivity or magnetic permeability reduces $\delta$, provided the relevant material parameters remain applicable.

For a thick conductor the tangential surface electrical impedance is $Z_s=(1-i)/(\sigma \delta )$ with the stated time convention and consistent field orientation. Its real part $1/(\sigma \delta )$ represents dissipative resistance per square. Current crowding near the surface raises the AC resistance of a wire when $\delta$ is small compared with its radius; thin strands and laminated magnetic cores reduce suitable eddy-current losses. When $\sigma /\gg \omega ϵ$, the full complex propagation constant must replace the good-conductor approximation. In a magnetic metal the magnetic permeability can itself be nonlinear and frequency dependent.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

Two equivalent second-law statements. The Clausius statement of the second law forbids a cyclic device whose sole net effect is transferring heat from a colder reservoir to a hotter one. A refrigerator can make that transfer when supplied with work, so the sole-effect qualification matters. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole net effect is drawing heat from a single reservoir and converting all of it into work. A heat engine can produce work when it also rejects heat to a colder reservoir.

The two statements are equivalent by combining a supposed violating device with an ordinary reversible engine or refrigerator: one can cancel one reservoir exchange and obtain a violation of the other statement. The following original schematic shows the forbidden sole effects, with arrows indicating energy flow.

Microscopic fluctuations. The second law is a macroscopic statistical law. A finite fluctuation can transfer a small heat amount $Q$ from cold to hot, with reservoir entropy change $Q/T_h-Q/T_c<0$. Such an individual event does not imply that a cyclic engine can extract systematic useful work from equilibrium. Entropy-decreasing trajectories become exceedingly improbable when the entropy involved is large compared with $k_B$. Under suitable microscopic reversibility conditions, fluctuation relations quantify the relative probabilities of positive and negative entropy production and give nonnegative mean total entropy production. Thus a deterministic no-fluctuation reading has a limit at small scales or short times; the statistical second law remains applicable.

Maximum heating rate. Interpret the quoted $0.5\mathrm{kW}$ as the maximum work-input rate ${\dot{W}}_{\max }$ to an ideal reversible heat pump. This interpretation is needed to obtain a heating rate larger than its electrical input; the wording does not explicitly distinguish input power from delivered heat. A reversible cycle satisfiesTherefore the heating coefficient of performance isWith $T_h=293\mathrm{K}$ and $T_c=273\mathrm{K}$,The additional heat comes from the outdoor reservoir; it is not produced by converting work into more energy than was supplied.

Lowest outdoor temperature. Let $\Delta T=T_h-T_c>0$. The largest available heating rate is ${\dot{W}}_{\max }{T}_h/\Delta T$, while the house loses $\lambda \Delta T$. Equilibrium can be maintained whenAt the limiting outdoor temperature,HenceThis is the indoor-temperature limit of a Carnot heat pump. It assumes the ideal pump retains its stated input-power capability over the operating range. Real heat exchangers require finite temperature differences and incur irreversibility, so the reversible result is an ideal upper performance bound.

Conventional heater. At this limiting outdoor temperature, the required delivered heating power is simply the house's heat loss:This is heat delivered to the house. The fuel-input power would be $3.83/\eta \mathrm{kW}$ for heater efficiency $\eta$. At the earlier outdoor temperature of $273\mathrm{K}$, the required delivered heat would instead be $0.1(293-273)=2.00\mathrm{kW}$; the final question's reference temperature must be kept clear.

If $0.5\mathrm{kW}$ is read literally as a maximum delivered heat rate rather than a work-input rating, the alternative answers are ${\dot{Q}}_{h,\max }=0.5\mathrm{kW}$, $T_{c,\min }=293-0.5/0.1=288\mathrm{K}$, and a conventional heater delivering $0.5\mathrm{kW}$ at that limit. Such a pump could not balance the stated house loss at $273\mathrm{K}$. Both readings follow consistently once the ambiguous power rating is defined; the Carnot calculation above uses the work-input reading.

The Poynting vector is the local electromagnetic energy flux density:Here $\mathbf{E}$ is the electric field and $\mathbf{H}$ the magnetic field intensity. For an oriented area element $d\mathbf{S}$, $\mathbf{N}\cdot d\mathbf{S}$ is the electromagnetic power crossing that element in its positive normal direction. Thus ${\int }_S\mathbf{N}\cdot d\mathbf{S}$ is outward power through a closed surface, in watts. The Poynting vector has units $\mathrm{W}{\mathrm{m}}^{-2}$, and its direction describes local electromagnetic energy transport.

Use the divergence identity for a cross product, followed by Faraday's law of induction and the Ampère-Maxwell equation:With the electromagnetic energy density of part (b), these giveIntegrate over a fixed volume $V$ and apply the divergence theorem, choosing $d\mathbf{S}$ outward:This is Poynting's theorem. The outward electromagnetic energy flux equals minus the power transferred to matter and minus the rate at which electromagnetic energy accumulates inside. For the dissipative electrical conductor in this question, the matter term is Ohmic heating.

The Ohmic heating per unit volume is $\mathbf{E}\cdot \mathbf{J}=J^2/\sigma$, uniform in this wire. Multiplying by its volume gives the dissipated powerwhere the electrical resistance is $R_{\mathrm{wire}}=L/(\sigma \pi a^2)$. The result is in watts; over a time interval $\Delta t$ the dissipated energy is $P_{\mathrm{loss}}\Delta t$.

On the curved wire surface, the outward normal is $\hat{\mathbf{r}}$ and the Poynting vector has the constant radial value $N_r(a)=-I^2/(2\sigma {\pi }^2{a}^3)$. Its outward flux isThe Poynting vector is radial, so it has zero normal component on the two end faces in this approximation. This is therefore also the outward flux through the complete wire surface. Equivalently, positive electromagnetic power $P_{\mathrm{loss}}$ flows inward.

The fields are steady, so the stored electromagnetic energy does not change. Poynting's theorem consequently requires the incoming power to equal the Ohmic heating in part (e), exactly as found. The energy reaches the wire through the surrounding electromagnetic field. A local check is $\nabla \cdot \mathbf{N}=r^{-1}{\partial }_r(r{N}_r)=-I^2/(\sigma {\pi }^2{a}^4)=-\mathbf{E}\cdot \mathbf{J}$.

An incident electromagnetic wave generally produces reflected and transmitted waves because the two media have different electromagnetic response. Consider a flat interface between homogeneous, linear, lossless media with permittivities ${ϵ}_j$ and magnetic permeabilities ${\mu }_j$. Their wave speeds and wave impedances areFor each plane electromagnetic wave, $\mathbf{H}=\hat{\mathbf{k}}\times \mathbf{E}/Z_j$. With no free surface charge or current, the interface conditions for electromagnetic fields require continuous tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{D},\mathbf{B}$. Equality of the phase at every point on the boundary gives equal angular frequencies and tangential wavevectors. Hence the reflection angle equals the incidence angle and Snell's law iswhere $n_j=c/v_j$ and each angle is measured from the normal.

At normal incidence use the same tangential axis for all electric amplitudes. The two independent boundary equations areSolving gives the amplitude Fresnel equationsFor nonmagnetic media, these become $r=(n_1-n_2)/(n_1+n_2)$ and $t=2n_1/(n_1+n_2)$. Reflection into a medium of larger refractive index therefore reverses the electric amplitude's sign at normal incidence.

The normal Poynting vector determines the reflected and transmitted fractions of incident power. At normal incidence,It is the power fractions, rather than the squared electric amplitudes alone, that add to unity.

At oblique incidence separate transverse electric polarization ($s$, electric field perpendicular to the incidence plane) from transverse magnetic polarization ($p$, electric field in that plane). For nonmagnetic media, writing $c_i=\cos {\theta }_i$ and $c_t=\cos {\theta }_t$, the Fresnel equations areHere choose a common $s$ direction and, for each ray, the $p$ direction as $\hat{\mathbf{s}}\times \hat{\mathbf{k}}$. This makes the reflected $p$ basis reverse its tangential direction at normal incidence, explaining why the displayed $r_p$ then has the opposite sign to $r_s$. Physical reflectance is independent of this sign convention. For either polarization with propagating transmission,For $p$ polarization, zero reflection occurs at the Brewster angle $\tan {\theta }_B=n_2/n_1$. If $n_1>n_2$, incidence above the critical angle for total internal reflection $\sin {\theta }_c=n_2/n_1$ produces an evanescent field in total internal reflection and $\mathcal{R}=1$ for an isolated lossless interface. A lossy second medium requires complex response coefficients and an absorption term in the energy balance.

The complete cycle must be a reversible thermodynamic process: no friction, viscous dissipation, unrestrained expansion, or other source of entropy. Each moving boundary must be arbitrarily close to mechanical equilibrium. Heat transfer to or from a bath must occur across an infinitesimal temperature difference, and the insulated stages must be genuinely adiabatic processes. There must also be no direct heat leak between the heat baths.

For an engine exchanging heat only with these two heat baths, the Second law of thermodynamics givesand thusPractical engines generate entropy through finite temperature differences, friction and other dissipative effects. They therefore fall below the Carnot cycle limit. Approaching reversible heat exchange generally requires vanishingly slow operation for a finite heat-transfer apparatus, limiting useful power.

The equilibrium photon gas is isotropic, so the energy density per unit solid angle in a narrow angular-frequency interval is $u(T,\omega )d\omega /(4\pi )$. For rays making angle $\theta$ with an outward surface normal, a surface element $dA$ is crossed in time $dt$ by rays initially in a volume $cdtdA\cos \theta$. Only the outward hemisphere contributes. ThusThe hemispherical restriction supplies a factor $1/2$ and the directional projection supplies another $1/2$. A small aperture in a cavity therefore emits this spectral power per unit area. A blackbody at the same temperature emits the same amount: otherwise radiation exchange between it and the aperture would violate thermal equilibrium.

Integrating Planck's law over all positive angular frequencies gives the total emitted heat flux densitySet $x=\hslash \omega /(k_{B}T)$. ThenThe remaining integral is independent of temperature, proving $M\propto T^4$. Its value follows by expanding $(e^x-1{)}^{-1}={\sum }_{m=1}^{\infty }{e}^{-mx}$ and integrating termwise:Consequently the Stefan–Boltzmann law isThis is emitted power alone. A body in a radiation environment also absorbs incoming radiation, which must be included when calculating net radiative cooling.

Kirchhoff's law of thermal radiation equates spectral absorptivity and spectral emissivity at the same wavelength and direction. A surface that is a poor absorber in the thermal infrared is also a poor emitter there; a coating cannot be assigned its solar absorption independently of its thermal emission spectrum.

(i) All-wavelength black paint. It absorbs the sunlight and also radiates maximally at the spacecraft's own temperature. With the isothermal-sphere assumptions of (d), the limiting black-surface distance isThe spacecraft can operate at $R\ge 50r$ in this model. A surface already assumed to be a blackbody is not improved by another black coating; relative to option (ii), however, this paint permits closer operation.

(ii) Black at short wavelengths and grey in the infrared. By Wien's displacement law, changing the black-body radiation temperature from $6000\mathrm{K}$ to $600\mathrm{K}$ shifts the wavelength peak from about $0.5\mu \mathrm{m}$ to about $5\mu \mathrm{m}$. Most solar power is therefore in the highly absorbing short-wavelength band, whereas most spacecraft emission lies beyond $2\mu \mathrm{m}$, where the spectral emissivity is reduced. The coating retains most of the heating but suppresses much of the cooling. Under the intended two-band grey body approximation, option (ii) makes the spacecraft hotter at the same distance and requires a larger minimum distance than black paint.

To express the comparison without inventing a grey absorptivity, let ${\bar{\alpha }}_S$ be radiometric absorptance averaged over the solar spectrum and ${\overline{ϵ}}_T$ be emissivity averaged over the spacecraft's own thermal spectrum. The radiative equilibrium equation becomesAt the failure threshold,For an ideal two-band coating take a constant radiometric absorptance $g<1$ beyond ${\lambda }_c=2\mu \mathrm{m}$, and unity below it. If $f(T)$ is the fraction of the power emitted by a blackbody below ${\lambda }_c$, Planck's law givesSince $f(T_S)>f(600\mathrm{K})$, this model has ${\bar{\alpha }}_S>{\overline{ϵ}}_{600}$ and hence $R_{\min }>50r$. The paper supplies neither a numerical value of $g$ nor a detailed spectrum, so no unique numerical distance for the grey paint follows. If its long-wavelength absorption is not approximately grey, the spectrum must be integrated explicitly rather than inferred from a peak alone. A coating designed to improve close approach would instead reflect much of the solar radiation while retaining strong thermal infrared emission.

The instantaneous mechanical power supplied by a force iswhere v is the velocity of its point of application in the chosen frame.

Power is the rate of energy transfer:Net mechanical power equals the rate of kinetic-energy change in the work–energy description.

For drum-frequency dimensional scaling, let $f\propto r^p{\sigma }^q{T}^s$, where $\sigma$ is surface mass density. The dimensions are $[\sigma ]=M{L}^{-2}$ and $[T]=M{T}^{-2}$, so matching powers gives $s=1/2$, $q=-1/2$ and $p=-1$:The dimensionless normal mode constant $C$ is unchanged for the same circular shape and boundary condition. The doubled tension does not offset the doubled radius.