= Solution
\b[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.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2019/paper-2-c11-second-law.png]
{title=Forbidden sole effects in the Clausius and Kelvin–Planck formulations of the second law}
\b[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.
\b[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 satisfies
$$
\frac{\dot Q_h}{T_h}=\frac{\dot Q_c}{T_c},\qquad
\dot W=\dot Q_h-\dot Q_c.
$$
Therefore the heating <coefficient of performance> is
$$
\mathrm{COP}_h=\frac{\dot Q_h}{\dot W}=\frac{T_h}{T_h-T_c}.
$$
With $T_h=293\,\mathrm K$ and $T_c=273\,\mathrm K$,
$$
\boxed{\dot Q_{h,\max}=(0.5)\frac{293}{20}=7.325\,\mathrm{kW}\simeq7.33\,\mathrm{kW}.}
$$
The additional <heat> comes from the outdoor reservoir; it is not produced by converting <work> into more <energy> than was supplied.
\b[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 when
$$
\lambda\Delta T\leq\frac{\dot W_{\max}T_h}{\Delta T},
\qquad
\Delta T^2\leq\frac{\dot W_{\max}T_h}{\lambda}.
$$
At the limiting outdoor <temperature>,
$$
\Delta T_{\max}=\sqrt{\frac{(0.5\,\mathrm{kW})(293\,\mathrm K)}{0.1\,\mathrm{kW\,K^{-1}}}}
=38.275\,\mathrm K.
$$
Hence
$$
\boxed{T_{c,\min}=254.725\,\mathrm K\simeq255\,\mathrm K
\quad( -18.4\,{}^\circ\mathrm C).}
$$
This 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.
\b[Conventional heater.] At this limiting outdoor <temperature>, the required delivered heating <power> is simply the house's <heat> loss:
$$
\boxed{\dot Q_{\mathrm{gas}}=\lambda\Delta T_{\max}=3.83\,\mathrm{kW}.}
$$
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.
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