 Solution

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.