Question bank — Heat transfer — conduction (Fourier's law k), convection, radiation (Stefan-Boltzmann σT⁴)

True in sensation, not in fact — both sit at room temperature. Metal has a high thermal conductivity $k$ so it conducts heat out of your hand faster, which your nerves read as "colder"; you feel heat-flow rate $H$, not temperature.

False on its own, true with help. Left alone, heat only flows hot→cold (Second law of thermodynamics); a fridge or heat pump can move it the other way, but only by doing external work.

True. In $H=kA\Delta T/\Delta x$, thickness $\Delta x$ sits in the denominator, so double $\Delta x$ → half $H$. Physically, heat crosses by a relay of molecular collisions; a wall twice as thick means twice the chain of hand-offs and twice the total resistance $R=\Delta x/(kA)$, so the flow is throttled by half.

True — they are thermal resistances in parallel, and parallel paths add conductances ($1/R$), so total conductance doubles and $H$ doubles for the same $\Delta T$.

False. "Black body" means it absorbs all incident radiation and is the best possible emitter; the Sun is nearly a black body yet glows white-hot. Colour is about the visible slice of a hot spectrum — see Black body radiation.

False. Emission never stops for any $T>0$ K; at thermal equilibrium the body emits $e\sigma A{T}^4$ and absorbs the identical $e\sigma A{T}_0^4$, so the net exchange $P_{\text{net}}=e\sigma A(T^4-T_0^4)$ is zero while both flows continue.

True — this is Kirchhoff's law: at a given wavelength emissivity $e$ equals absorptivity, so a matte black surface both soaks up and radiates efficiently, while shiny silver does neither well.

False. Convection requires bulk flow of a fluid; a rigid solid cannot circulate, so a metal bar moves heat by conduction only. Note: inside that solid, heat still travels microscopically — via lattice vibrations (phonons) and free-electron drift — but that hand-off is conduction, not convection, because no matter moves in bulk.

True. If more heat entered a layer than left it, energy would pile up and temperatures would keep changing — contradicting "steady". So $H$ is identical through each series layer, like one current through series resistors.

Single root error: $T$ must be absolute (kelvin), not Celsius. The correct value is $5800^{\circ }\text{C}=6073$ K. Because $T^4$ raises the absolute temperature to a power, plugging in $5800$ (a Celsius number) is not "close enough" — it is a different quantity entirely and underestimates the power.

Wrong — dropping the minus breaks direction. Heat flows down the gradient (toward lower $T$), so where $dT/dx<0$ the minus makes the current $H$ come out positive; without it you would predict heat climbing uphill, violating the Second law of thermodynamics. Only drop it when deliberately computing a magnitude with $\Delta T=T_{\text{hot}}-T_{\text{cold}}$.

Wrong — that adds conductances, which only applies to parallel paths. In series the same current $H$ must cross both slabs, so their resistances add: $R_{\text{tot}}=\Delta x_1/(k_{1}A)+\Delta x_2/(k_{2}A)$, then $H=\Delta T/R_{\text{tot}}$. Adding conductances here would let a thick insulator magically increase flow, which is nonsense.

The vacuum removes the medium, so it kills conduction and convection, which both need matter — it does nothing to radiation, which is waves and crosses vacuum freely. Radiation is stopped by a separate trick: silvered walls with low emissivity $e$. Conflating the two features misses why a thermos needs both.

Wrong by a lot. Power $\propto T^4$, so the factor is $(600/300{)}^4=2^4=16$, not $2$. The trap is treating a fourth-power law like a linear one; the whole point of Stefan–Boltzmann is that small temperature rises give huge power jumps.

Not necessary — it uses a difference $T_s-T_{\text{fluid}}$, and a difference is numerically identical in °C and K (the $273$ offsets cancel). Kelvin is only forced when you raise an absolute temperature to a power, as in $T^4$.

Misleading — insulation depends on resistance $R=\Delta x/(kA)$, not on $k$ alone. A pane is very thin ($\Delta x$ tiny), which shrinks $R$ and makes $H=\Delta T/R$ large. Real insulation comes from the trapped air gap in double glazing, where $k_{\text{air}}$ is far smaller — see Thermal conductivity k.

It comes from integrating Planck's law (the blackbody spectrum) over all wavelengths; the total area under that curve scales as $T^4$. It is not chosen by hand — it falls out of the physics of Black body radiation.

Radiation is electromagnetic waves needing no medium, so it crosses the vacuum of space; sound is a pressure wave needing matter to travel through, and space has essentially none.

Still air has tiny $k\approx 0.025$, so pure conduction is agonizingly slow. Convection physically transports hot air packets across the room via buoyant circulation, moving energy far faster than molecule-to-molecule hand-off.

Both obey the same linear "flow = drive / resistance" structure, so series/parallel combination rules from Electrical resistance Ohm's law carry over directly, letting you reason about composite walls without re-deriving anything.

The atmosphere is transparent to the Sun's short-wavelength radiation but partly opaque to the Earth's long-wavelength (infrared) re-emission, trapping outgoing energy and raising equilibrium temperature — the mechanism behind the Greenhouse effect.

When $T\gg T_0$, the $T_0^4$ term is orders of magnitude smaller (e.g. $300^4$ is ~4800× smaller than $2500^4$), so it barely shifts the answer — a legitimate 80/20 estimate, not a rule of physics.

A matte black surface has high emissivity $e$ near 1, maximizing radiated power $P=e\sigma A{T}^4$; a shiny low-$e$ surface would radiate poorly and shed less heat.

Zero. $P=e\sigma A{T}^4$ gives $P=0$ when $T=0$. This is a limiting/degenerate case — no thermal agitation of charges means no thermal radiation.

Yes — it emits $e\sigma A{T}^4$ and absorbs the equal amount, so emission never ceases; only the net flow $P_{\text{net}}=e\sigma A(T^4-T_0^4)=0$ vanishes at equilibrium.

Not in steady state through a single uniform rod — a nonzero $\Delta T$ forces nonzero $H$. But if $k\to 0$ (a perfect insulator) or $A\to 0$, $H\to 0$; a near-vacuum gap approximates this by killing conduction entirely.

$H=kA\Delta T/\Delta x\to \infty$. An infinitely thin wall offers zero resistance, so heat passes freely — the mathematical limit that explains why thin, high-$k$ contacts (like a metal spoon) feel so effective.

Zero — $P_{\text{net}}=0\cdot \sigma A(T^4-T_0^4)=0$. It neither emits nor absorbs radiation, so it can only exchange heat by conduction or convection.

Yes — when $T<T_0$ (body colder than surroundings), $T^4-T_0^4<0$, so $P_{\text{net}}<0$: the body gains radiant energy on balance, e.g. a cold object warming in a warm room.