Question bank — Temperature — thermal equilibrium, thermometers, scales

False — energy also depends on how much stuff there is; a huge lake at 20 °C holds vastly more thermal energy than a hot needle at 500 °C. Temperature is intensive, energy is extensive.

True — equal temperature means they are in thermal equilibrium, so the net heat flow is zero (see Zeroth Law of Thermodynamics).

False — the scales have different zeros. Going from 20 °C to 40 °C takes the gas from 293 K to 313 K, nowhere near double, so "doubling" is only meaningful on the Kelvin scale.

True — Celsius and Kelvin degrees are the same size (100 divisions between ice and steam on both); the +273.15 offset cancels for any difference.

False — outdoors both sit at air temperature; metal just conducts heat out of your hand faster, so touch reports conductivity, not temperature.

True — the straight line of $P$ versus $t_C$ hits $P=0$ at $-273.15$ °C, and since an ideal gas cannot have negative pressure, that extrapolated crossing marks the hard floor $0$ K.

False — it measures its own temperature after reaching equilibrium with you; the Zeroth Law is what lets us then claim that reading is your temperature too.

False — different liquids expand slightly non-linearly and disagree between the fixed points; only dilute (ideal) gases agree, which is why the gas thermometer defines the absolute scale.

The $+32$ is an origin shift, valid only for absolute readings, not differences. A change converts as $\Delta t_F=\frac{9}{5}(15)=27$ °F.

It only gives an absolute scale when the property $X$ genuinely goes to zero at $T=0$; a mercury column has finite length at absolute zero, so this proportional form is properly reserved for gas pressure or volume, which do vanish there.

The Zeroth Law says the opposite — $A$ and $B$ are already in equilibrium with each other, so no net heat flows between them.

A linear law $T=aX+b$ has two unknowns; one point cannot fix both slope and intercept. The single-fixed-point SI definition works only because it also fixes the intercept at zero by decree ($T\to 0$ as $X\to 0$).

The origin shift was dropped; you must add 32, giving $98.6$ °F. Skipping the +32 confuses an interval with an actual reading.

Classically motion stops, but quantum mechanics leaves a residual zero-point energy; absolute zero is the state of minimum energy, not literally zero energy (link to Kinetic Theory of Gases).

Because all dilute gases behave universally at low pressure and agree between the fixed points, whereas different liquids expand non-linearly and disagree, so the gas thermometer removes the choice-of-substance ambiguity.

It guarantees that thermal equilibrium is transitive, so all mutually-equilibrated systems share one consistent number; without transitivity, "the temperature of an object" would not be a well-defined property.

Picture drawing a ruler on the thermometer: one mark alone tells you nothing about spacing, but two marks let you cut the gap into equal steps and extend those steps outward — that is exactly what "linear" means, a fixed step size all the way along. It is the simplest, most reproducible ruler you can build, and any smoothly-rising property can be defined to be linear in $T$; the price is that different substances then disagree slightly between the marks.

Look at the extrapolated line in the $P$-versus-$t_C$ figure: it reaches zero pressure at $-273.15$ °C, and an ideal gas cannot push with negative pressure, so there is no lower state to reach — this is a true floor, not an arbitrary scale origin that could dip negative.

Heat transferred depends on internal energy content, not temperature; the pool's enormous mass carries far more thermal energy despite its lower temperature (see Heat and Internal Energy).

The two scales are straight lines with different slopes, so they can only meet once. Setting the readings equal, $x=\frac{9}{5}x+32$, we move the $\frac{9}{5}x$ across: $x-\frac{9}{5}x=32$, i.e. $-\frac{4}{5}x=32$, so $x=-\frac{5}{4}\times 32=-40$. Geometrically that is the single point where the C-line and F-line cross (see the C-vs-F figure below).

$0$ K — the proportional definition $T=T_{\text{tr}}P/P_{\text{tr}}$ forces $T=0$ at $P=0$, which is absolute zero and physically unreachable for a real gas.

They have reached thermal equilibrium, so their temperatures are now equal and no net heat flows between them — but this describes the settled state, not the whole history; heat may well have flowed earlier before things stopped changing.

No — thermal equilibrium is defined for thermal contact only (heat can pass, but no work or matter exchange); allowing those would introduce other imbalances.

No — for a change $\Delta T=\Delta t_C$ exactly, because the offset cancels in any difference; the +273.15 applies only to converting an absolute reading.

No finite one — they differ by a constant 273.15 everywhere, so their straight lines are parallel and never cross (unlike C and F, whose different slopes let them meet at $-40$).

For an ideal gas it holds — a gas can only push. But condensed phases (a stretched liquid in a capillary, sap in tall trees) can sit in metastable negative-pressure states where they are under tension; that is a real, subtle effect and does not affect the ideal-gas extrapolation used to define absolute zero.