Zeroth law — transitivity of thermal equilibrium: Worked examples

This page is the "throw everything at it" workbook for the Zeroth law. We first list every kind of situation the law can hand you, then work each one until nothing is left unseen.

The scenario matrix

Below, " $\sim$ " means in thermal equilibrium with (no net heat flow, macroscopic properties settled). We write $T_{A}$ for the empirical temperature label of body $A$ ; the law guarantees $A \sim B ⟺ T_{A} = T_{B}$ .

Cell	Case class	What is tested	Example
C1	Positive transitivity	$A \sim C, B \sim C \Rightarrow A \sim B$	Ex 1
C2	One match fails	$A \sim C$ but $B \neq \sim C$ ⇒ predict flow direction	Ex 2
C3	The "both fail" trap	$A \neq \sim C, B \neq \sim C$ ⇒ no conclusion	Ex 3
C4	Degenerate / reflexive	body with itself; identical copies	Ex 4
C5	Limiting / chain	four+ bodies, transitivity iterated	Ex 5
C6	Real-world word problem	thermometer used indirectly	Ex 6
C7	Intensive vs extensive twist	equal $T$ ≠ equal energy	Ex 7
C8	Exam twist (Second-law confusion)	direction of flow is NOT Zeroth law	Ex 8

Every cell gets its own fully worked example below.

Ex 1 — Cell C1: the core positive case

Forecast: Guess now — will heat flow from $A$ to $B$ , from $B$ to $A$ , or neither?

Read the two facts. Mercury stopped moving in $A$ at mark 5 $\Rightarrow$ $A \sim C$ , with $T_{A} = 5$ . It also stopped at mark 5 in $B$ $\Rightarrow$ $B \sim C$ , with $T_{B} = 5$ . Why this step? A settled column = zero net heat flow = equilibrium by definition.
Apply transitivity. $A \sim C$ and $B \sim C$ $\Rightarrow$ $A \sim B$ (the Zeroth law). Why this step? This is literally the law's statement — the "same friend" rule.
Translate to the label. $T_{A} = 5 = T_{B} \Rightarrow T_{A} = T_{B}$ . Why this step? Equal thermometer reading means equal empirical temperature.

Answer: No net heat flow — they are already in equilibrium.

Verify — read the figure below. The figure draws the three bodies as a triangle: $A$ (teal) and $B$ (plum) sit at the top, thermometer $C$ (orange) at the bottom. The two solid orange edges are the facts you measured — " $A \sim C$ " and " $B \sim C$ ", both stamped "mark 5". The dashed black edge across the top is the induced conclusion " $A \sim B$ ": you never drew it by measurement, the Zeroth law drew it for you. Read the dashed line as the visual form of step 2. Sanity check on units: mark 5 is the same scale value in both dips, so subtracting gives $5 - 5 = 0$ net drive.

Figure — Zeroth law — transitivity of thermal equilibrium

Figure s01 — the matchmaker triangle: two solid measured edges to $C$ force the dashed $A - B$ edge.

Ex 2 — Cell C2: one match, one mismatch

Forecast: Which block cools down?

Extract labels. $A \sim C$ at mark 7 $\Rightarrow$ $T_{A} = 7$ . $D \sim C$ at mark 3 $\Rightarrow$ $T_{D} = 3$ . Why this step? Each settled reading is a valid temperature label.
Compare. $T_{A} = 7 \neq = 3 = T_{D}$ , so $A \neq \sim D$ — they are not in equilibrium. Why this step? $A \sim D ⟺ T_{A} = T_{D}$ ; unequal labels $\Rightarrow$ not equilibrated.
Decide the direction. Heat physically flows from the higher label toward the lower: $A (7) ⇝ D (3)$ — here $⇝$ is the physical-flow arrow, not the logical $\Rightarrow$ . Why this step? ⚠ This direction fact is the Second law, not the Zeroth law. The Zeroth law only told us "not in equilibrium"; it stays silent on direction.

Answer: Heat flows from $A$ to $D$ until both reach a common intermediate reading between 3 and 7.

Verify — why $T_{f}$ lies strictly between 3 and 7. Two reasons, both physical:

Energy conservation (no heat leaks to the room): whatever heat leaves $A$ enters $D$ . Writing the final common temperature as $T_{f}$ , the heat lost by the hot block equals the heat gained by the cold block, $m_{A} c_{A} (T_{A} - T_{f}) = m_{D} c_{D} (T_{f} - T_{D})$ — every symbol here is one of $T_{A} = 7$ , $T_{D} = 3$ , or the unknown $T_{f}$ ; there is no other quantity. Since the hotter body only ever loses and the colder only ever gains, $T_{f}$ cannot rise above the hottest start ( $7$ ) nor drop below the coldest start ( $3$ ).
Monotonic heat flow (Second law): flow continues only while a temperature gap remains, and always shrinks that gap, so it halts inside the interval, never outside. Hence $3 < T_{f} < 7$ . If both bodies had equal $m c$ (equal heat capacity), solving $m c (7 - T_{f}) = m c (T_{f} - 3)$ gives the midpoint $T_{f} = \frac{7 + 3}{2} = 5$ .

Ex 3 — Cell C3: the "both fail" trap

Forecast: Are $A$ and $B$ in equilibrium, or not, or unknown?

State what the law covers. The Zeroth law only fires on the positive case: two " $\sim C$ " facts. Here we have two " $\neq \sim C$ " facts. Why this step? Transitivity is a rule about equalities, not inequalities.
Show the law gives no verdict. From $A \neq \sim C$ and $B \neq \sim C$ alone, the law concludes nothing. Why this step? "Not equal to a third thing" does not force two things apart or together.
But use the actual readings. We were told both read 9 $\Rightarrow$ $T_{A} = 9 = T_{B}$ $\Rightarrow$ $A \sim B$ . Why this step? The equilibrium comes from their own equal labels, not from the failed comparison to $C$ .

Answer: $A$ and $B$ are in equilibrium (both at 9), even though both fail against $C$ . The friend's reasoning ("can't say anything") is correct only about the failed-C route; the direct readings settle it.

Verify: $T_{A} - T_{B} = 9 - 9 = 0 \Rightarrow$ equilibrium. This is exactly the [!mistake] from the parent note ("both NOT in equilibrium with C ⇒ differ") shown to be false by a concrete counter-case.

Ex 4 — Cell C4: degenerate & reflexive inputs

Forecast: Any surprises for these "trivial" inputs?

(a) Reflexivity. A body cannot have net heat flow with itself $\Rightarrow$ $A \sim A$ always. Why this step? This is the reflexive leg that (with symmetry + transitivity) makes $\sim$ an equivalence relation.
(b) Two identical readings. $A \sim C$ at 4 and $B \sim C$ at 4 $\Rightarrow$ by C1-style transitivity, $A \sim B$ . Why this step? "Identical" is not a special rule — it is just the ordinary equal-label case.
Edge check. What if $C$ itself is one of the bars? If bar $A$ is the thermometer, then " $A \sim A$ " is reflexivity again — no contradiction. Why this step? Degenerate labelling (reference = subject) still obeys the relation.

Answer: (a) Yes, trivially (reflexive). (b) Yes (equal labels).

Verify: Reflexive: net flow $= 0$ with itself by definition. Label check: $4 - 4 = 0$ .

Ex 5 — Cell C5: chain of four bodies (limiting/iterated)

Forecast: Can transitivity "hop" across an intermediate we never directly linked to $A$ ?

First hop. $A \sim C, B \sim C \Rightarrow A \sim B$ . Why this step? Zeroth law on $C$ as matchmaker.
Second hop. Now use $A \sim B$ and $B \sim D$ $\Rightarrow$ $A \sim D$ . Why this step? Apply the law again with $B$ as the new matchmaker. Transitivity chains.
Class view. $A, B, C, D$ all land in one equivalence class — one shared label $T = 6$ . Why this step? Any equivalence relation partitions everything into classes; a connected chain sits in a single class.

Answer: Yes, $A \sim D$ ; all four share $T = 6$ .

Verify — read the figure below. The figure lays out the four circles $A, B, C, D$ , each stamped "mark 6". The three solid edges are the given facts ( $A \sim C$ , $B \sim C$ , $B \sim D$ ). The dashed black edge from $A$ straight to $D$ is the two-hop conclusion — trace it and you can see the relay: $A \to C \to B \to D$ collapses into one direct link. The dotted plum loop encircling all four is the single equivalence class they now share. Symbolically: $T_{A} = T_{C} = 6$ , $T_{B} = T_{C} = 6$ , $T_{B} = T_{D} = 6 \Rightarrow T_{A} = T_{D} = 6$ ; difference $6 - 6 = 0$ .

Figure s02 — three measured links (solid) chain into the $A - D$ conclusion (dashed); the dotted loop is the one class at $T = 6$ .

Ex 6 — Cell C6: real-world word problem

Forecast: Is the indirect claim justified?

Model the bodies (with the calibration caveat). Patient $A$ , patient $B$ , and a properly calibrated thermometer $C$ whose reading is a single-valued function of true surface temperature. Why this step? Word problems become Zeroth-law problems the moment you spot the common third reference — but only after checking that "same reading = same temperature" actually holds for this instrument.
Two equal reads $\Rightarrow$ two equal temperatures. Each settled reading $37.0^{\circ} C$ maps to the same true skin temperature, so effectively $A \sim C$ and $B \sim C$ in temperature. Why this step? Under calibration, equal instrument output means equal $T$ — the input the law needs.
Conclude indirectly. Equal $T_{A}$ and equal $T_{B}$ (both $37.0^{\circ} C$ ) $\Rightarrow$ $A \sim B$ : same skin temperature, never touched directly. Why this step? This is the practical payoff of the law — the whole reason thermometers are trustworthy comparators.

Answer: Yes — under standard calibration, $T_{A} = T_{B} = 37.0^{\circ} C$ , and the indirect equality is exactly what the Zeroth law licenses.

Verify: $37.0 - 37.0 = 0.0^{\circ} C$ difference. Units check: both readings are in $^{\circ} C$ on the same instrument, so the subtraction is dimensionally valid.

Ex 7 — Cell C7: intensive-vs-extensive twist

Forecast: Equal temperature — does that force equal internal energy?

Equilibrium is fine. $A \sim C$ at 30, $B \sim C$ at 30 $\Rightarrow$ $A \sim B$ . No heat flows if joined. Why this step? Zeroth law, C1 pattern — the equilibrium claim is correct.
Separate the two quantities. Temperature is an intensive property (independent of amount); internal energy is extensive (scales with amount). Why this step? See Intensive vs extensive properties — this is the exact distinction being abused.
Quantify the gap. Take water, $c \approx 4186 J kg^{- 1} K^{- 1}$ . Pool $\approx 5 \times 1 0^{4} kg$ ; cup $\approx 0.25 kg$ . Internal energy relative to $0^{\circ} C$ : $U = m c Δ T$ with $Δ T = 30 K$ .
- Pool: $U_{A} = (5 \times 1 0^{4}) (4186) (30) = 6.279 \times 1 0^{9} J$ .
- Cup: $U_{B} = (0.25) (4186) (30) = 3.1395 \times 1 0^{4} J$ . Why this step? Concrete numbers show equal $T$ with wildly unequal $U$ .

Answer: Equilibrium: yes. Equal heat content: no — the pool holds about $2 \times 1 0^{5}$ times more internal energy despite the identical reading.

Verify: The ratio $U_{A} / U_{B} = (m_{A} c Δ T) / (m_{B} c Δ T)$ — the shared $c$ and $Δ T$ cancel — so it equals the mass ratio $\frac{5 \times 1 0 ^{4}}{0.25} = 2 \times 1 0^{5}$ . This matches the direct division $\frac{6.279 \times 1 0 ^{9}}{3.1395 \times 1 0 ^{4}} = 2 \times 1 0^{5}$ , so the two energy figures and the ratio are mutually consistent. Numbers checked in VERIFY.

Ex 8 — Cell C8: exam twist (Second-law confusion)

Forecast: Is attributing the flow direction to the Zeroth law correct?

Isolate what the Zeroth law actually asserts. Only transitivity of the zero-flow (equilibrium) situation — it never mentions direction. Why this step? Re-read the definition: it is an "if... in equilibrium... then... in equilibrium" statement. No "flows from."
Assign the direction fact to its owner. "Hot $⇝$ cold" (physical-flow arrow) is a consequence of the Second law of thermodynamics. Why this step? Direction of spontaneous heat flow is an entropy (Second-law) result, not a Zeroth-law one.
Verdict. The claim's physics is true but its attribution is wrong $\Rightarrow$ statement as written is False. Why this step? Exams penalise wrong attribution even with a true-sounding sentence.

Answer: False — the direction of heat flow belongs to the Second law; the Zeroth law only concerns equilibrium (zero flow).

Verify: Logical check — the Zeroth law is a symmetric relation ( $A \sim B \Leftrightarrow B \sim A$ ), so it cannot encode an asymmetric "hot $⇝$ cold" direction. A symmetric relation carries no arrow of preference.

Recall Quick self-test across the matrix

Both read equal on the same thermometer — equilibrium? ::: Yes (C1). One reads 7, other reads 3 — which law gives the flow direction? ::: The Second law, not the Zeroth (C2/C8). Both fail to match a third body — are they necessarily different? ::: No — no conclusion from failed matches; check their own readings (C3). Equal temperature ⇒ equal internal energy? ::: No — $T$ intensive, $U$ extensive (C7). Can transitivity chain through several bodies? ::: Yes — apply the law repeatedly, one class (C5).

Connections

Zeroth law — transitivity of thermal equilibrium (parent)
Temperature and its measurement
Thermometers and temperature scales
Thermal equilibrium and heat
Second law of thermodynamics (direction of flow)
Equivalence relations (Mathematics)
Intensive vs extensive properties