Zeroth law — transitivity of thermal equilibrium: Worked examples

1.7.2 · D3 · Physics › Thermodynamics › Zeroth law — transitivity of thermal equilibrium

Yeh page Zeroth law ke liye ek "sab kuch daal do" workbook hai. Pehle hum har tarah ki situation list karte hain jo yeh law de sakta hai, phir har ek ko tab tak work out karte hain jab tak kuch bhi unseen na rahe.

The scenario matrix

Neeche, " $\sim$ " ka matlab in thermal equilibrium with hai (koi net heat flow nahi, macroscopic properties settle ho gayi hain). Hum body $A$ ke empirical temperature label ke liye $T_{A}$ likhte hain; law guarantee karta hai $A \sim B ⟺ T_{A} = T_{B}$ .

Cell	Case class	Kya test ho raha hai	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

Har cell ko apna fully worked example neeche milega.

Ex 1 — Cell C1: the core positive case

Forecast: Abhi guess karo — kya $A$ se $B$ mein heat flow hogi, $B$ se $A$ mein, ya kuch bhi nahi?

Do facts padho. Mercury $A$ mein mark 5 par ruk gaya $\Rightarrow$ $A \sim C$ , $T_{A} = 5$ ke saath. Woh $B$ mein bhi mark 5 par ruka $\Rightarrow$ $B \sim C$ , $T_{B} = 5$ ke saath. Yeh step kyun? Ek settled column = zero net heat flow = equilibrium by definition.
Transitivity apply karo. $A \sim C$ aur $B \sim C$ $\Rightarrow$ $A \sim B$ (Zeroth law). Yeh step kyun? Yeh literally law ka statement hai — "same friend" rule.
Label mein translate karo. $T_{A} = 5 = T_{B} \Rightarrow T_{A} = T_{B}$ . Yeh step kyun? Equal thermometer reading ka matlab equal empirical temperature hai.

Answer: Koi net heat flow nahi — woh already equilibrium mein hain.

Verify — neeche figure dekho. Figure teen bodies ko ek triangle ki tarah draw karta hai: $A$ (teal) aur $B$ (plum) upar hain, thermometer $C$ (orange) neeche. Do solid orange edges woh facts hain jo tumne measure kiye — " $A \sim C$ " aur " $B \sim C$ ", dono par "mark 5" stamped hai. Upar dashed black edge induced conclusion " $A \sim B$ " hai: tumne ise measurement se nahi khaincha, Zeroth law ne khaincha. Dashed line ko step 2 ki visual form ki tarah padho. Units par sanity check: mark 5 dono dips mein same scale value hai, toh subtract karne par $5 - 5 = 0$ net drive milta hai.

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: Kaun sa block cool down hoga?

Labels extract karo. $A \sim C$ at mark 7 $\Rightarrow$ $T_{A} = 7$ . $D \sim C$ at mark 3 $\Rightarrow$ $T_{D} = 3$ . Yeh step kyun? Har settled reading ek valid temperature label hai.
Compare karo. $T_{A} = 7 \neq = 3 = T_{D}$ , toh $A \neq \sim D$ — woh equilibrium mein nahi hain. Yeh step kyun? $A \sim D ⟺ T_{A} = T_{D}$ ; unequal labels $\Rightarrow$ not equilibrated.
Direction decide karo. Heat physically higher label se lower ki taraf flow karti hai: $A (7) ⇝ D (3)$ — yahan $⇝$ physical-flow arrow hai, logical $\Rightarrow$ nahi. Yeh step kyun? ⚠ Yeh direction fact Second law hai, Zeroth law nahi. Zeroth law ne sirf bataya "equilibrium mein nahi hai"; direction ke baare mein woh chup rehta hai.

Answer: Heat $A$ se $D$ ki taraf flow hoti hai jab tak dono 3 aur 7 ke beech ek common intermediate reading tak nahi pahunch jaate.

Verify — $T_{f}$ strictly 3 aur 7 ke beech kyun hai. Do reasons, dono physical:

Energy conservation (room mein koi heat leak nahi): jo bhi heat $A$ se nikali woh $D$ mein gayi. Final common temperature ko $T_{f}$ likhte hain, hot block se heat lost = cold block mein heat gained, $m_{A} c_{A} (T_{A} - T_{f}) = m_{D} c_{D} (T_{f} - T_{D})$ — yahan har symbol $T_{A} = 7$ , $T_{D} = 3$ , ya unknown $T_{f}$ mein se ek hai; koi aur quantity nahi. Kyunki hotter body sirf lose karti hai aur colder sirf gain karti hai, $T_{f}$ hottest start ( $7$ ) se upar nahi ja sakta aur coldest start ( $3$ ) se neeche nahi ja sakta.
Monotonic heat flow (Second law): flow sirf tab continue hoti hai jab temperature gap ho, aur woh gap ko hamesha shrink karti hai, toh woh interval ke andar ruk jaati hai, kabhi bahar nahi. Isliye $3 < T_{f} < 7$ . Agar dono bodies ka $m c$ equal ho (equal heat capacity), toh $m c (7 - T_{f}) = m c (T_{f} - 3)$ solve karne par midpoint $T_{f} = \frac{7 + 3}{2} = 5$ milta hai.

Ex 3 — Cell C3: the "both fail" trap

Forecast: Kya $A$ aur $B$ equilibrium mein hain, ya nahi, ya unknown?

Law kya cover karta hai yeh batao. Zeroth law sirf positive case par fire karta hai: do " $\sim C$ " facts. Yahan humhare paas do " $\neq \sim C$ " facts hain. Yeh step kyun? Transitivity equalities ke baare mein rule hai, inequalities ke baare mein nahi.
Dikhao ki law koi verdict nahi deta. Sirf $A \neq \sim C$ aur $B \neq \sim C$ se, law kuch nahi conclude karta. Yeh step kyun? "Kisi teesri cheez ke equal nahi" do cheezein alag ya saath hone ko force nahi karta.
Lekin actual readings use karo. Humein bataya gaya tha ki dono 9 padh rahi hain $\Rightarrow$ $T_{A} = 9 = T_{B}$ $\Rightarrow$ $A \sim B$ . Yeh step kyun? Equilibrium unke apne equal labels se aata hai, $C$ ke saath failed comparison se nahi.

Answer: $A$ aur $B$ equilibrium mein hain (dono 9 par), chahe dono $C$ ke against fail ho jaayein. Friend ka reasoning ("kuch nahi keh sakte") sirf failed-C route ke baare mein sahi hai; direct readings ise settle karte hain.

Verify: $T_{A} - T_{B} = 9 - 9 = 0 \Rightarrow$ equilibrium. Yeh exactly woh [!mistake] hai parent note se ("dono $C$ ke saath NOT in equilibrium $\Rightarrow$ differ karti hain") jo ek concrete counter-case se galat dikhaya gaya hai.

Ex 4 — Cell C4: degenerate & reflexive inputs

Forecast: Kya inhe "trivial" inputs mein koi surprise hai?

(a) Reflexivity. Ek body khud ke saath net heat flow nahi kar sakti $\Rightarrow$ $A \sim A$ hamesha. Yeh step kyun? Yeh reflexive leg hai jo (symmetry + transitivity ke saath) $\sim$ ko equivalence relation banata hai.
(b) Do identical readings. $A \sim C$ at 4 aur $B \sim C$ at 4 $\Rightarrow$ C1-style transitivity se, $A \sim B$ . Yeh step kyun? "Identical" koi special rule nahi hai — yeh sirf ordinary equal-label case hai.
Edge check. Agar $C$ khud bars mein se ek ho? Agar bar $A$ hi thermometer hai, toh " $A \sim A$ " phir se reflexivity hai — koi contradiction nahi. Yeh step kyun? Degenerate labelling (reference = subject) phir bhi relation ko obey karta hai.

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

Verify: Reflexive: net flow $= 0$ khud ke saath by definition. Label check: $4 - 4 = 0$ .

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

Forecast: Kya transitivity ek intermediate ke upar "hop" kar sakti hai jise humne kabhi directly $A$ se link nahi kiya?

Pehla hop. $A \sim C, B \sim C \Rightarrow A \sim B$ . Yeh step kyun? $C$ ko matchmaker ki tarah use karke Zeroth law.
Doosra hop. Ab $A \sim B$ aur $B \sim D$ use karo $\Rightarrow$ $A \sim D$ . Yeh step kyun? Law ko phir se apply karo $B$ ko naye matchmaker ki tarah. Transitivity chain hoti hai.
Class view. $A, B, C, D$ sab ek equivalence class mein aate hain — ek shared label $T = 6$ . Yeh step kyun? Koi bhi equivalence relation sab kuch classes mein partition karta hai; ek connected chain ek hi class mein hoti hai.

Answer: Haan, $A \sim D$ ; charon $T = 6$ share karte hain.

Verify — neeche figure dekho. Figure chaar circles $A, B, C, D$ ko lay out karta hai, har ek par "mark 6" stamped hai. Teen solid edges given facts hain ( $A \sim C$ , $B \sim C$ , $B \sim D$ ). $A$ se seedha $D$ tak dashed black edge do-hop conclusion hai — ise trace karo aur tum relay dekh sakte ho: $A \to C \to B \to D$ ek direct link mein collapse ho jaata hai. Sab charon ko encircle karta dotted plum loop woh single equivalence class hai jo woh ab share karte hain. 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 — teen measured links (solid) $A - D$ conclusion (dashed) mein chain hote hain; dotted loop $T = 6$ par ek class hai.

Ex 6 — Cell C6: real-world word problem

Forecast: Kya indirect claim justified hai?

Bodies ko model karo (calibration caveat ke saath). Patient $A$ , patient $B$ , aur ek properly calibrated thermometer $C$ jiska reading true surface temperature ka single-valued function hai. Yeh step kyun? Word problems Zeroth-law problems ban jaate hain jab tum common third reference spot karte ho — lekin sirf check karne ke baad ki "same reading = same temperature" is instrument ke liye actually hold karta hai.
Do equal reads $\Rightarrow$ do equal temperatures. Har settled reading $37.0^{\circ} C$ same true skin temperature map karti hai, toh effectively $A \sim C$ aur $B \sim C$ temperature mein. Yeh step kyun? Calibration ke under, equal instrument output ka matlab equal $T$ hai — woh input jo law ko chahiye.
Indirectly conclude karo. Equal $T_{A}$ aur equal $T_{B}$ (dono $37.0^{\circ} C$ ) $\Rightarrow$ $A \sim B$ : same skin temperature, directly kabhi touch nahi kiya. Yeh step kyun? Yahi law ka practical payoff hai — practically reason hai ki thermometers trustworthy comparators hain.

Answer: Haan — standard calibration ke under, $T_{A} = T_{B} = 37.0^{\circ} C$ , aur indirect equality exactly wahi hai jo Zeroth law allow karta hai.

Verify: $37.0 - 37.0 = 0.0^{\circ} C$ difference. Units check: dono readings same instrument par $^{\circ} C$ mein hain, toh subtraction dimensionally valid hai.

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

Forecast: Equal temperature — kya woh internal energy ko equal force karta hai?

Equilibrium theek hai. $A \sim C$ at 30, $B \sim C$ at 30 $\Rightarrow$ $A \sim B$ . Join karne par heat nahi flow karegi. Yeh step kyun? Zeroth law, C1 pattern — equilibrium claim sahi hai.
Do quantities alag karo. Temperature ek intensive property hai (amount se independent); internal energy extensive hai (amount ke saath scale karti hai). Yeh step kyun? Intensive vs extensive properties dekho — yahi exact distinction hai jo abuse ki ja rahi hai.
Gap quantify karo. Water lo, $c \approx 4186 J kg^{- 1} K^{- 1}$ . Pool $\approx 5 \times 1 0^{4} kg$ ; cup $\approx 0.25 kg$ . $0^{\circ} C$ ke relative internal energy: $U = m c Δ T$ with $Δ T = 30 K$ .
- Pool: $U_{A} = (5 \times 1 0^{4}) (4186) (30) = 6.279 \times 1 0^{9} J$ .
- Cup: $U_{B} = (0.25) (4186) (30) = 3.1395 \times 1 0^{4} J$ . Yeh step kyun? Concrete numbers dikhate hain equal $T$ with wildly unequal $U$ .

Answer: Equilibrium: haan. Equal heat content: nahi — pool identical reading ke bawajood lagbhag $2 \times 1 0^{5}$ guna zyaada internal energy rakhta hai.

Verify: Ratio $U_{A} / U_{B} = (m_{A} c Δ T) / (m_{B} c Δ T)$ — shared $c$ aur $Δ T$ cancel ho jaate hain — toh yeh mass ratio $\frac{5 \times 1 0 ^{4}}{0.25} = 2 \times 1 0^{5}$ ke equal hai. Yeh direct division $\frac{6.279 \times 1 0 ^{9}}{3.1395 \times 1 0 ^{4}} = 2 \times 1 0^{5}$ se match karta hai, toh do energy figures aur ratio mutually consistent hain. Numbers VERIFY mein check kiye.

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

Forecast: Kya flow direction ko Zeroth law se attribute karna sahi hai?

Isolate karo ki Zeroth law actually kya assert karta hai. Sirf zero-flow (equilibrium) situation ki transitivity — yeh kabhi direction mention nahi karta. Yeh step kyun? Definition phir se padho: yeh ek "agar... equilibrium mein... toh... equilibrium mein" statement hai. Koi "flows from" nahi.
Direction fact ko uske owner ko assign karo. "Hot $⇝$ cold" (physical-flow arrow) Second law of thermodynamics ka consequence hai. Yeh step kyun? Spontaneous heat flow ki direction ek entropy (Second-law) result hai, Zeroth-law nahi.
Verdict. Claim ki physics sahi hai lekin uski attribution galat hai $\Rightarrow$ jaise likha hai statement False hai. Yeh step kyun? Exams galat attribution ko penalise karte hain chahe sentence sahi lagta ho.

Answer: False — heat flow ki direction Second law ki hai; Zeroth law sirf equilibrium (zero flow) se concerned hai.

Verify: Logical check — Zeroth law ek symmetric relation hai ( $A \sim B \Leftrightarrow B \sim A$ ), toh yeh asymmetric "hot $⇝$ cold" direction encode nahi kar sakta. Ek symmetric relation mein preference ka koi arrow nahi hota.

Recall Matrix ke across quick self-test

Dono same thermometer par equal padh rahe hain — equilibrium? ::: Haan (C1). Ek 7 padh raha hai, doosra 3 — kaun sa law flow direction deta hai? ::: Second law, Zeroth nahi (C2/C8). Dono teesri body se match nahi karte — kya woh necessarily alag hain? ::: Nahi — failed matches se koi conclusion nahi; unki apni readings check karo (C3). Equal temperature ⇒ equal internal energy? ::: Nahi — $T$ intensive, $U$ extensive (C7). Kya transitivity kai bodies ke through chain ho sakti hai? ::: Haan — law ko repeatedly apply karo, ek 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 (flow ki direction)
Equivalence relations (Mathematics)
Intensive vs extensive properties