Zeroth law — transitivity of thermal equilibrium: Exercises

1.7.2 · D4 · Physics › Thermodynamics › Zeroth law — transitivity of thermal equilibrium

Shuru karne se pehle, teen symbols jo hum baar baar use karenge — ek baar earn karo, har jagah use karo:

Intuition "

$A \sim B ⟺ T_{A} = T_{B}$ " kyun — dono directions, ek picture ke saath Is claim mein do arrows hain, aur dono apna kaam karte hain:

( $\Rightarrow$ ) equilibrium se equal numbers milte hain. Zeroth law ke according, jo bhi bodies ek doosre ke saath equilibrium mein hain woh ek equivalence class banati hain. Hum poori class ko ek number $T$ assign karte hain. Toh $A \sim B$ (same class) hona by construction $T_{A} = T_{B}$ force karta hai — yeh direction ek definition hai.
( $\Leftarrow$ ) equal numbers se equilibrium milta hai. Yeh substantive direction hai aur iske liye calibration honest hona chahiye: hum sirf woh thermometer allow karte hain jiska reading true hotness ka strictly monotonic function ho (zyada reading $⟺$ zyada garam). Toh "same reading" kabhi "different hotness" hide nahi kar sakti — different hotness different readings force kar deti. Isliye equal readings $\Rightarrow$ same class $\Rightarrow A \sim B$ . Monotonicity woh promise hai jo "only if" ko sach banati hai. (Hum kyun monotonicity pe L4.1 mein aur kyun ek honest scale exist karti hai woh transitivity par depend karti hai, L5.1 mein wapas aayenge.)

Neeche wali figure class ko ek shelf ki tarah dikhati hai: ek shelf par sabka label same hota hai; ek monotone rod "kaun si shelf" ko "kaun sa number" mein badal deti hai.

Figure — Zeroth law — transitivity of thermal equilibrium

Level 1 — Recognition

(Kya tum rule spot kar ke use sahi se state kar sakte ho?)

L1.1

Zeroth law ko apne shabdon mein state karo, aur batao ki iske teen relational properties (reflexive / symmetric / transitive) mein se kaun sa isko directly supply karta hai.

Recall Solution

Statement. Agar $A \sim C$ aur $B \sim C$ , toh $A \sim B$ . Property supplied. Transitivity — "chaining" property. Reflexivity ( $A \sim A$ ) aur symmetry ( $A \sim B \Rightarrow B \sim A$ ) obvious/free hain; transitivity woh physical content hai jise law ke roop mein assert karna pada.

L1.2

Ek thermometer $C$ cup $A$ mein aur cup $B$ mein dono baar same mark read karta hai. $A$ ko $B$ se touch karaye bina, Zeroth law kya conclude karne deti hai?

Recall Solution

$C$ par dono baar same mark ka matlab hai $A \sim C$ aur $B \sim C$ with $T_{A} = T_{C} = T_{B}$ . Transitivity se, $A \sim B$ : unhe ek saath daalo toh koi net heat flow nahi hogi. Yahi practical payoff hai — comparison ek reference $C$ ke through, kabhi directly nahi.

L1.3

True ya False: "Zeroth law batati hai ki heat hot se cold ki taraf flow karti hai." Justify karo.

Recall Solution

False. Flow ki direction (hot → cold) Second law ka kaam hai. Zeroth law sirf zero-flow situations (equilibrium) aur unki transitivity describe karti hai. Yeh direction par bilkul silent hai.

Level 2 — Application

(Rule ko concrete setups mein lagao.)

L2.1

Ek thermometer par blocks read kiye: $T_{A} = 5$ , $T_{B} = 5$ , $T_{D} = 7$ (arbitrary "levels"). Predict karo kya hoga jab tum (a) $A$ – $B$ , (b) $A$ – $D$ ko touch karoge. Har step ke liye, mark karo ki kaun sa law use kar rahe ho.

Figure teen blocks ke liye reading-ladder dikhati hai — equal rungs vs. ek gap.

Recall Solution

(a) $T_{A} = T_{B} \Rightarrow A \sim B$ (Zeroth law, calibrated ladder par equal readings ke through): koi net heat flow nahi hogi. (b) $T_{D} = 7 \neq = 5 = T_{A}$ , toh equal readings nahi $\Rightarrow A \neq \sim D$ (Zeroth law sirf yeh batati hai ki woh equilibrium mein nahi hain — heat flow karegi, lekin law kis direction mein flow hogi yeh nahi batata). Kis direction mein flow hogi? Zyada garam $D$ se thande $A$ mein. ⚠️ Yeh directional claim Second law hai, Zeroth nahi. Zeroth law akele " $A \neq \sim D$ , isliye kuch flow hogi" tak ruk jaati hai. Jab flow khatam hogi dono 5 aur 7 ke beech ek common value read karenge — aur us waqt, phir se Zeroth law se, $A \sim D$ .

L2.2

Iron block $A$ $8 0^{\circ} C$ par hai, water body $B$ (ek swimming pool) $8 0^{\circ} C$ par hai. Kya woh thermal equilibrium mein hain? Kya unki internal energy same hai?

Recall Solution

Equilibrium: haan — equal temperature ka matlab $A \sim B$ hai; unhe join karne par koi net heat flow nahi hogi. Same energy: nahi — internal energy extensive hai (matter ki quantity ke saath scale hoti hai); pool mein bohot zyada hai. Temperature intensive hoti hai. Equilibrium equal $T$ ke baare mein hai, equal energy ke baare mein nahi.

L2.3

$A \sim C$ aur $B \sim C$ . Blank fill karo aur use ki gayi property ka naam batao: $A \sim B$ isliye $T_{A}__T_{B}$ .

Recall Solution

Blank " $=$ " hai: $T_{A} = T_{B}$ . Chain: $A \sim C \Rightarrow T_{A} = T_{C}$ ; $B \sim C \Rightarrow T_{B} = T_{C}$ ; equality transitive hai toh $T_{A} = T_{B}$ , isliye $A \sim B$ . Property: transitivity (Zeroth law), numerical equality ki transitivity se mirror hoti hai.

Level 3 — Analysis

(Reason karo ki arguments kyun hold karte hain ya toot jaate hain.)

L3.1

Ek student claim karta hai: " $A \neq \sim C$ aur $B \neq \sim C$ , isliye $A \neq \sim B$ ." Ek concrete counterexample do.

Recall Solution

Maano $T_{A} = T_{B} = 9 0^{\circ} C$ aur $T_{C} = 2 0^{\circ} C$ . Toh $A \neq \sim C$ aur $B \neq \sim C$ (dono $C$ se disagree karte hain), phir bhi $T_{A} = T_{B}$ toh $A \sim B$ . Zeroth law sirf positive case par fire hoti hai ( $A \sim C$ aur $B \sim C$ ). "Equilibrium mein nahi" transitive nahi hoti.

L3.2

Temperature ko ek well-defined single number hone ke liye relation $\sim$ symmetric kyun hona chahiye? Definition se argue karo.

Recall Solution

Ek poore equivalence class ko ek number $T$ attach karne ke liye, membership ordering par depend nahi karni chahiye: agar $A$ ka class $B$ ke saath share hota hai, toh $B$ ka class $A$ ke saath share hona chahiye. Agar $\sim$ symmetric nahi hota, toh " $A$ , $B$ ke class mein hai lekin $B$ , $A$ ke class mein nahi" ek relationship ke liye do alag labels coexist karne deta — body $\to$ number map ill-defined ho jaata. Jab heat flow ruk jaati hai, koi preferred direction nahi hoti, isliye $\sim$ hai symmetric, aur label consistent hai.

L3.3

Maano ek imaginary universe hai jahan equilibrium transitive nahi hai: $A \sim C$ , $B \sim C$ , lekin $A \neq \sim B$ . Explain karo ki wahan koi thermometer kyun kaam nahi kar sakta.

Recall Solution

Ek thermometer kisi body ki state ko ek number ke roop mein report karta hai aur bodies ko apne through compare karta hai ( $C$ ). Agar $A$ aur $B$ dono $C$ ki reading match karte hain phir bhi ek doosre ko match nahi karte, toh "same reading" ka matlab "same hotness" nahi raha. Reading ka $A$ vs $B$ ke baare mein koi reliable matlab nahi — poora device bekar hai. Transitivity exactly woh property hai jo ek mediated comparison ko valid banati hai.

Neeche ki figure padho: do green double-arrows ( $A - C$ aur $B - C$ ) closed loops hain — woh equilibria hold karte hain. $A$ aur $B$ ke beech red dashed double-arrow broken draw ki gayi hai: is broken universe mein woh close karne se mana kar deti hai, isliye $A$ aur $B$ ko koi consistent label pin nahi kiya ja sakta. Hamara real universe is picture ko forbid karta hai — yahi Zeroth law ka exactly content hai.

Level 4 — Synthesis

(Law ko doosre tools ke saath combine karo.)

L4.1

Ek brass rod ko length-based thermometer ki tarah use kiya jaata hai: iska length $L = L_{0} (1 + α θ)$ hai jahan $θ$ temperature hai $^{\circ} C$ mein, $L_{0} = 1.000 m$ at $0^{\circ} C$ , aur $α = 1.9 \times 1 0^{- 5}^{\circ} C^{- 1}$ . Cup $A$ mein rod $1.000247 m$ measure karta hai; cup $B$ mein bhi $1.000247 m$ measure karta hai. (a) Har cup ka temperature find karo. (b) Zeroth-law conclusion state karo. (c) Rod ki length temperature ki tarah kaam kyun karti hai?

Figure $L$ ko $θ$ ke against plot karta hai — ek seedhi, hamesha upar jaati line — taaki tum dekh sako ki kyun har length exactly ek temperature se map hoti hai.

Recall Solution

(a) $L = L_{0} (1 + α θ)$ ko $θ$ ke liye solve karo: $θ = \frac{L / L _{0} - 1}{α} = \frac{1.000247/1.000 - 1}{1.9 \times 1 0 ^{- 5}} = \frac{2.47 \times 1 0 ^{- 4}}{1.9 \times 1 0 ^{- 5}} = 13. 0^{\circ} C .$ Dono cups same length dete hain, isliye $θ_{A} = θ_{B} = 13. 0^{\circ} C$ . (b) Equal thermometer reading $\Rightarrow A \sim C$ aur $B \sim C$ same $T$ par $\Rightarrow$ Zeroth law se $A \sim B$ : mix karne par koi net heat flow nahi hogi. (c) Length kyun kaam karti hai — monotonicity argument. Graph dekho: $L (θ) = L_{0} (1 + α θ)$ ka slope $L_{0} α > 0$ har jagah hai, isliye yeh strictly increasing hai — zyada garam hamesha matlab zyada lamba, thanda hamesha matlab chhota, koi repeats nahi. Yahi "monotonic" ka matlab hai aur yahi exactly woh hai jo ek temperature scale ko chahiye: (i) har length ek temperature se correspond karti hai (line kabhi flat nahi hoti, isliye do temperatures ek length share nahi karte), aur (ii) lengths ka order hotness ke order se match karta hai (isliye "same length" honestly "same hotness" matlab rakhta hai — notation box se " $\Leftarrow$ " arrow). Koi bhi state property jiska hotness ke against strictly-rising graph ho — mercury height, electrical resistance, gas pressure — equally well kaam kar sakti. Agar graph kabhi flat ho jaaye ya neeche aaye, toh do alag hotnesses ek reading share kar lengi aur scale jhooth bolegi.

L4.2

Do thermometers — mercury ( $C_{1}$ ) aur brass rod ( $C_{2}$ ) — dono ko same water bath $W$ ke saath equilibrium mein laaya jaata hai. Mercury "level 40" read karta hai, rod $13. 0^{\circ} C$ read karta hai. Ek naya cup $A$ mercury ko "level 40" deta hai. Zeroth law ko do baar use karte hue, rod $A$ mein kya read karega?

Recall Solution

Chain 1: $A \sim C_{1}$ (dono "level 40") aur $W \sim C_{1}$ (dono "level 40") $\Rightarrow A \sim W$ . Chain 2: $A \sim W$ aur $C_{2} \sim W$ (rod $W$ par $13. 0^{\circ} C$ par calibrate hai) $\Rightarrow A \sim C_{2}$ . Toh rod, $A$ mein rakha jaaye, toh usi reading par equilibrium reach karega jo $W$ par tha: $13. 0^{\circ} C$ . Do alag thermometers agree karte hain kyunki transitivity $W$ ko unke beech mediate karne deti hai.

Level 5 — Mastery

(Argument khud banao, edge cases bhi shamil.)

L5.1

Scratch se prove karo ki agar $\sim$ reflexive, symmetric aur transitive hai, toh sets $[X] = {Y : Y \sim X}$ (isko $X$ ki class kaho) ya toh identical hain ya disjoint — isliye har body exactly ek class mein aati hai, aur hum use ek number $T$ assign kar sakte hain.

Recall Solution

Do classes $[X]$ aur $[Y]$ lo aur maano woh overlap karte hain: koi $Z$ hai jiske liye $Z \sim X$ aur $Z \sim Y$ .

Koi bhi $W \in [X]$ lo, toh $W \sim X$ . $Z \sim X$ aur symmetry se $X \sim Z$ ; transitivity se $W \sim Z$ milta hai (via $W \sim X \sim Z$ ). Phir $W \sim Z$ aur $Z \sim Y$ se $W \sim Y$ milta hai, toh $W \in [Y]$ . Isliye $[X] \subseteq [Y]$ .
Usi argument se $X, Y$ swap karke, $[Y] \subseteq [X]$ . Toh $[X] = [Y]$ . Isliye koi bhi do classes jo ek bhi member share karti hain equal hain; warna disjoint hain. Reflexivity ( $X \sim X$ ) guarantee karta hai $X \in [X]$ , isliye koi body chhoot nahi jaati. Har body exactly ek class mein belong karti hai $\Rightarrow$ har class ko ek number $T$ attach karo. Woh number temperature hai. ∎

L5.2

Degenerate / limiting cases. Har ek discuss karo: (a) ek body jo khud ke saath equilibrium mein hai; (b) do bodies jinke temperatures mein infinitesimal amount $ϵ \to 0$ ka difference hai; (c) finite heat capacity wali ek body jo thermometer $C$ ki tarah use hoti hai — real, non-ideal case.

Recall Solution

(a) Reflexivity: $A \sim A$ trivially — class hamesha apna member contain karti hai; yahi hai jo $[A]$ ko non-empty rehne deta hai aur partition ko sab kuch cover karne deta hai.

(b) $ϵ \to 0$ limit — kyun decree se nahi, continuously hota hai. Jab $ϵ$ se alag temperature wali do bodies touch karti hain, jo net heat flow hoti hai woh us difference se driven hoti hai: Second law ke according flow rate tab zero ho jaati hai jab difference zero ho jaata hai (koi gradient nahi, koi drive nahi). Concretely, settle hone se pehle exchange hone wali heat ki matra gap $ϵ$ ke saath scale karti hai (chhota gap $\Rightarrow$ chhota exchange). Jaise $ϵ \to 0$ hota hai exchange heat $\to 0$ continuously — koi sudden jump nahi, kyunki driving cause smoothly zero tak shrink hota hai. Toh "barely different" smoothly "exactly equal" mein grade ho jaata hai: exactly $T_{A} = T_{B}$ aur $A \sim B$ boundary par. Hidden assumption sirf yeh hai ki heat flow temperature difference ka ek smooth response hai (koi threshold nahi, koi jump nahi) — ordinary bodies ke liye sach hai.

(c) Finite heat capacity — measurement disturb karti hai. Ek real thermometer $C$ ki non-zero heat capacity hoti hai, isliye $C \sim A$ reach karne mein $A$ se kuch heat leni ya $A$ ko deni padti hai, jisse $A$ ka temperature $C$ ki taraf nudge hota hai. Figure dekho: final common reading $A$ ki original value aur $C$ ki starting value ke beech land karti hai, heat capacities se weighted. Practice mein fix: $C$ ko $A$ ke comparison mein tiny banao (chhoti heat capacity), taki nudge negligible ho — reading $\to$ $A$ ka true temperature. L5.2 ke "textbook limit" ka ideal thermometer isi picture ka $C$ -heat-capacity- $\to 0$ end hai.

L5.3

Ek one-paragraph experiment design karo jo Zeroth law ko falsify kare, aur batao ki kaun sa result falsification count hoga.

Recall Solution

Reference $C$ aur do bodies $A, B$ lo. $A$ ko $C$ ke saath equilibrium mein laao (reading ruk jaaye), record karo; alag se $B$ ko usi $C$ ke saath usi reading par equilibrium mein laao. Ab — zaruri hai ki dono ko room ke saath re-equilibrate na hone do — $A$ aur $B$ ko sensitive heat-flow detection ke saath directly join karo. Prediction (Zeroth law): zero net heat flow. Falsification: $A$ aur $B$ ke beech ek reproducible, non-zero net heat flow jabki dono $C$ se match kiye hon. Aisa koi result kabhi observe nahi hua — isliye " $T$ exist karti hai" ko thermometry ki poori neev banana safe hai.

Recall Quick self-check

Kis level ne test kiya ki " $\neq \sim$ transitive nahi hai"? ::: Level 3 (L3.1). L4.1 mein dono cups ka temperature kya tha? ::: $13. 0^{\circ} C$ . Equivalence-class partition proof ka output kya hai? ::: Har class ke liye ek single number (temperature). L2.1 mein actually kaun sa law kehta hai ki heat hot→cold flow karti hai? ::: Second law — Zeroth law sirf kehti hai "equilibrated nahi, isliye kuch flow hogi." Rod ke $L (θ)$ graph ki kaun si property ise valid thermometer banati hai? ::: Strict monotonicity (hamesha rising) — ek length per temperature, order preserved.

Connections

Zeroth law — transitivity of thermal equilibrium (parent)
Temperature and its measurement
Thermometers and temperature scales
Thermal equilibrium and heat
Equivalence relations (Mathematics)
Intensive vs extensive properties
Second law of thermodynamics (direction of heat flow — not Zeroth)