Zeroth law — transitivity of thermal equilibrium: Question bank

1.7.2 · D5 · Physics › Thermodynamics › Zeroth law — transitivity of thermal equilibrium

True ya false — justify karo

True/False:

A \sim C

aur

B \sim C

milke

A \sim B

force karte hain.

True — yahi Zeroth law hai (transitivity). Ek hi reference

C

share karna

A

aur

B

ko same equivalence class mein daal deta hai, isliye

T_{A} = T_{C} = T_{B}

True/False:

A \neq \sim C

aur

B \neq \sim C

milke

A \neq \sim B

force karte hain.

False — ye law sirf positive case par kaam karta hai.

A

aur

B

dono

90^{\circ} C

par ho sakte hain (matlab

A \sim B

) jabki

C

20^{\circ} C

par baitha ho; dono

C

ke against fail karte hain phir bhi aapas mein agree karte hain.

True/False: Zeroth law batata hai ki heat hot se cold ki taraf flow karta hai.

False — flow ki direction Second law hai. Zeroth law sirf zero-flow (equilibrium) situation describe karta hai aur ye nahi batata ki jab systems agree nahi karte to heat kis taraf jaata hai.

True/False: Thermal equilibrium mein do bodies ka internal energy equal hona chahiye.

False — equilibrium ka matlab equal temperature (intensive) hai, equal energy (extensive) nahi. Ek pool aur ek cup dono

30^{\circ} C

par bilkul alag amounts of internal energy rakhte hain.

True/False: Relation "

\sim

" symmetric hai.

True — jab heat flow ruk jaata hai to koi preferred direction nahi hoti, isliye

A \sim B

se turant

B \sim A

milta hai. Symmetry un teen properties mein se ek hai jo

\sim

ko equivalence relation banati hai.

True/False: Ek body apne aap ke saath thermal equilibrium mein hoti hai.

True — reflexivity,

A \sim A

. Apni hi state ki copy ke against rakha gaya system koi net heat exchange nahi karta; ye "obvious" property hai phir bhi list karna zaroori hai.

True/False: Zeroth law First aur Second laws ke baad discover hua tha.

True — ye logically pehle aane ke bawajood baad mein (Fowler, 1930s) formalise hua; zyada basic hone ke bawajood last mein naam milna hi wajah hai ki ise number zero mila.

True/False: Agar ek thermometer do cups mein same mark dikhata hai, to cups equilibrium mein hain chahe kabhi touch na kiye hon.

True — thermometer shared system

C

ka role nibhata hai; matching marks

A \sim C

aur

B \sim C

dete hain, aur transitivity direct contact ke bina

A \sim B

predict karta hai.

True/False: "Thermal contact" ka matlab hai do systems ko aapas mein mix karna.

False — contact ka matlab hai wo heat exchange kar sakte hain alag rehte hue (jaise mercury aur chai ke beech glass). Mixing ek alag process hai; equilibrium matter ko alag rakh ke judge ki jaati hai.

Error dhundho

Flaw dhundho: "Kyunki

A \sim C

aur

C \sim B

hai, aur equilibrium transitive hai, heat

A

B

ki taraf flow karta hai."

Transitivity ka conclusion hai

A \sim B

— matlab koi net heat flow nahi. Flow direction ke baare mein kuch nahi nikalta; ye reasoning Second law ko chupke se andar le aati hai.

Flaw dhundho: "

A

aur

B

ka internal energy same hai, isliye

A \sim B

Equal energy ka matlab equal temperature nahi hota. Equilibrium

T_{A} = T_{B}

se decide hoti hai, jo ek intensive property hai; energy extensive hai aur alag ho sakti hai jabki temperatures phir bhi match na karein.

Flaw dhundho: "Zeroth law temperature define karta hai, isliye

T

ka numerical formula hona chahiye."

Ye temperature ki existence ek consistent label ke roop mein define karta hai, formula nahi. Ye guarantee karta hai ki ek function

T (P, V)

exist karta hai jaise

A \sim B ⟺ T_{A} = T_{B}

; actual numbers assign karna scales ka kaam hai.

Flaw dhundho: "Kyunki

\sim

reflexive aur symmetric hai, ye ek equivalence relation hai."

Tum transitivity miss kar rahe ho — jo ki Zeroth law khud hai. Sirf reflexive + symmetric kaafi nahi hai; teeno properties milke systems ko classes mein partition karti hain.

Flaw dhundho: "Ek thermometer chai ka heat content measure karta hai."

Thermometer chai ke saath equilibrium mein aata hai aur uska temperature report karta hai, heat/energy content nahi. Ek teaspoon aur ek bathtub same temperature par same reading denge, chahe energy bilkul alag ho.

Flaw dhundho: "

A \sim B

verify hua mercury ruk ke, isliye

A

aur

B

mein equal mercury temperature hai."

Mercury ruk jaana dikhata hai ki thermometer (

C

) har cup ke saath alag-alag equilibrium mein aaya. Jo cheez

A

aur

B

ko link karti hai wo shared reading hai, mercury ka cups mein maujud hona nahi.

Why questions

Temperature ko equivalence class ka "common label" kyun kehte hain?

Kyunki ek class ka har member mutual equilibrium mein hota hai, hum poori class ko ek number attach kar sakte hain; wo number

T

hai, jo saare members mein shared hota hai aur classes ke beech alag hota hai.

Zeroth law thermometer ko kaam karne deta hai, kaise?

Ye guarantee karta hai ki do objects ke against matching reading (har ek

A \sim C

) ka matlab hai wo objects aapas mein bhi agree karte hain, isliye ek instrument un bodies ko compare kar sakta hai jinhein wo kabhi saath touch nahi karta.

Transitivity ke bina

T

ek single-valued label ke roop mein exist kyun nahi kar sakta?

Transitivity ke bina,

A \sim C

aur

B \sim C

allow kar sakte the

A \neq \sim B

— phir

A

B

C

sab ek hi

T

nahi rakh sakte, aur symbol

T

ambiguous ho jaata.

Temperature intensive kyun hai extensive nahi?

Kyunki ye ek equilibrium state label karta hai, total amount nahi: equilibrium mein ek body ko aadha karne par har aadha same temperature par rehta hai, jabki uski energy aadhi ho jaati hai.

Zeroth law First aur Second laws se "logically prior" kyun hai?

First aur Second laws energy aur entropy ke baare mein temperature ke terms mein baat karte hain, isliye temperature pehle define honi chahiye — aur wahi definition exactly Zeroth law deta hai.

Edge cases

Edge case: system

C

ek broken (stuck) column wala thermometer hai — kya transitivity argument abhi bhi hold karta hai?

Nahi — ek stuck column kisi bhi body ke saath actually equilibrium nahi reach kar sakta, isliye "

A \sim C

" kabhi truly establish nahi hota aur transitivity chain ka koi valid link nahi hota.

Edge case: teen bodies pairwise same temperature par — kitne equivalence classes?

Ek class jisme teeno hain. Har pair

A \sim B

satisfy karta hai, isliye wo sab ek single class mein merge ho jaate hain jo ek temperature value rakhti hai.

Edge case:

A

50^{\circ} C

B

50^{\circ} C

, lekin

C

20^{\circ} C

— Zeroth law kya kehta hai?

A

–

B

ke baare mein $C$ ke through kuch nahi, kyunki koi bhi

C

ke saath equilibrium mein nahi hai. Alag se

T_{A} = T_{B}

abhi bhi

A \sim B

deta hai; law bas yahan use hone wala tool nahi hai.

Edge case: kya Zeroth law us waqt apply hota hai jab do bodies touch karti hain?

Nahi — relation "

\sim

" ke liye equilibrium reach hona zaroori hai (koi net flow nahi, properties settle ho gayi hon). Transient heat exchange ke dauran bodies abhi

\sim

se related nahi hain.

Edge case: ek body jo khud do portions mein split hai — kya wo equilibrium mein hain?

Haan, reflexivity se: identical states koi net heat exchange nahi karte, isliye har portion doosre ke saath aur poore ke saath equilibrium mein hai.

Recall Ek-line self-test

Agar tum explain kar sako ki " $A \neq \sim C$ aur $B \neq \sim C$ " tumhe kuch kyun nahi batata, to tum is topic ka sabse zyada miss hone wala point samajh gaye ho. Relation sirf positive case mein conclusions force karta hai ::: Do bodies $C$ se disagree karte hue bhi identical temperatures par ho sakti hain, isliye " $C$ ke saath not-equilibrium" hona $A$ vs $B$ ke baare mein koi information nahi deta.

Connections

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