Visual walkthrough — Latent heat — phase transitions

1.7.5 · D2 · Physics › Thermodynamics › Latent heat — phase transitions

Is page ko padhne se pehle aapne Latent heat — phase transitions (parent) padha hoga. Yahan hum uski central formula ko bilkul scratch se, pictures ke saath banate hain.

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Step 0 — Woh do sawaal jo heat answer kar sakti hai

Kisi bhi formula se pehle yeh samjho: jab tum kisi substance mein heat energy daalte ho, woh energy hamesha in do mein se ek kaam karti hai:

• Job A — molecules ko tez karo. Tez molecules = zyada temperature. Yeh kinetic energy ki kahani hai.
• Job B — molecules ko alag karo. Speed mein koi badlaav nahi, bas molecules ke beech ki pakad todna. Yeh potential energy ke roop mein store hoti hai.

Toh heat ya toh cheezein garam kar sakti hai (Job A) ya rearrange kar sakti hai (Job B). Neeche har step mein bas yeh poochha ja raha hai: abhi kaun sa kaam ho raha hai?

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Step 1 — Woh picture jisse sab shuru hota hai: the heating curve

KYA. Hum temperature (vertical axis, $°C$ mein) ko dali gayi heat (horizontal axis, joules mein) ke against plot karte hain, jab hum $1$ block water-stuff ko thandi ice se garam steam tak steadily heat karte hain.

Graph se shuru KYU? Kyunki is curve ki shape — kahan yeh chadhti hai aur kahan flat hoti hai — yahi derivation hai. Har chadhai Job A hai; har flat stretch Job B hai. Shape padho, aur formula khud likh jaata hai.

PICTURE. Neeche wali staircase dekho. Iske paanch parts hain: climb, flat, climb, flat, climb. Ise apni ungli se left se right trace karo.

Dhyan do: heat hamesha aa rahi hai (hum heating kabhi band nahi karte), phir bhi flat parts par temperature hilne se inkar kar deti hai. Yahi saara raaz hai. Ab hum paanch pieces ko alag alag explain karenge, kyunki — aur yahi key insight hai — total heat bas paanch pieces ka sum hai:

$Q_{\text{total}}=Q_1+Q_2+Q_3+Q_4+Q_5$

Har $Q$ staircase ke ek piece ki heat hai. Hum inhe ek ek karke banate hain.

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Step 2 — Ek climbing piece: ice ko garam karna ($Q_1$)

KYA. Ice freezing se neeche shuru hoti hai, maano $-10°C$ par. Hum ise $0°C$ tak heat karte hain. Yeh graph par pehli chadhai hai.

Yahi formula KYU aur koi nahi? Temperature badal rahi hai, toh Job A ho raha hai — heat kinetic energy ban rahi hai. Woh tool jo heat ko temperature change se jodata hai woh hai specific heat capacity $c$: yeh sawaal ka jawab deta hai "kitne joules se unit mass ek degree badhti hai?"

PICTURE. Laal arrow slope par chadhta hai; temperature ke har degree rise par heat ka ek fixed chunk lagta hai.

$Q_1=\underset{\begin{array}{c}\text{mass of}\\ \text{the block (kg)}\end{array}}{\underbrace{m}}\underset{\begin{array}{c}\text{J to warm 1 kg}\\ \text{of ice by 1}°\text{C}\end{array}}{\underbrace{c_{\text{ice}}}}\underset{\begin{array}{c}\text{temperature rise}\\ =0-(-10)=10°\text{C}\end{array}}{\underbrace{\Delta T}}$

• $m$ — zyada cheez, zyada heat chahiye. Block double karo toh $Q_1$ bhi double.
• $c_{\text{ice}}$ — solid ice ke liye "price per degree per kilogram".
• $\Delta T$ — slope par kitna upar chadhte hain. Yahan $10°C$.

Agar ice pehle se hi $0°C$ par ho, toh $\Delta T=0$ aur $Q_1=0$ — yeh step simply khatam ho jaata hai. (Yahi situation parent ke Worked Example 1 mein hai.)

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Step 3 — Pehla flat piece: melting ($Q_2$)

KYA. Ab ice $0°C$ par hai. Heating jaari rakho — lekin temperature badhna band ho jaati hai. Hum pehle flat plateau par aa gaye hain. Yahan ice paani banta hai.

Chadhai KYU ruk jaati hai? Job A suspend ho gaya hai; Job B ne baar le liya hai. Ab aane wala har joule molecules ko unki rigid lattice se nikalne mein kharch ho raha hai, unhe tez karne mein nahi. Kyunki temperature = average kinetic energy, aur kinetic energy nahi badal rahi, thermometer bilkul still rehta hai.

Naya formula KYU? Kyunki yahan $\Delta T=0$ hai, formula $mc\Delta T$ dega $Q=0$ — saaf galat hai, kyunki heat toh aa rahi hai! Humein ek aisa formula chahiye jo phase change measure kare, temperature change nahi. Woh hai latent heat of fusion $L_f$.

PICTURE. Amber flat line: heat andar aa rahi hai (arrow), temperature flat (dashed level line), bonds dhile pad rahe hain (lattice → liquid).

$Q_2=\underset{\begin{array}{c}\text{mass being}\\ \text{melted (kg)}\end{array}}{\underbrace{m}}\underset{\begin{array}{c}\text{J to melt 1 kg}\\ \text{at 0}°\text{C, no }\Delta T\end{array}}{\underbrace{L_f}}$

• Koi $\Delta T$ nahi aata — temperature yahan variable nahi hai. Bas kitna mass phase change karta hai, yahi matter karta hai.
• $L_f\approx 3.34\times 10^5$ J/kg water ke liye. Yahi amber plateau ki length hai.

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Step 4 — Ek climbing piece: paani ko garam karna ($Q_3$)

KYA. Ab saari ice $0°C$ par paani ban gayi hai. Heating jaari rakho aur temperature phir se chadhti hai — is baar $100°C$ tak. Doosri chadhai.

$mc\Delta T$ par wapas KYU? Temperature phir se badal rahi hai, toh Job A resume ho gaya. Lattice pehle hi toot chuka hai; extra heat ab liquid molecules ko tez karta hai. Hum $c_{\text{water}}$ use karte hain (ice se alag price-per-degree — liquid water ko garam karna zyada mushkil hai).

PICTURE. Step 2 jaisa hi climbing arrow, lekin lamba ($100°$ ki rise) aur zyada steep price ($c_{\text{water}}>c_{\text{ice}}$).

$Q_3=\underset{\text{mass (kg)}}{\underbrace{m}}\underset{\begin{array}{c}\text{J per kg per}°\text{C}\\ \text{for }liquid\text{ water}\end{array}}{\underbrace{c_{\text{water}}}}\underset{100-0=100°\text{C}}{\underbrace{\Delta T}}$

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Step 5 — Doosra flat piece: boiling ($Q_4$)

KYA. Paani $100°C$ tak pahuncha. Temperature phir flat ho jaati hai — doosra plateau. Yahan liquid gas (steam) banta hai.

Yeh plateau itna lamba KYU hai? Melting ne sirf lattice ko dhila kiya tha; molecules abhi bhi touch kar rahe the. Boiling mein molecules ko poori tarah alag karna padta hai aur surrounding atmosphere ko peechhe dhakkelna padta hai (dekho First law of thermodynamics). Poori separation mein bahut zyada energy lagti hai, toh yeh flat stretch bahut bada hota hai.

Phir latent KYU? Wahi reasoning jaise Step 3 mein: $\Delta T=0$, phase badal raha hai, toh $Q=mL$ — ab vaporisation ki latent heat $L_v$ ke saath.

PICTURE. Bahut lamba amber plateau — melting wale se lagbhag $7\times$ bada — molecules poori tarah alag ud rahe hain.

$Q_4=\underset{\text{mass (kg)}}{\underbrace{m}}\underset{\begin{array}{c}\text{J to boil 1 kg}\\ \approx 22.6\times 10^5\text{ J/kg}\end{array}}{\underbrace{L_v}}$

Numbers compare karo: $L_v\approx 22.6\times 10^5$ versus $L_f\approx 3.34\times 10^5$ — vaporisation lagbhag saat guna bada hai. Yeh ratio master curve par donon plateaus ki width mein draw hota hai.

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Step 6 — Aakhri climbing piece: steam ko superheating karna ($Q_5$)

KYA. Ab sab steam hai $100°C$ par. Ise aur garam karo, maano $110°C$ tak. Final climb.

Ek baar aur $mc\Delta T$ KYU? Temperature badh rahi hai ⇒ Job A ⇒ specific heat. Steam ka apna alag price-per-degree $c_{\text{steam}}$ hota hai.

PICTURE. Staircase ke top-right par chhoti si aakhri chadhai.

$Q_5=\underset{\text{mass (kg)}}{\underbrace{m}}\underset{\begin{array}{c}\text{J per kg per}°\text{C}\\ \text{for steam}\end{array}}{\underbrace{c_{\text{steam}}}}\underset{110-100=10°\text{C}}{\underbrace{\Delta T}}$

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Step 7 — Pieces ko add karna: master formula

KYA. Ab hum paanch saari heats ko stack karte hain. Kyunki pieces ek ke baad ek hote hain aur energy simply accumulate hoti hai, hum inhe add karte hain.

Addition KYU kaam karta hai. Heat conserved aur additive hoti hai — poore safar ki heat har leg ki heat ka sum hoti hai. Koi overlap nahi, koi double counting nahi: har leg ya toh chadhti hai (Job A) ya flat hoti hai (Job B), kabhi dono nahi.

$Q_{\text{total}}=\underset{Q_1:\text{warm ice}}{\underbrace{m{c}_{\text{ice}}\Delta T}}+\underset{Q_2:\text{melt}}{\underbrace{m{L}_f}}+\underset{Q_3:\text{warm water}}{\underbrace{m{c}_{\text{water}}\Delta T}}+\underset{Q_4:\text{boil}}{\underbrace{m{L}_v}}+\underset{Q_5:\text{warm steam}}{\underbrace{m{c}_{\text{steam}}\Delta T}}$

Numeric check (parent Worked Example 2, $m=0.05$ kg, $-10°C\to 110°C$):

Leg	Term	Value (J)
$Q_1$ warm ice	$0.05\cdot 2100\cdot 10$	$1050$
$Q_2$ melt	$0.05\cdot 3.34\times 10^5$	$16700$
$Q_3$ warm water	$0.05\cdot 4186\cdot 100$	$20930$
$Q_4$ boil	$0.05\cdot 22.6\times 10^5$	$113000$
$Q_5$ warm steam	$0.05\cdot 2010\cdot 10$	$1005$
Total		$\approx 1.527\times 10^5$

Boiling leg ($Q_4=113$ kJ) sab par bhaari hai — bilkul wahi jo lamba amber plateau predict karta tha.

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Step 8 — Degenerate aur edge cases (kabhi mat chaunko)

Real problems mein aksar poori staircase nahi hoti. Har stub ko handle karo:

• Ice pehle se $0°C$ par hai → $Q_1=0$ (pehli chadhai hata do). Step 3 se shuru karo. Yahi Worked Example 1 hai.
• $25°C$ par paani mein ruko → $Q_1,Q_2$ rakho aur $Q_3$ ka kuch hissa; $Q_4,Q_5$ tak kabhi mat pahuncho. Staircase ko wahan truncate karo jahan ruko.
• Sirf partial melting (mixing problems, Calorimetry — method of mixtures) → $Q_2=m_{\text{ice}}{L}_f$ mein, woh mass $m_{\text{ice}}$ nikalo jo available heat melt kar sakti hai: $m_{\text{ice}}=Q/L_f$. Worked Example 3 mein, $31395/(3.34\times 10^5)\approx 94$ g melt hota hai.
• Cooling / freezing → staircase ko ulta chalao: heat release hoti hai, har $Q$ minus sign ke saath aata hai. Freezing exactly wahi $m{L}_f$ wapas deta hai jo melting ne absorb kiya tha.
• Bilkul plateau temperature par lekin koi heat nahi → kuch nahi badlega; tum flat par baithe raho jab tak heat aaye ya jaaye nahi.

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Ek-picture summary

Upar sab kuch ek annotated staircase mein compress — do tools (climbs par $mc\Delta T$, flats par $mL$), paanch legs, aur har flat ka reason.

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Socho ek bheed of kids frozen ek huddle mein hain, sab haath pakde hue. Pehle tum unhe energy dete ho aur woh zyada tezi se kaanpte hain — yahi hai ice ka warm hona, temperature chadh rahi hai (Step 2). Phir kuch ajeeb hota hai: tum energy dete rehte ho, lekin woh tez kaanpna band kar dete hain — balki woh ise ek doosre ke haath chhodne mein use karte hain. "Temperature" thari rehti hai jab tak woh azaad ho jaate hain (melting, Step 3). Jab sabke haath azaad ho jaate hain (ab woh ek milling liquid hain), nayi energy unhe tezi se bhaane lagaati hai — paani warm ho raha hai (Step 4). Phir sabse bada pause: woh bahut zyada energy lagaate hain poori tarah alag hone aur ek room mein bikhar jaane mein (boiling, Step 5) — isliye flat stretch itna lamba hai. Aakhir mein bikre hue kids (steam) phir se tez ho jaate hain (Step 6). Har "tez bhaago" leg ($mc\Delta T$) aur har "haath chhodna / ud jaana" leg ($mL$) ki energy add karo, aur total milta hai — master formula (Step 7). Ek leg miss karna aisa hai jaise bhool jaao ki kids ko alag hone aur udne ke beech abhi bhi warm hona tha.

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Connections

• Specific heat capacity — $Q=mc\Delta T$ tool jo har climbing leg par use hota hai.
• Calorimetry — method of mixtures — partial-melting aur cooling edge cases (Step 8).
• Kinetic theory of gases — temperature = average kinetic energy KYU hai (Step 0).
• Evaporation vs boiling — surface vs bulk vaporisation, dono $L_v$ use karte hain.
• First law of thermodynamics — boiling atmosphere par kaam bhi KYU karta hai (Step 5).
• Phase diagrams — yeh paanch legs $P$–$T$ space mein kahan hain.