Worked examples — Latent heat — phase transitions

1.7.5 · D3 · Physics › Thermodynamics › Latent heat — phase transitions

Yeh parent topic ka ek problem clinic hai. Hum yahaan theory dobara nahi padhayenge — balki hum har tarah ki situation ko dhundhenge jo ek latent-heat problem mein ho sakti hai, har ek ko ek box mein rakhenge, aur har box ke liye ek example solve karenge. Jab tum finish karo, toh koi bhi phase-change problem tumhe surprise nahi karegi.

Shuru karne se pehle, yeh do tools apne paas rakh lo — sirf yahi do formulas kaam aati hain:

Water ke constants jo hum baar baar use karenge (inhe ek baar yaad kar lo):

Symbol	Meaning	Value
$c_{\text{ice}}$	ice ki specific heat	$2100$ J/kg·K
$c_w$	liquid water ki specific heat	$4186$ J/kg·K
$c_{\text{steam}}$	steam ki specific heat	$2010$ J/kg·K
$L_f$	latent heat of fusion (melt/freeze)	$3.34\times 10^5$ J/kg
$L_v$	latent heat of vaporisation (boil/condense)	$22.6\times 10^5$ J/kg

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The scenario matrix

Har latent-heat problem in case classes mein se ek hoti hai. Neeche ke examples mein woh cell label ki gayi hai jise woh cover karta hai, aur milke yeh sab cells ko hit karte hain.

#	Case class	Ise kya alag banata hai	Example
A	Pure plateau, heat andar	mass already transition temp par hai, sirf $mL$	Ex 1
B	Pure plateau, heat bahar (freezing/condensing)	sign flip hota hai: heat release hoti hai, $L$ same magnitude	Ex 2
C	Multi-region climb (slope→flat→slope→…)	kinks ke paas kai terms ka sum	Ex 3
D	Mixing / calorimetry, kuch nahi bachta	heat lost = heat gained, sab use ho gayi	Ex 4
E	Mixing jahan ek substance pehle khatam ho jaati hai	check karna zaroori hai ki reaction kitni door tak jaati hai	Ex 5
F	Degenerate / limiting input (zero mass, exact-boundary energy)	edge cases jo formulas ko trip karate hain	Ex 6
G	Real-world word problem	sweating / evaporation se cooling	Ex 7
H	Exam-style twist: final temperature unknown	tumhe dhundna padega ki mixture kahaan settle hoga	Ex 8

Do badi ideas jo sab ke andar chupi hain:

1. Heat flow ki direction sign decide karti hai, lekin $L$ hamesha same size ka hota hai (melting absorb karta hai $L$, freezing release karta hai same amount).
2. Tum kabhi bhi final state guess nahi karte — tum compute karte ho ki kitni energy available hai aur dekhte ho ki woh melt–warm–boil ladder par kitni door tak jaati hai.

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Example 1 — Cell A: pure plateau, heat in

Forecast: Sirf ek term hogi ya kai? Aage padhne se pehle joules ka andaza lagao.

1. State identify karo. Ice apne melting point $0°C$ par exactly baith rahi hai. Yeh step kyun? Agar substance already transition temperature par hai, toh koi slope nahi chadhna — har joule seedha plateau ($mL$) mein jaata hai.
2. Tool chunno. Phase change ⇒ use karo $Q=m{L}_f$. Yeh step kyun? Melting ke dauraan temperature nahi badlti, isliye $\Delta T=0$ aur $Q=mc\Delta T$ galat tarike se zero dega. Hum plateau formula par switch karna zaroori hai.
3. Plug in karo. $Q=m{L}_f=0.400\times 3.34\times 10^5=1.336\times 10^5\text{ J}=133.6\text{ kJ}$

Verify: Units: $\text{kg}\times \text{J/kg}=\text{J}$ ✓. Parent topic ke Ex 1 mein $0.200$ kg ke liye $66.8$ kJ mila; mass double karne par heat double honi chahiye — aur $2\times 66.8=133.6$ kJ ✓.

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Example 2 — Cell B: pure plateau, heat OUT (freezing)

Forecast: Same mass ke liye melting energy se bada, chhota, ya equal?

1. Reverse trip ko pehchano. Freezing, melting ka ulta safar hai. Yeh step kyun? Latent heat symmetric hoti hai: freezing par release hone wali energy, same mass ke liye melting mein absorb hone wali energy ke equal hoti hai. Sirf direction flip hoti hai.
2. Magnitude compute karo. $|Q|=m{L}_f=0.150\times 3.34\times 10^5=5.01\times 10^4\text{ J}=50.1\text{ kJ}$
3. Sign lagao. Heat paani ke dwara release hoti hai, isliye paani ke viewpoint se $Q=-5.01\times 10^4$ J. Yeh step kyun? Sign bookkeeping (dekho First law of thermodynamics) tab matter karti hai jab tum ise baad mein doosre flows mein add karo; magnitude wahi hai jo freezer ko pump out karni hai.

Verify: Same $m$ aur $L_f$ jaise ek melting problem mein ⇒ same magnitude, opposite sign — exactly wahi jo symmetry demand karti hai ✓. Units J ✓.

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Example 3 — Cell C: full multi-region climb

Forecast: Kaun sa single term sabse bada bill hoga?

Heating-curve figure dekho — hum paanch regions cross karte hain, har kink par safar ko todke.

1. Ice ko warm karo $-20°C\to 0°C$ (slope, $\Delta T=20$): $Q_1=m{c}_{\text{ice}}\Delta T=0.030\times 2100\times 20=1260\text{ J}$ Yeh step kyun? Temperature badh rahi hai ⇒ slope tool $mc\Delta T$.
2. Melt karo $0°C$ par (plateau): $Q_2=m{L}_f=0.030\times 3.34\times 10^5=10020\text{ J}$ Yeh step kyun? Pehle kink par curve flat ho jaata hai — phase change, toh $mL$ par switch karo.
3. Paani ko warm karo $0°C\to 100°C$ (slope, $\Delta T=100$): $Q_3=m{c}_w\Delta T=0.030\times 4186\times 100=12558\text{ J}$ Yeh step kyun? Classic galti yahi hai ki yeh skip ho jaata hai: liquid water ko boil karne se pehle abhi bhi 100 degree chadhne padenge.
4. Boil karo $100°C$ par (plateau): $Q_4=m{L}_v=0.030\times 22.6\times 10^5=67800\text{ J}$
5. Steam ko warm karo $100°C\to 120°C$ (slope, $\Delta T=20$): $Q_5=m{c}_{\text{steam}}\Delta T=0.030\times 2010\times 20=1206\text{ J}$
6. Sum karo. $Q_{\text{total}}=1260+10020+12558+67800+1206=92844\text{ J}\approx 9.28\times 10^4\text{ J}$

Verify: Boiling term ($67800$ J) sabse bada hai — "Vaporise = Vast" se match karta hai ✓. Saare slopes $mc\Delta T$ use karte hain, dono plateaus $mL$ use karte hain ✓. Units J poore mein ✓.

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Example 4 — Cell D: mixing with nothing left over

Forecast: Paani ki apni mass se zyada ya kam?

1. Energy conservation likho (dekho Calorimetry — method of mixtures): $\underset{m{c}_w\Delta T}{\underbrace{\text{warm water dwara heat lost}}}=\underset{m_{\text{ice}}{L}_f}{\underbrace{\text{melting ice dwara heat gained}}}$ Yeh step kyun? Insulated container mein koi energy escape nahi hoti — warm side ka loss, cold side ke gain ke equal hota hai.
2. Paani ke cooling $25°C\to 0°C$ se release hone wali heat: $Q=0.300\times 4186\times 25=31395\text{ J}$
3. Equal set karo aur ice mass ke liye solve karo: $m_{\text{ice}}=\frac{Q}{L_f}=\frac{31395}{3.34\times 10^5}=0.0940\text{ kg}=94.0\text{ g}$ Yeh step kyun? Ice sirf melt hoti hai ($0°C$ par rehti hai), isliye usse milne wala har joule latent hai — $L_f$ se divide karo.

Verify: $94$ g ice $300$ g paani se melt hoti hai — paani mein kaafi energy hai lekin $L_f$ bada hai, isliye yeh apni mass ka ek tehai se kam melt karti hai. Reasonable ✓. Units: $\text{J}÷(\text{J/kg})=\text{kg}$ ✓.

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Example 5 — Cell E: one substance runs out first

Forecast: Poori tarah melt hua pond, ya ice-water slush bacha?

1. Available heat ka budget banao. Warm water zyada se zyada $0°C$ tak gir sakta hai, release karke: $Q_{\text{avail}}=m_w{c}_w\Delta T=0.050\times 4186\times 30=6279\text{ J}$ Yeh step kyun? Paani $0°C$ se neeche nahi ja sakta khud freeze hue bina, isliye yeh woh maximum energy hai jo ice draw kar sakti hai.
2. SAARI ice melt karne ke liye heat chahiye: $Q_{\text{needed}}=m_{\text{ice}}{L}_f=0.100\times 3.34\times 10^5=33400\text{ J}$ Yeh step kyun? Hum demand vs supply compare karte hain. Yeh crucial check hai jo runs-out case ko pakadta hai.
3. Compare karo. $Q_{\text{avail}}=6279$ J $<$ $Q_{\text{needed}}=33400$ J. Kaafi nahi — sirf kuch ice melt hogi. Yeh step kyun? Jab supply < demand, phase change ruk jaata hai; mixture plateau temperature $0°C$ par bachi hui ice ke saath settle hota hai.
4. Actually kitni ice melt hoti hai? $m_{\text{melted}}=\frac{Q_{\text{avail}}}{L_f}=\frac{6279}{3.34\times 10^5}=0.01880\text{ kg}=18.8\text{ g}$
5. Final state: temperature $=0°C$; ice remaining $=100-18.8=81.2$ g, $50+18.8=68.8$ g paani mein float kar rahi hai.

Verify: Melt hua mass ($18.8$ g) total ice ($100$ g) se kaafi kam hai, "not enough heat" se consistent ✓. Energy check: $18.8\text{ g}\times L_f=0.0188\times 3.34\times 10^5=6279$ J $=Q_{\text{avail}}$ ✓.

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Example 6 — Cell F: degenerate & boundary inputs

Forecast: Kya formulas break hoti hain, ya gracefully behave karti hain?

Part (a):

1. $Q=m{L}_f=0\times 3.34\times 10^5=0$ J. Yeh step kyun? Zero mass ka matlab zero substance hai jiska phase change ho — formula $0$ return karta hai, jo physically sahi hai, bug nahi.

Part (b): 2. Saari ice melt karne ki energy: $Q_{\text{melt}}=0.030\times 3.34\times 10^5=10020$ J. Yeh step kyun? Hum supplied energy ko plateau ki total demand se compare karte hain. 3. Compare karo. Supplied $=10020$ J exactly equal hai melt demand ke. Yeh step kyun? Yeh razor's-edge boundary hai: melting finish karne ke liye exactly enough, resulting water ko warm karne ke liye zero bacha. 4. Final state: $0.030$ kg liquid water exactly $0°C$ par — har joule plateau par kharch, temperature abhi nahi badhi.

Verify: (a) koi bhi formula times zero mass zero hai ✓. (b) $10020=0.030\times 3.34\times 10^5$ ✓; melting ke baad kuch nahi bachta, isliye $\Delta T=0$ aur paani $0°C$ par rehta hai ✓.

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Example 7 — Cell G: real-world word problem (sweat cooling)

Forecast: Kuch joules, ya tens of thousands?

1. Phase change spot karo. Liquid sweat → water vapour: yeh vaporisation hai (dekho Evaporation vs boiling). Yeh step kyun? Evaporation ek phase change hai, isliye plateau tool $Q=m{L}_v$ apply hota hai — bhale hi yeh $100°C$ se neeche, surface par hota ho.
2. Heat carried away compute karo: $Q=m{L}_v=0.020\times 2.43\times 10^6=4.86\times 10^4\text{ J}=48.6\text{ kJ}$ Yeh step kyun? Evaporating molecules tumhari skin se yeh latent energy chura lete hain, isliye tumhe thanda feel hota hai.

Verify: Sirf $20$ g se $48.6$ kJ — bahut bada, kyunki $L_v$ bahut bada hota hai. Isliye sweating itna effective coolant hai ✓. Units: $\text{kg}\times \text{J/kg}=\text{J}$ ✓.

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Example 8 — Cell H: exam twist, final temperature unknown

Forecast: Mixture $0°C$ se neeche, $0°C$ par, ya $0°C$ se upar hogi? Koi ice bachegi?

1. Melt-first check. Saari ice melt karne ke liye energy chahiye: $Q_{\text{melt}}=0.040\times 3.34\times 10^5=13360\text{ J}$ Warm water $0°C$ tak cool hote hue kitni de sakta hai: $Q_{\text{water}\to 0}=0.200\times 4186\times 50=41860\text{ J}$ Yeh step kyun? Temperature dhundhne se pehle, confirm karo ki ice poori tarah melt hoti hai. Kyunki $41860>13360$, saari ice melt hoti hai aur energy bachti hai — isliye final temperature $0°C$ se upar hogi.
2. Mixed final temperature $T_f$ ke liye balance set up karo. Melting ke baad, hamare paas $0.040$ kg melt-water hai (starting at $0°C$) jo warm ho raha hai, aur $0.200$ kg original water cool ho raha hai: $\underset{\text{melting par kharch}}{\underbrace{Q_{\text{melt}}}}+\underset{\text{melt-water ko warm karo}}{\underbrace{m_{\text{ice}}{c}_w(T_f-0)}}=\underset{\text{original water se heat}}{\underbrace{m_w{c}_w(50-T_f)}}$ Yeh step kyun? Energy conservation mein har term: ice ka energy budget hai "latent to melt" plus "sensible to warm up to $T_f$," jo original water ke $50°C$ se $T_f$ tak cool hone se pay hota hai.
3. Numbers substitute karo ($c_w=4186$): $13360+0.040\times 4186\times T_f=0.200\times 4186\times (50-T_f)$ $13360+167.44T_f=41860-837.2T_f$
4. $T_f$ ke liye solve karo: $167.44T_f+837.2T_f=41860-13360$ $1004.64T_f=28500⟹T_f=28.37°C$

Verify: $T_f$, $0$ aur $50°C$ ke beech hai — physically sensible ✓. Left side: $13360+167.44\times 28.37=13360+4750=18110$ J; right side: $837.2\times (50-28.37)=837.2\times 21.63=18110$ J — balanced ✓.

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Recall Which matrix cell am I in? (decision checklist)

Har cell ke liye guiding question reveal karo. Substance already transition temp par hai, sirf phase change ho raha hai ::: Cell A/B — pure plateau, $Q=mL$ (sign = direction). Ek solid state se hot gas tak kai regions mein heating ::: Cell C — slopes ($mc\Delta T$) aur plateaus ($mL$) ka sum. Do bodies mix ki gayi, bataya gaya ki woh exactly ek boundary temperature par pahunchti hain ::: Cell D — heat lost = heat gained, unknown mass ke liye solve karo. Do bodies mix ki gayi, unsure hai ki phase change complete bhi hota hai ya nahi ::: Cell E — pehle available vs needed heat compare karo. Zero mass, ya energy exactly ek plateau demand ke equal ::: Cell F — formulas phir bhi hold karte hain; "nothing left over" boundary dekho. Evaporation se real cooling ::: Cell G — surface vaporisation, $Q=m{L}_v$. Final temperature di nahi gayi ::: Cell H — pehle melt-check, phir $T_f$ ke liye energy balance.

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Connections

• Latent heat — phase transitions — woh parent topic jinhe yeh examples drill karte hain.
• Specific heat capacity — $Q=mc\Delta T$ tool jo har slope par use hota hai.
• Calorimetry — method of mixtures — Cells D, E, H ki "heat lost = heat gained" backbone.
• Kinetic theory of gases — kyon plateaus fixed temperatures par hote hain.
• Evaporation vs boiling — sweat-cooling problem (Cell G) ke peeche physics.
• First law of thermodynamics — heat in vs out ke sign conventions (Cell B).
• Phase diagrams — har transition temperature kahaan se aati hai.