Latent heat — phase transitions — Hinglish notes

1.7.5 · Physics › Thermodynamics

Latent heat KYA hai?

Dhyan do $L_{v} ≫ L_{f}$ water ke liye — KYUN? Boiling mein molecules ko poori tarah alag kiya jaata hai intermolecular attraction ke against (aur atmosphere ko bhi push back karna padta hai), jabki melting sirf ek rigid lattice ko dhila karti hai ek still-touching liquid mein. Total separation ke liye bahut zyada energy chahiye.

Phase change ke dauran temperature constant KYU rehta hai?

Yahi is subtopic ka sabse gehri "KYU" hai: heat energy kinetic (→ temperature) aur potential (→ phase) mein split hoti hai. Phase transition ke dauran yeh sab potential hai.

Heating curve kaise padhein (total heat ki derivation)

Jab tum $0° C$ se neeche ki ice ko steadily heat karte ho $100° C$ se upar steam tak, tum 5 regions cross karte ho. Total heat bas inka sum hai:

$Q_{total} = warm ice m c_{ice} Δ T + melt m L_{f} + warm water m c_{water} Δ T + boil m L_{v} + warm steam m c_{steam} Δ T$

Yeh form kyun? Sloped regions (temperature badhti hai) $Q = m c Δ T$ use karte hain; flat plateaus (phase change, $Δ T = 0$ ) $Q = m L$ use karte hain. Har region independent hai — energy simply add hoti hai.

Figure — Latent heat — phase transitions

Worked Example 1 — ice melting

Q: $200$ g ice jo pehle se $0° C$ pe hai usse melt karne ke liye kitni heat chahiye? ( $L_{f} = 3.34 \times 1 0^{5}$ J/kg)

$Q = m L_{f} = 0.200 \times 3.34 \times 1 0^{5} = 6.68 \times 1 0^{4} J \approx 66.8 kJ$

Yeh step kyun? Ice pehle se $0° C$ pe hai, toh koi $m c Δ T$ term nahi — har joule seedha fusion mein jaati hai.

Worked Example 2 — full heating curve

Q: $50$ g ice ko $- 10° C$ se $110° C$ steam tak heat karo. Data: $c_{ice} = 2100$ , $c_{water} = 4186$ , $c_{steam} = 2010$ J/kg·K; $L_{f} = 3.34 \times 1 0^{5}$ , $L_{v} = 22.6 \times 1 0^{5}$ J/kg. $m = 0.05$ kg.

Step	Formula	Value (J)	Kyun
Warm ice $- 10 \to 0$	$m c_{ice} (10)$	$1050$	sloped, $Δ T = 10$
Melt at $0° C$	$m L_{f}$	$16700$	plateau, $Δ T = 0$
Warm water $0 \to 100$	$m c_{w} (100)$	$20930$	sloped, $Δ T = 100$
Boil at $100° C$	$m L_{v}$	$113000$	plateau, $Δ T = 0$
Warm steam $100 \to 110$	$m c_{s} (10)$	$1005$	sloped, $Δ T = 10$

$Q_{total} = 1050 + 16700 + 20930 + 113000 + 1005 \approx 1.527 \times 1 0^{5} J$

Yeh step kyun? Humne journey ko curve ke har kink pe split kiya; boiling plateau dominate karta hai (113 kJ) kyunki $L_{v}$ bahut bada hai.

Worked Example 3 — calorimetry (mixing)

Q: $300$ g water at $25° C$ mein $0° C$ pe ice daalke thanda kiya jaata hai. Kitni ice poori tarah pighalti hai agar water $0° C$ tak cool ho jaaye? ( $L_{f} = 3.34 \times 1 0^{5}$ , $c_{w} = 4186$ )

Water dwara release ki gayi heat: $Q = m c_{w} Δ T = 0.300 \times 4186 \times 25 = 31395 J$ Yeh ice pighalti hai: $m_{ice} = \frac{Q}{L _{f}} = \frac{31395}{3.34 \times 1 0 ^{5}} \approx 0.094 kg = 94 g$

Yeh step kyun? Energy conservation: water ka heat lost = ice ka heat gained (latent). Water $m c Δ T$ deta hai; ice usse $m L_{f}$ ke roop mein "kharach" karti hai.

Common mistake Common galtiyan (steel-manned)

1. "Heat add karne se temperature hamesha badhti hai." Kyun sahi lagta hai: Roz ki zindagi mein heating = garam hona. Fix: Phase change ke dauran saari heat potential energy ban jaati hai (bond-breaking), toh $T$ flat rehta hai. Plateau dhundho!

2. Melting/boiling ke dauran $Q = m c Δ T$ use karna. Kyun sahi lagta hai: Wahi formula ek second pehle use kiya tha. Fix: Wahan $Δ T = 0$ hai → yeh $Q = 0$ dega, jo bakwaas hai. Plateaus pe $Q = m L$ use karo.

3. Fusion aur vaporisation ke beech liquid ko warm karna bhool jaana. Kyun sahi lagta hai: Tum sochte ho "ice → water → steam, do L terms." Fix: Water ko abhi bhi $0 \to 100° C$ climb karna hoga — woh ek alag $m c Δ T$ slope hai.

4. $L_{f}$ aur $L_{v}$ ko mix up karna. Fix: Vaporisation molecules ko poori tarah alag karta hai → $L_{v}$ bahut bada hota hai.

Recall Feynman: 12-saal ke bacche ko samjhao

Socho ek group of kids frozen huddle mein haath pakde hue hain (ice). Unhe chodne aur ghoomne (liquid) ke liye tumhe energy deni hogi — lekin jab woh chhod rahe hote hain, us energy se woh tez nahi daudte, toh "temperature" same rehta hai. Sabke free hone ke baad hi nayi energy unhe tez daudati hai. Unhe saari directions mein fly off karne ke liye (steam) ek HUGE push chahiye poori tarah alag karne ke liye — isliye boiling mein melting se bahut zyada energy lagti hai.

Flashcards

"Latent" ka matlab kya hai aur heat ko yeh naam kyun diya gaya?

Latin mein hidden ka matlab; heat bina kisi temperature change ke absorb/release hoti hai, toh thermometer se "hidden" hai.

Latent heat equation units ke saath batao.

Q = m L

, jahan

Q

J mein,

m

kg mein,

L

J/kg mein.

Phase change ke dauran temperature constant kyun rehta hai?

Saari added heat potential energy ban jaati hai (intermolecular bonds todhna), kinetic energy nahi, toh average KE — aur isliye temperature — unchanged rehta hai.

Water ke liye

L_{v}

itna bada kyun hai

L_{f}

se?

Vaporisation molecules ko attraction ke against poori tarah alag karta hai (aur atmosphere pe kaam bhi karta hai); fusion sirf lattice ko touching liquid mein dhila karta hai.

Flat plateau pe kaun sa formula, slope pe kaun sa?

Plateau (ΔT=0):

Q = m L

. Slope (T changes):

Q = m c Δ T

Water ke liye

L_{f}

aur

L_{v}

ki approximate value?

L_{f} \approx 3.34 \times 1 0^{5}

J/kg,

L_{v} \approx 22.6 \times 1 0^{5}

J/kg.

200 g ice ko 0°C pe melt karne ki heat?

Q = 0.2 \times 3.34 \times 1 0^{5} = 66.8

kJ.

Heating curve mein do flat regions physically kya represent karti hain?

Melting (fusion plateau at 0°C) aur boiling (vaporisation plateau at 100°C water ke liye 1 atm pe).

Connections

Specific heat capacity — $Q = m c Δ T$ partner jo sloped regions pe use hota hai.
Calorimetry — method of mixtures — mixing problems ke liye energy conservation.
Kinetic theory of gases — constant- $T$ plateaus ke peeche KE↔temperature link explain karta hai.
Evaporation vs boiling — surface vaporisation bhi latent heat use karti hai (cooling effect).
First law of thermodynamics — $L_{v}$ mein internal-energy rise + atmosphere ko push back karne ka work dono shamil hain.
Phase diagrams — yahan in transitions ko $P$ – $T$ space mein dekha jaata hai.