Latent heat — phase transitions: Exercises — Hinglish deep dive

1.7.5 · D4 · Physics › Thermodynamics › Latent heat — phase transitions

Figure — Latent heat — phase transitions

Upar wala figure tumhara map hai: har slope ka cost hai $m c Δ T$ , har flat step ka cost hai $m L$ . Baar baar isko dekhte raho — har problem bas yehi hai: "main is staircase ke kaun se pieces climb kar raha hoon?"

Level 1 — Recognition

Goal: sahi formula identify karo aur ek substitution mein answer nikalo.

L1.1

Q: $150$ g ice bilkul $0° C$ par rakhee hai. Use poora pighlaane ke liye kitni heat chahiye?

Recall Solution

Ice already melting point par hai, isliye koi warming term nahi — yeh ek pure plateau hai, use karo $Q = m L_{f}$ . $Q = m L_{f} = 0.150 \times 3.34 \times 1 0^{5} = 5.01 \times 1 0^{4} J = 50.1 kJ$ Figure mein kaisa dikhta hai: figure mein flat orange "melt" step, $m = 0.150$ kg ke saath climb kiya.

L1.2

Q: $80$ g paani jo $100° C$ par hai, use $100° C$ par steam mein convert karne mein kitni heat lagegi?

Recall Solution

Dono start aur end $100° C$ par hain → temperature change nahi hoti → vaporisation plateau, use karo $Q = m L_{v}$ . $Q = m L_{v} = 0.080 \times 22.6 \times 1 0^{5} = 1.808 \times 1 0^{5} J \approx 181 kJ$

L1.3

Q: Ek student $0° C$ par ice ke ek tukde mein $Q = 3.34 \times 1 0^{4}$ J dalta hai aur woh poori pighal jaati hai, kuch bhi bacha nahi. Ice ka mass kya tha?

Recall Solution

Plateau formula $Q = m L_{f}$ ko mass ke liye rearrange karo: $m = \frac{Q}{L _{f}} = \frac{3.34 \times 1 0 ^{4}}{3.34 \times 1 0 ^{5}} = 0.100 kg = 100 g$

Level 2 — Application

Goal: ek calculation mein do ya zyada terms; har kink par journey ko split karo.

L2.1

Q: $200$ g ice ko $- 15° C$ se warm karke $0° C$ par paani banao (yaani, pehle warm karo, phir pighlaao).

Recall Solution

Do pieces: slope (ice warm karo) phir plateau (pighlaao).

Ice warm karo $- 15 \to 0$ : $Δ T = 15$ K, $Q_{1} = m c_{ice} Δ T = 0.200 \times 2100 \times 15 = 6300 J$
$0° C$ par pighlaao: $Q_{2} = m L_{f} = 0.200 \times 3.34 \times 1 0^{5} = 66800 J$
Total: $Q = Q_{1} + Q_{2} = 6300 + 66800 = 73100 J = 73.1 kJ$ Figure mein kaisa dikhta hai: staircase ka pehla grey slope + pehla orange plateau.

L2.2

Q: $250$ g paani ko $20° C$ se all the way steam tak $100° C$ par le jaane mein kitni heat chahiye?

Recall Solution

Slope (paani warm karo $20 \to 100$ ) phir plateau (boil karo).

Paani warm karo: $Δ T = 80$ K, $Q_{1} = m c_{w} Δ T = 0.250 \times 4186 \times 80 = 83720 J$
$100° C$ par boil karo: $Q_{2} = m L_{v} = 0.250 \times 22.6 \times 1 0^{5} = 565000 J$
Total: $Q = 83720 + 565000 = 648720 J \approx 649 kJ$ Dhyan do, boil term warming term ko bahut zyada dominate karta hai — $L_{v}$ bahut bada hota hai.

L2.3

Q: $100° C$ par steam, mass $30$ g, condense hoti hai aur resulting paani phir $40° C$ tak thanda hota hai. Kitni heat release hoti hai?

Recall Solution

"Downhill" jaane par heat release hoti hai, lekin magnitudes ke liye same formulas use hote hain.

$100° C$ par condense karo (plateau, $L_{v}$ release karta hai): $Q_{1} = m L_{v} = 0.030 \times 22.6 \times 1 0^{5} = 67800 J$
Paani cool karo $100 \to 40$ : $Δ T = 60$ K, $Q_{2} = m c_{w} Δ T = 0.030 \times 4186 \times 60 = 7534.8 J$
Total released: $Q = 67800 + 7534.8 = 75334.8 J \approx 75.3 kJ$

Level 3 — Analysis

Goal: ek mixture mein energy conservation (method of mixtures) — heat lost = heat gained.

L3.1

Q: $400$ g paani $30° C$ par hai, usme $0° C$ ki ice daali jaati hai, aur woh $0° C$ par settle hota hai, kuch ice poori tarah pighal jaati hai. Kitni mass ki ice pighi?

Recall Solution

Energy conservation: warm paani se lost heat = ice ke melting mein gained heat (latent heat ke roop mein).

Paani ka heat release hoga $30 \to 0$ cool hone par: $Q = m_{w} c_{w} Δ T = 0.400 \times 4186 \times 30 = 50232 J$
Yeh ice pighlaata hai $Q = m_{ice} L_{f}$ se: $m_{ice} = \frac{Q}{L _{f}} = \frac{50232}{3.34 \times 1 0 ^{5}} \approx 0.1504 kg \approx 150 g$

L3.2

Q: Ek calorimeter mein $250$ g paani $60° C$ par hai. Usme $50$ g ice $0° C$ par daali jaati hai. Maano sari ice pighal jaati hai, toh final temperature nikalo (calorimeter ki apni heat capacity ignore karo).

Recall Solution

Final temperature ko $T$ maano. Do cheezein heat gain karti hain (ice pighalti hai, phir woh meltwater $0 \to T$ tak warm hota hai); ek cheez heat lose karti hai (original paani $60 \to T$ tak thanda hota hai). Gained = lost set karo.

Ice side se heat gained: $Q_{gain} = melt m_{i} L_{f} + meltwater warm karo m_{i} c_{w} (T - 0)$ Warm paani ka heat lost: $Q_{lost} = m_{w} c_{w} (60 - T)$ Equal set karo: $m_{i} L_{f} + m_{i} c_{w} T = m_{w} c_{w} (60 - T)$ Numbers ( $m_{i} = 0.05$ , $m_{w} = 0.25$ ): $(0.05) (3.34 \times 1 0^{5}) + (0.05) (4186) T = (0.25) (4186) (60 - T)$ $16700 + 209.3 T = 62790 - 1046.5 T$ $1255.8 T = 46090 \Rightarrow T = 36.7° C$ Sanity check: $T$ , $0$ aur $60° C$ ke beech hai, isliye humara "sari ice pighal jaati hai" wala assumption consistent hai.

Level 4 — Synthesis

Goal: full heating curves aur multi-region journeys with sign/direction bookkeeping.

L4.1

Q: $40$ g ice ko $- 20° C$ se all the way steam tak $120° C$ par heat karo. Total heat do aur batao kaun sa single step dominate karta hai.

Recall Solution

Paanch regions — slope, plateau, slope, plateau, slope ( $m = 0.040$ kg):

Step	Formula	Value (J)
Ice warm karo $- 20 \to 0$ ( $Δ T = 20$ )	$m c_{ice} (20)$	$1680$
$0° C$ par pighlaao	$m L_{f}$	$13360$
Paani warm karo $0 \to 100$ ( $Δ T = 100$ )	$m c_{w} (100)$	$16744$
$100° C$ par boil karo	$m L_{v}$	$90400$
Steam warm karo $100 \to 120$ ( $Δ T = 20$ )	$m c_{steam} (20)$	$1608$

$Q_{total} = 1680 + 13360 + 16744 + 90400 + 1608 = 123792 J \approx 1.238 \times 1 0^{5} J$ Dominant step: boiling plateau ( $90.4$ kJ, total ka ~73%) kyunki $L_{v}$ ab tak ki sabse badi quantity hai.

L4.2

Q: Tum $60$ g ice ko $- 10° C$ se start karte hue fixed $50, 000$ J heat supply karte ho. Final state aur temperature kya hai?

Recall Solution

Energy ko staircase ke along spend karo aur dekho $50000$ J humein kitni door le jaati hai ( $m = 0.060$ ).

Ice warm karo $- 10 \to 0$ : chahiye $m c_{ice} (10) = 0.060 \times 2100 \times 10 = 1260$ J. Remaining: $50000 - 1260 = 48740$ J.
Saari ice pighlaao: chahiye $m L_{f} = 0.060 \times 3.34 \times 1 0^{5} = 20040$ J. Remaining: $48740 - 20040 = 28700$ J.
Paani ko $100° C$ ki taraf warm karo: poori climb ke liye chahiye $m c_{w} (100) = 0.060 \times 4186 \times 100 = 25116$ J. Hamare paas $28700$ J hai — kaafi zyada, paani $100° C$ reach karta hai. Remaining: $28700 - 25116 = 3584$ J.
Kuch paani boil karo: saara $60$ g boil karne ke liye chahiye $m L_{v} = 135600$ J, lekin sirf $3584$ J bacha hai. Toh sirf kuch hissa boil hoga: $m_{boiled} = \frac{3584}{22.6 \times 1 0 ^{5}} = 1.586 \times 1 0^{- 3} kg \approx 1.59 g$ Final state: paani aur steam $100° C$ par coexist karte hain — lagbhag $1.59$ g boil off ho gaya, baaki $\approx 58.4$ g liquid water hai. Temperature $100° C$ par stuck hai (energy plateau ke beech mein hi khatam ho gayi).

Level 5 — Mastery

Goal: partial phase changes, equilibrium jahan sab kuch nahi pighalta, aur reasoning ki kaun sa side "jeet" ta hai.

L5.1

Q: $100$ g ice $0° C$ par, $60$ g paani $50° C$ mein daali jaati hai. Kya saari ice pighalti hai? Agar nahi, toh final temperature aur baaki bachi ice ki mass nikalo.

Recall Solution

Pehle "sari ice pighalti hai" assumption test karo — available energy aur required energy compare karo.

Warm paani $0° C$ tak cool hone mein kitni heat de sakta hai: $Q_{available} = m_{w} c_{w} Δ T = 0.060 \times 4186 \times 50 = 12558 J$
Saari ice pighlaane ke liye heat chahiye: $Q_{needed} = m_{i} L_{f} = 0.100 \times 3.34 \times 1 0^{5} = 33400 J$ Kyunki $12558 < 33400$ , saari ice nahi pighialti. System $0° C$ par settle hoga (ice + water coexist).
Available $12558$ J se kitni ice actually pighalti hai: $m_{melt} = \frac{Q _{available}}{L _{f}} = \frac{12558}{3.34 \times 1 0 ^{5}} = 0.0376 kg = 37.6 g$
Ice baaki: $100 - 37.6 = 62.4 g ice, final temperature 0° C$

L5.2

Q: $100° C$ par kitni steam, $500$ g paani $20° C$ par condense hoke use $80° C$ tak raise kar sakti hai? (Condensed steam paani ban jaati hai aur $80° C$ tak cool bhi hoti hai.)

Recall Solution

Maano $m_{s}$ = steam ki mass. Woh do stages mein heat release karti hai (condense hona, phir $100 \to 80$ cool hona); thanda paani usse absorb karta hai $20 \to 80$ warm hoke.

Thande paani ka heat gained: $Q_{gain} = m_{w} c_{w} Δ T = 0.500 \times 4186 \times 60 = 125580 J$
Steam ka heat released: $Q_{rel} = m_{s} L_{v} + m_{s} c_{w} (100 - 80) = m_{s} (22.6 \times 1 0^{5}) + m_{s} (4186) (20)$ $= m_{s} (2.26 \times 1 0^{6} + 83720) = m_{s} (2.34372 \times 1 0^{6})$
Gained = released set karo: $m_{s} = \frac{125580}{2.34372 \times 1 0 ^{6}} = 0.05359 kg \approx 53.6 g$ Insight: kyunki condensing steam apna bada $L_{v}$ dump karti hai, steam ki chhoti mass paani ki badi mass ko garam kar sakti hai — yahi reason hai ki steam burn, boiling water burn se kahin zyada dangerous hota hai.

Connections

Specific heat capacity — upar har sloped step mein.
Calorimetry — method of mixtures — L3–L5 ka energy-balance backbone.
Kinetic theory of gases — kyon plateaus constant temperature par rehte hain.
Evaporation vs boiling — L5.2 ka steam-burn intuition.
First law of thermodynamics — $L_{v}$ energy actually kahan jaati hai.
Phase diagrams — inhi transitions ki $P$ – $T$ picture.