JEE/NEET Physics · Thermal Properties of Matter series · Part 2 of 4 · All parts →
- Specific heat c: energy to raise 1 kg by 1 °C — water’s huge 4186 J/(kg·°C)
- Q = mcΔT — the sensible-heat bill (temperature visibly changes)
- Latent heat L: energy for phase change at CONSTANT temperature — Q = mL
- Water’s values: c = 4186; L_fusion = 3.36×10⁵; L_vapour = 2.26×10⁶ J/kg
- Mixing problems: heat lost by hot = heat gained by cold (calorimetry’s one law)
Boil a kettle and the water climbs to 100 °C, then sits there for minutes while energy pours in — all of it secretly converting water to steam. Phase changes have hidden bills, and they’re enormous. Part 2 of the Thermal Properties of Matter series.
- Specific heat: the warming bill
- Latent heat: the phase-change bill
- The heating graph’s flat stairs
- Calorimetry: the conservation ledger
- Solved examples
- Common mistakes
- This physics in your daily life
- Practice set
- Recap
Specific Heat: The Warming Bill
Different substances charge different rates for warming. Water is expensive: 4186 J per kg per degree — five times sand’s rate, which is why beaches scorch while the sea stays cool. Q = mcΔT prices any temperature change.
Latent Heat: The Phase-Change Bill
| Letter | What it means (plain words) | Value / unit |
|---|---|---|
| c | specific heat capacity | J/(kg·°C); water 4186 |
| L | latent heat (fusion or vaporisation) | J/kg; water: L_f 3.36×10⁵, L_v 2.26×10⁶ |
| Q | heat supplied or removed | joules |
The Heating Graph’s Flat Stairs
Graph ice→water→steam against energy input: three sloped climbs (ice warming, water warming, steam warming) separated by two long flats (melting at 0°, boiling at 100°). The flats are longer than they look — vaporising 1 kg of water takes 5.4× more energy than heating it from 0 to 100 °C!
Calorimetry: The Conservation Ledger
Hot thing meets cold thing in an insulated vessel: heat lost = heat gained. Every mixing problem is one equation: m_h c_h (T_h − T_f) = m_c c_c (T_f − T_c), plus mL terms for any phase changes along the way.
Solved Examples
Q = 2 × 4186 × 80 ≈ 6.7×10⁵ J.
✔
Answer: ≈670 kJ
Equal masses, same c → 60 °C, the simple average.
Unequal masses would weight toward the heavier side. ✔
Answer: 60 °C
Ice can absorb melting 0.05×3.36×10⁵ = 16,800 J; warm water can give 0.5×4186×40 = 83,720 J — enough to melt it all.
Then (0.55 kg, mixed) balances: 16,800 + 0.55×4186×(T−0) = 83,720 → T ≈ 29 °C, all liquid.
Always compare ‘melting budget’ vs ‘available heat’ FIRST. ✔
Answer: ≈29 °C, all water
- Skipping the latent step. Ice→steam needs THREE terms: mc(0−(−x)) + mL_f + mcΔT… forgetting L is the classic error.
- Using c for phase changes. During melt/boil, temperature is fixed: only mL counts, c multiplies zero.
- Final temperature assumed. In ice problems, always test whether the ice fully melts — the answer may be a slush mixture at 0 °C.
- Negative sign chaos. Heat lost is a positive number: write magnitudes on each side of the ledger, not signed values.
This Physics in Your Daily Life
- Sweating cools you — evaporation’s latent heat is drawn FROM your skin: 2.26 MJ per litre of sweat, the body’s best air-conditioner.
- Pressure cookers — raise boiling point to ~120 °C: food cooks faster because water can finally get hotter than 100 °C.
- Coastal climates stay mild — the sea’s huge c buffers temperature swings: water is the planet’s thermal flywheel.
- Ice packs and cold drinks — melting ice absorbs 336 kJ/kg while staying at 0 °C: latent heat is a chemical-free cold battery.
- Steam burns worse than boiling water — the steam deposits its 2.26 MJ/kg vaporisation bill directly on your skin: respect the latent.
A theatre full of seated people asked to stand: they rise, one row at a time, using energy — yet everyone ends merely ‘standing’, no taller. Melting is atoms leaving their assigned seats (crystal lattice) for the aisles (liquid chaos): the energy buys freedom, not speed. Only when all are standing can the crowd start ‘jiggling’ hotter.
Heat 1 kg ice at 0° → 1 kg water at 0° costs 336,000 J with ZERO temperature change; boiling it away costs 2,260,000 J more — the same energy that would heat the water from 0 to 540 °C if it stayed liquid. The flats dwarf the slopes.
The energy-temperature graph: rising slopes (warming, Q = mcΔT) and flat plateaus (melting, boiling — Q = mL). The boiling plateau is ~5.4× the whole 0–100 climb. Read the graph left to right and you’ve read every kettle’s life story.
Practice set (answers hidden — try first)
(NEET-level) Q to heat 1 kg water by 10 °C ≈
(JEE Main-level) Energy to melt 0.2 kg ice at 0 °C:
(NEET-level) During boiling, water’s temperature:
(Concept) Evaporation cools the liquid because:
(JEE Main-level) 0.1 kg steam at 100° condenses on 1 kg water at 20°: final T ≈
- Q = mcΔT for warming; Q = mL for phase change
- water: c = 4186, L_f = 3.36×10⁵, L_v = 2.26×10⁶ J/kg
- boiling beats warming: 5.4× the 0–100 bill
- calorimetry: heat lost = heat gained
- check whether ice fully melts first
- 🔁 specific vs latent heat
- 🔁 heating graph shape
- 🔁 mixing ledger method
- 🧠 Chant: ‘sensible slope, latent flat’.
- 🧠 Steam warning: ‘vapour’s bill lands on your skin’.
- 🏠 Daily: sweating = evaporative cooling at 2.26 MJ/L.
- 🏠 Daily: coastal mildness = water’s huge c.
Quick revision
- Specific heat c: energy to raise 1 kg by 1 °C — water’s huge 4186 J/(kg·°C)
- Q = mcΔT — the sensible-heat bill (temperature visibly changes)
- Latent heat L: energy for phase change at CONSTANT temperature — Q = mL
- Water’s values: c = 4186; L_fusion = 3.36×10⁵; L_vapour = 2.26×10⁶ J/kg
- Mixing problems: heat lost by hot = heat gained by cold (calorimetry’s one law)
- Specific heat: the warming bill
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