Simulation · Physics · Class 11
Heating curve: where the heat goes
From the lesson Latent heat in Thermal Properties of Matter. Change the values and watch what happens.
The idea behind it
NCERT §10.8.1
- During a change of state, heat goes in (or out) but the temperature does not change. The heat per unit mass needed for the change is the latent heat L, so Q = mL. Its SI unit is J kg⁻¹, and it depends on pressure.
- L_f, the latent heat of fusion, applies to solid ↔ liquid; L_v, the latent heat of vaporisation, applies to liquid ↔ gas. The same amount of heat is released when the change runs the other way.
- For water at standard pressure: L_f = 3.33 × 10⁵ J kg⁻¹ and L_v = 22.6 × 10⁵ J kg⁻¹. Melting 1 kg of ice at 0 °C takes 3.33 × 10⁵ J; turning 1 kg of water at 100 °C into steam takes 22.6 × 10⁵ J.
- Steam at 100 °C scalds more badly than water at 100 °C: as it condenses on skin it gives up its latent heat of vaporisation as well.
- Melting point, L_f, boiling point, L_v (L in 10⁵ J kg⁻¹): ethanol −114 °C, 1.0, 78 °C, 8.5; gold 1063 °C, 0.645, 2660 °C, 15.8; lead 328 °C, 0.25, 1744 °C, 8.67; mercury −39 °C, 0.12, 357 °C, 2.7; nitrogen −210 °C, 0.26, −196 °C, 2.0; oxygen −219 °C, 0.14, −183 °C, 2.1; water 0 °C, 3.33, 100 °C, 22.6.
- A temperature–heat graph for heating ice to steam has sloping parts (temperature rising, Q = msΔT) and flat parts (state changing, Q = mL). The slope of a sloping part is 1/(ms), so ice rises faster than water for the same heat.
- Worked example: a 0.15 kg block of ice at 0 °C goes into 0.30 kg of water at 50 °C; the mixture settles at 6.7 °C. Heat lost by water = 0.30 × 4186 × 43.3 = 54376.14 J; heat to warm the melted ice = 0.15 × 4186 × 6.7 = 4206.93 J; so 0.15 L_f = 50169.21 J and L_f ≈ 3.34 × 10⁵ J kg⁻¹.
- Worked example: turning a 3 kg ice block that starts at −12 °C into steam at 100 °C, with s_ice = 2100 J kg⁻¹ K⁻¹, s_water = 4186 J kg⁻¹ K⁻¹, L_f = 3.35 × 10⁵ J kg⁻¹, L_steam = 2.256 × 10⁶ J kg⁻¹. Warm ice 75600 J, melt 1005000 J, warm water 1255800 J, boil 6768000 J; total 9104400 J ≈ 9.1 × 10⁶ J.
- In that example boiling alone takes about three-quarters of the total heat, far more than warming the water from 0 °C to 100 °C.
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