Simulation · Physics · Class 11
A lake in winter
From the lesson Anomalous expansion of water and thermal stress in Thermal Properties of Matter. Change the values and watch what happens.
A lake in winterPhysics · Class 11
The idea behind it
NCERT §10.5
- Water does not follow the usual rule between 0 °C and 4 °C: it contracts on heating in this range. From 4 °C upwards it expands on heating like other liquids.
- So water has its largest density at 4 °C. Its volume is least there and the density falls on either side.
- Consequence for lakes and ponds in winter: water cooled at the surface sinks until the whole body reaches 4 °C. Colder water then stays on top, and the lake freezes from the surface downwards.
- The ice layer and the cold water above the depths act as insulators, so the water at the bottom stays near 4 °C and water life survives the winter. If water behaved normally, lakes would freeze from the bottom up.
- Thermal stress: a rod whose ends are held fixed cannot expand when heated, so its supports compress it. The compressive strain equals the expansion that was prevented, αΔT.
- Stress = Y × strain = YαΔT. For a steel rail 5 m long and 40 cm² in section, warmed by 10 °C with α = 1.2 × 10⁻⁵ K⁻¹ and Y = 2 × 10¹¹ N m⁻²: strain 1.2 × 10⁻⁴, stress 2.4 × 10⁷ N m⁻², force 2.4 × 10⁷ × 40 × 10⁻⁴ ≈ 10⁵ N. This is why rails and bridges have expansion gaps.
- A blacksmith fixes an iron ring on the wooden rim of a horse-cart wheel. At 27 °C the ring's diameter is 5.231 m and the rim's is 5.243 m; with α = 1.20 × 10⁻⁵ K⁻¹ the ring must be heated to about 218 °C. It fits hot and grips the rim tightly as it cools.
- Working for the ring: ΔL = 5.243 − 5.231 = 0.012 m = α L ΔT, so ΔT = 0.012/(1.20 × 10⁻⁵ × 5.231) ≈ 191.2 K, and T = 27 + 191.2 ≈ 218 °C.