Due to cold weather a 1 m water pipe of cross-sectional area 1 cm² is filled with ice at -10°C. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 kΩ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33 × 10⁵ J kg⁻¹, specific heat of ice = 2 × 10³ J kg⁻¹ and density of ice = 10³ kg/m³

Solution

Physics sir · 4 steps

The ice must first warm to 0 °C and then melt, and the heater has to supply both lots of heat.

1. Mass of ice. Volume = 1 m × 1 × 10⁻⁴ m² = 10⁻⁴ m³. Mass m = 10³ × 10⁻⁴ = 0.1 kg.
2. Total heat needed. Warming from -10°C to 0°C needs 0.1 × 2 × 10³ × 10 = 2000 J. Melting needs 0.1 × 3.33 × 10⁵ = 33300 J. Together that is 35300 J.
3. Heater power. Joule heating gives P = I² R = (0.5)² × 4000 = 1000 W.

   t = (m c Δ T + m L)/(I² R) = (35300 J)/(1000 W) = 35.3 s

4. So. The minimum time is 35.3 s, which is option (b).

The idea to keep. Ice below 0 °C takes two heats: mc Δ T to bring it up to 0 °C, then mL to melt it. Add them, then divide by the power to get the time.

Answer from the NTA final answer key. Solution by PrepWiser.