Question

At 0°C the latent heat of the ice<-->liquid transition is 3.34 × 10⁵ J/kg. Clean water can be cooled a few degrees below 0°C without freezing on an ordinary time-scale, even though ice would have lower G. This non-equilibrium liquid state typically remains until some disturbance (e.g. a bubble) triggers the freezing.

1) What is the entropy difference between 4 kg of liquid water and 4 kg of ice at 0°C?

2) The specific heat of liquid water is c_pw= 4187 J/K-kg. The specific heat of ice is c_pi= 2108 J/K-kg. At T = -3.6 °C, what is the entropy difference between the mass of supercooled liquid water in part (1) and the same mass of ice? @ -3.6 °C: S_liquid-S_ice = ?

3) At -3.6 °C, what is the latent heat of the transition supercooled water--> ice per kg?

Answer 1

1. Given

Latent heat of ice (L) = 3.34 x 10^5 J/ kg

Mass of liquid water (m) = 4 kg

Difference in entropy(Q) = m *L = 3.34 x 10^5 * 4= - 1.34 x 10^6 J

2. Heat lost by water at 0 °C to water at – 3.6 °C

            H1 = m x c * change in temperature = 4 * 4187 * 3.6 = 60293 J

           Heat lost by ice at 0 °C to ice at – 3.6 °C

            H2 = 4 * 2108 * 3.6 = 30355 J

           Entropy difference between the mass of supercooled liquid water and the same mass of ice = - 1.34 x 10^6 – 30355 + 60293 = - 1310062 J

3 .Latent heat of transition (L’) = 1310062 / 4 = 327515.5.5 J/Kg