When a tiny ice crystal is dropped into supercooled water, the water starts to freeze immediately. Determine the mass of ice formed from 1 kg of water supercooled to -8 degree C. (Take specific heat of water = 4200 J/kg/C and latent heat of fusion = 336000 J/kg.)

40 g
60 g
80 g
100 g

Correct answer: 100 g

Solution

Warming the supercooled water from -8 C to 0 C releases heat that freezes a mass m_ice, giving m_ice = (1 * 4200 * 8)/336000 = 0.1 kg = 100 g.

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