Question:
100 g of water is supercooled to -10oC. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?
Solution:
Mass of water = 100 g
At -10oC the mixture has water and ice
Heat required by the mixture is ms∆t = (100)(1)(0-(-10) = 1000 Cal
Therefore, the mass of the mixture, m = Q/L = 12.5 g
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