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  • 100 g of water is supercooled to -10^oC. 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?

100 g of water is supercooled to -10^{\circ}C. 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?

Answers (1)

Given: Mass of water = 100g

Ice mixes with water at -10^{\circ}CNow, the heat required by --10^{\circ}C ice to 0^{\circ} C ice = ms\Delta t

                        = 100 \times 1 \times [0 - (-10)]

Thus, Q = 1000cal

Thus,

m = \frac{Q}{L}

            =\frac{1000}{80}

            = 12.5 gm

Thus, there is 12.5 gm of water and ice in the mixture and hence its temperature remains 0^{\circ} C.

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