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?
Mass of water = 100 gm Change in temperature = 0 – (-10) = 10ºC Specific heat of water Sw = 1 Cal/g/ºc Latent heat of fusion L W f u s i o n LfusionW = 80 cal/g Heat required by water from – 10ºC to 0ºC Q = ms∆T = 100 × 1 × 10 Q = 1000 cal Let ice melted be m gm, Q = mL or m = Q L QL = 1000 80 100080 = 12.5gm.
