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100 g of water is supercooled to - `10^@C.` At this point, due to same disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperautre of the resultant mixture and how much mass would freeze ?`[s_w = 1 cal//g// .^@ C and L_(Fusion)^(w) = 80 cal//g]`

Answer» Heat required to bring 100 g super cooled water from `- 10^@`C to `0^@C` where ice is formed, ` = ms Delta T =100 xx 1 xx[0 - (-10)] = 1000 cal.` Thus the temperature of the resultant mixture (i.e. water and ice) `= 0^@C` If m gram of water freezes into ice. then `m xx 80 = 1000 cal` or `m = (1000)/(80) g = 12.5 g`


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