100g of water is supercooled to -10°C. At this
point, due to some disturbance mecha-nised
or otherwise some of it suddenly freezes to
ice. What will be the temperature of the
resultant mixture and how much mass would
freeze?
s. = I cal/g/Cand L" Fusion = 80 cal/g
1)0°C, 12.5g 2) 0°C, 2.5g
3) -10°C, 2.5g 4) -10°C, 12.5g
Answers
Answered by
4
Explanation:
sorry I don't know answer
Answered by
2
Answer:
option :1
Explanation:
Heat need to freeze 100g supercooled water = ms(detaT)
Temperature of the rsultant mixture = 0 degrees
80cal heat freezes = 1g
1000cal heat freezes = 1000/80 =12.5g
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