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Question
100 g of water is supercooled to –10°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?
`[S_w = 1cal/g/^circC and L_(Fusion)^w = 80cal/g]`
Solution
According to the problem, the mass of water (m) = 100 g
Change in temperature ΔT = `0 - (-10) = 10^circC`
Specific heat of water (Sw) = 1 cal/g/°C
Latent heat of fusion of water `L_(fusion)^w` = 80 cal/g
Amount of heat required to change the temperature of 100 g of water at – 10°C to 0°C,
`Q = mS_wΔT`
= `100 xx 1 xx [0 - (- 10)]`
= 1000 cal
Let m gram of ice be melted.
∴ `Q = mL`
or `m = Q/L = 1000/80` = 12.5 g
As a small mass of ice is melted, thus the temperature of the resultant mixture will remain 0°C which contains 12.5 g of ice and 87.5 g of water.
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