Q. 205.0( 1 Vote )

# 100 g of water is supercooled to –10° C. At this point, due to some disturbance mechanized or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?[Sw = 1 cal/g/° C and Lw fusion = 80 cal/ g]

Here, it’s given

Mass of water, m=100g

Cooled temperature, T1=-10°C

Specific heat capacity of water, Sw= 1 Cal/g/°C

Latent heat of fusion, Lw=80 Cal/g

Let the mass which would freeze be m’.

Ice forms at 0°C.

So, the temperature of the resultant mixture will be T2 = 0°C

We know,

Hence, the temperature of the resultant mixture is 0°C and the mass which would freeze is 12.5g.

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