Question

100 g of water is supercooled to −10ºC. 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?

[ S_w = 1 cal/g/ºC amd \(L_{Fusion}^W\) = 80 cal/g]

Answer 1

Given : Mass of water = 100 gm

Change in temperature = 0 – (-10) = 10ºC

Specific heat of water S_w = 1 Cal/g/ºc

Latent heat of fusion \(L_{fusion}^W\) = 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 = \(\frac{Q}{L}\) = \(\frac{1000}{80}\) = 12.5gm.