question_answer 42)

                  100 g of water is supercooled to \[-{{10}^{o}}C\]. At this point, due to some disturbance mechanised or otherwise some of it sudden freezes to ice. What will be die temperature of the resultant mixture and how much may would freeze?                 \[[{{S}_{\omega }}=\,1\,\,cal/g{{/}^{o}}C\,\,\text{and}\,\,{{L}^{w}}_{Fusion}=80\,cal/g]\]                

Answer:

                  Heat need to freeze 100 g supercooled water                 \[=m{{S}_{w}}\Delta T\]                 Temperature of the resultant mixture \[={{0}^{o}}C\]                 80 cal heat freezes = 1 g                 1000 cal heat freezes \[=\frac{1000}{80}\,g\,=\,12.5\,g\]