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? [Sw=1cal/g/ and Lfusion= 80cal/g]
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? [Sw=1cal/g/ and Lfusion= 80cal/g]
This is a short answer type question as classified in NCERT Exemplar
Mass of water m =100
Change in temperature
Specific heat of water Sw= 1cal/g C
Latent heat of fusion of water Lfusion= 80cal/g
Heat required to bring water in super cooling from -10 to 0
Q= ms
Let m gram of ice be melted Q= mL
m=Q/
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This is a multiple choice answer as classified in NCERT Exemplar
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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven 2025
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