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/ ° C  and Lfusion= 80cal/g]

16 Views|Posted 8 months ago
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P
8 months ago

This is a short answer type question as classified in NCERT Exemplar

Mass of water m =100

Change in temperature ? T = 0 10 = 10 ° C

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 ° C

Q= ms ? t = 100 * * 10 = 1000 c a l

Let m gram of ice be melted Q= mL

m=Q/

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven 2025

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