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]
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1 Answer
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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/L=1000/80= 12.5g
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This is a multiple choice answer as classified in NCERT Exemplar
(b), (c), (d) When the hot milk in the table is transferred to the surroundings by conduction, convection and radiation.
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This is a multiple choice answer as classified in NCERT Exemplar
(a), (d) During the process AB temperature of the system is 00C . hence it represents phase change that is transformation of ice into water while temperature remains 00C.
BC represents rise in temperature of water from 00C to 1000C. now water starts converting into steam which is represented by CD.
Which is only represents by a, d options
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