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1 kg ice at 0 degree is mixed with 1 kg of water at 80 degree ,final temperature of mixture?(sp.heat of water-4200kj/kg)

sree lekshmi , 14 Years ago
Grade 12
anser 1 Answers
Sanandanan M

It is easier to explain things with CGS units..I mean calorie instead of Joule..

specific heat of water is 1 cal/g/K, so  ms(deltaT)=1000*1*80=80000 calories required for water to reach zero degree

latent heat of fusion of ice is 80cal/g, so mL=1000*80=80000 calories required for ice to reach water at zero degree

So, the mixture is water at zero degree.

Last Activity: 14 Years ago
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