10 g of ice at -10 degree Celsius is added to 10 g of water at 85 degree Celsius. What is the final temperature and amount of ice left in the system?

Please explain...

Specific heat of water is = 4.2 J/(g.K)

Specific heat of ice is = 2.1 J/(g.K)

Latent heat of ice is = 336 J/g

Heat released by 10 g of water at 85 ^{o}C in converting to water at 0^{ }^{o}C is,

H1 = msθ = (10)(4.2)(85) = 3570J

Heat gained by 10 g of ice at -10 ^{o}C in converting into ice at 0 ^{o}C is,

H2 = (10)(2.1)(10) = 210 J

Suppose, ‘x’ g of ice now melts. So, heat gained by this ‘x’ g of ice in converting into water at 0 ^{o}C is,

H3 = xL = 336x J

Now, H3 + H2 = H1

=> 336x + 210 = 3570

=> x = 10 g

Mass of ice remaining is = 10 – 10 = 0 , no ice will remain in the system.

Thus, in the resulting mixture the ice remaining is 0 g and the temperature of the mixture is 0 ^{o}C.

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