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One kilogram of ice at 0^(@)C is mixed w...

One kilogram of ice at `0^(@)C` is mixed with one kilogram of water at `80^(@)C`. The final temperature of the mixture is
(Take : specific heat of water `=4200 J kg^(-1) K^(-1)`, latent heat of ice `= 336 kJ//kg^(-1)`)

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One kilogram of ice at 0^(@)C is mixed with one kilogram of water at 80^(@)C . The final temperature of the mixture is (take: specific heat of water=4200 J" "kg^(-1)K^(-1) , latent theat of ice =336" kJ "kg^(-1) )

336 g of ice at 0^@C is mixed with 336 g of water at 80^@C . What is the final temperature of the mixture?

If 10 g of ice is added to 40 g of water at 15^(@)C , then the temperature of the mixture is (specific heat of water = 4.2 xx 10^(3) j kg^(-1) K^(-1) , Latent heat of fusion of ice = 3.36 xx 10^(5) j kg^(-1) )

If 10 g of ice is added to 40 g of water at 15^(@)C , then the temperature of the mixture is (specific heat of water = 4.2 xx 10^(3) j kg^(-1) K^(-1) , Latent heat of fusion of ice = 3.36 xx 10^(5) j kg^(-1) )

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1kg ice at -10^(@)C is mixed with 1kg water at 100^(@)C . Then final equilirium temperature and mixture content.

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50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)

50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)