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Statement I : If one gram of ice is mixed with 1 g of water at $80^{0} C$ , then the final temperature of mixture is zero.
Statement II : Latent heat of ice $80^{0} C$ cal/g.

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detailed solution

Correct option is B

Heat required by ice to melt = 1 ×80 = 80 calHeat lost by water in cooling from 800C to 00C = 1×1×(80-0) = 80 calFrom principle of calorimetry,Heat lost = Heat gainedThus, the final temperature of mixture will be 00C.

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Statement I : If one gram of ice is mixed with 1 g of water at 800C, then the final temperature of mixture is zero.Statement II : Latent heat of ice 800C cal/g.

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Statement I : If one gram of ice is mixed with 1 g of water at $80^{0} C$ , then the final temperature of mixture is zero.
Statement II : Latent heat of ice $80^{0} C$ cal/g.