15 g of ice at 0^ C is added to a vessel containing water at 40^ C . The mass of water and water equivalent of the vessel is 60 g. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be [

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15 g of ice at 0^ C is added to a vessel containing water at 40^ C . The mass of water and water equivalent of the vessel is 60 g. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be [ L_ ice = 80 , cal/g , S_ water = 1 , cal/g ]

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Correct answer: C. 16^ C

Let the final temperature of the mixture be T^ C . According to the principle of calorimetry, heat lost by the water and vessel equals the heat gained by the ice. Heat lost by water and vessel = m_ water S_ water (40 - T) Heat gained by ice = m_ ice L_ ice + m_ ice S_ water (T - 0) Equating the two: 60 1 (40 - T) = 15 80 + 15 1 T 2400 - 60T = 1200 + 15T 75T = 1200 T = 1200 75 = 16^ C Answer: 16^ C

15 g of ice at 0^ C is added to a vessel containing water at 40^ C . The mass of water and water equivalent of the vessel is 60 g. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be [ L_ ice = 80 , cal/g , S_ water = 1 , cal/g ]

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