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A quantity of ice at 0C0^{\circ} \mathrm{C} is added to 64.3 g64.3 \mathrm{~g} of water in a glass at 55C55^{\circ} \mathrm{C}. After the ice melted, the temperature of the water in the glass was 15C15^{\circ} \mathrm{C}. How much ice was added? The heat of fusion of water is 6.01 kJ/mol6.01 \mathrm{~kJ} / \mathrm{mol} and the specific heat is 4.18 J/(gC)4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right).

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A quantity of ice at 0°C\text{\textdegree{C}} is added to 64.3 g of water in a glass at 55°C\text{\textdegree{C}}. After the ice melted, the temperature of the water decreased to 15°C\text{\textdegree{C}}. Note that the ice melted and therefore it became part of the total water in the glass. That means the temperature of the ice increased to 15°C\text{\textdegree{C}}. We need to determine the mass of the added ice.

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