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I think that the standard form of this question assumes that the water+ice is initially at 0°C, so that there's no net warming of the water, only latent heat of melting. Conversely, one could compute what relative volumes of ice and water would be required to arrive at a final temperature above 0°C. It's also possible (actually: likely) for ice to start at some temperature below 0°C, typically around -12°C for home refrigerator/freezer units. So let's say we're starting with room-temperature water at 22°C, ice at -12°C, and a ratio of 50-50 water to ice, 100g total. Latent heat of fusion for water is 80 cal/g, and latent heat is 1 cal/g. We're heating 50g of ice by 12°C and then melting it. Additional heat, if necessary, is drawn from the environment. The ice absorbs 200,000 calories. Subtracting 200,000 calories from the 50g of water would cool it by 4000°C, not counting heat of fusion, or would pretty much freeze all of it solid, if we do. Or conversely, the water can release at most 1100 calories of heat energy before freezing. Working backwards, we find that our 50g of water would require only 12g of ice to cool it to 0°C, or a ratio of roughly 1g ice to 4.2g water. |