Since water reaches its peak density at about 4 °C, the actual outcome will vary slightly depending on the initial and the final temperature of the water once the ice has melted.
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.
No it wouldn't. The outcome will be the same regardless of the temperature, given the context of this problem. And the context is that the ice cubes are FLOATING, and that the temperature is hot enough for the ice to melt.
spindritf, not necessarily, we don't know the initial temperature of the water and on a hot day (as per the problem statement) the ice-water system will be absorbing some of the external energy.
In practical world, more likely that water will start at around 20-25°C mark and will be considerably cooler once the ice has melted but still above 4°C. So the water volume is likely to shrink a tiny bit due to the rising density and then some more due to ice and water evaporation.
But the effect might be barely noticeable and for some time water will stay level with the edges after ice melted.
The result of ice + water depends on the quantities and starting temperatures of both. You could have very cold ice in water resulting in freezing the entire volume. Or small amounts of warm ice in water resulting in water well above the freezing point.
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.