can a mixure of 1 liter of boiling water (100 degree C) and 1 liter of ice(0 degree C)?

No.

First of all, frozen water (ice) has has more energy taken out of it than the boiling water has had added to it....it is called the latent heat of fusion. When freezing water, you can cool down the liquid to zero degrees C but it will still remain a liquid. In order to make the water a solid, you need to take even more energy out.

In order to boil water into steam, you not only need to heat the water up to 100 degrees, but also you need to supply the latent heat of vaporization. Since the water in your question is still a liquid when being mixed with the ice, it has not yet had this extra energy supplied.

The ice has has more energy taken out (assuming equal mass), the equilibrium temperature would not be 50 degrees.

The latent heat of fusion does not equal the latent heat of vaporization by the way.

Also, the volumes of the water will differ depending on the state. 1 Liter of ice does not equal 1 liter of liquid water as the water expands as it freezes. It would be better to compare 1 kg of ice to 1 kg of boiling water. 1 Liter of water at room temperature has a mass of about 1 kg.

yep, really close. Do this in a styrofoam container -obviously a big one. otherwise a lot of the heat will escape rather than melt the ice.

now, the reason I say really close is because you really need the same mass of water and mass of ice. becasue ice is less dense than water this means you have slightly less ice (mass wise) than boiling water. This will tip the answer a little toward the 100 degrees. Probably by 1 or 2 degrees. So the answer, for the second guess would be about 52 degrees.

Third complication is that the specific heat of water varies with temperature. This means the amount of heat required to raise or lower the temperature of 1 gram of water 1 degree C is different depending if the water is hot or the water is cold -and actually if the water is solid (i.e. ice). So for a university level answer you would also need to take this into account and it would involve integrals.

okay, i see someone else is mentioning enthalpy. yep. i see that now. so skip the integrals. you should be able to determine the heat in 1 L of boiling water. You should then be able to determine the heat in one L of ice (was that ice at 0 degrees?) okay. Then when you add the water and the ice,... the heat should be added. So now we have almost 2 L of water with a certain amount of "heat" -we should be able to use some table to determine what temperature that would be.

whew!

No

* Heat is required to change water from a solid at 0 degrees to a liquid at 0 degrees.

* A liter of ice is not the same as a liter of water. The density of ice is less than water.

Find the enthalpy of the mass units of water at the two temperatures and find the average enthalpy value. Then look up the enthalpy value to see the temperature.

When mixed, the hot water will give up enough heat to cool down 40 degrees while the cold water absorbs enough heat to raise its temperature 40 degrees so the mixture will be at 50 degrees. If your hot water is at 90 C, then, since it isn't boiling, there will be no steam. Air that is cooled by its proximity to the cold water may cause water vapor in the hot air near the hot water to condense and become visible. As far as "exactly", we are talking theoretical numbers. In the real world as soon as you remove the heat from the hot water and/or remove the cold water from refrigeration, the temperatures will start to move to the temperature of the room where you are conducting this experiment.

Yes.. at the end you will have 2 liter of water (if ignore the condensation loss at boiling water) if in the condition that it is at the room temperature

No... the end temperature is depending on the environment temperature.

Ice is going to melt and 2 liter water is going to reach room temperature.

In a vacuum, you would get 1.67 liters of 87 degree C water.

The heat of vaporization is much higher than the heat of fusion, so things are not going to even out. Look in a basic chemistry or physics text.

the energy output from reducing the temperature of 1 liter of water from 100C to 0C is approximately 100 Kcal.

The energy required to melt 1 kilo of ice at 0C to 1 kilo of water at 0C is approximately 224 Kcal.

1 liter of ice is aproximately 0.9 kilo of water.

So the final result will be approximately 1.4 kilo of water and 0.5 kilo of ice at 0C. you can do the exact calculations

2 litres certainly not, because ice's density is less than that of water, so when your ice will have melted you'll have closer to just 1.9 litre of water.

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