Interesting problem (not homework)

[ideasrule]

Suppose I put 1 kg of -10-degree ice in a 1 m^3 vacuum chamber. How much heat do I have to add to turn all of the ice into vapor at 100 degrees Celsius?

This seems like a ridiculously simple problem, but I realized I've no idea how to solve it. First of all, ice sublimates in a vacuum, which complicates the problem a lot. Then there's the problem that when the water boils, the newly created gas has to do work against the existing pressure to expand. Any ideas?

 

[D H]

You can't. Think about what the pressure would be for 1 kg of water vapor occupying 1 cubic meter at 100 C. Then look at where that spot is on a water phase diagram.

 

[ideasrule]

Right, stupid me. What if there was 0.6 kg of ice? That would make the water vapor have almost exactly 1 atmosphere of pressure.

 

[D H]

0.58835 grams would be better.

WIth this amount of water, this means the water vapor is not going to be an ideal gas. On the plus side, you can make it so the water vapor takes a jaunt along right along the gas edge of the phase diagram.

 

[PJC01]

Interesting problem indeed! To solve this, we can use the concept of specific heat capacity and latent heat. Specific heat capacity is the amount of heat required to raise the temperature of a substance by 1 degree Celsius, while latent heat is the amount of heat required to change the state of a substance (such as from solid to liquid or from liquid to gas) without changing its temperature.

In this case, we have 1 kg of ice at -10 degrees Celsius, which means we need to add heat to raise its temperature to 0 degrees Celsius. The specific heat capacity of ice is 2.1 J/g°C, so we need 2.1 J/g°C x 1000 g x 10 degrees = 21,000 J of heat to raise the temperature of the ice to 0 degrees Celsius.

Once the ice reaches 0 degrees Celsius, it will start to melt into liquid water. To calculate the amount of heat required for this phase change, we use the latent heat of fusion for water, which is 334 J/g. This means we need 334 J/g x 1000 g = 334,000 J of heat to melt all of the ice into liquid water.

Now, we have 1 kg of liquid water at 0 degrees Celsius. To raise its temperature to 100 degrees Celsius, we need to add heat again. The specific heat capacity of water is 4.18 J/g°C, so we need 4.18 J/g°C x 1000 g x 100 degrees = 418,000 J of heat.

Finally, we have 1 kg of water at 100 degrees Celsius, and we need to turn it into vapor. This requires the latent heat of vaporization for water, which is 2260 J/g. So, we need 2260 J/g x 1000 g = 2,260,000 J of heat to turn all of the water into vapor.

Adding all of these values together, we get a total of 2,260,000 J + 418,000 J + 334,000 J + 21,000 J = 3,033,000 J of heat. This is the amount of heat we need to add to turn 1 kg of -10-degree ice into vapor at 100 degrees Celsius in a vacuum chamber.

However, as you mentioned, the pressure in the vacuum chamber will also affect this process. The gas molecules created from