# How much heat energy is required to bring 1 liter of water to the critical point?

This is not a homework question.

The critical point of water is 374 C and 22.06 MPa. At the critical point the latent heat of vaporization is 0 - why?

Assume the water starts at 100 C.

If you add the heat required to raise the temperature of water by 274 degrees - using 4200 J / kg, it doesn't add up to the heat required to vaporize water. Each degree of temperature requires 4200 J/kg so 374 - 100 = 274 * 4200 = 1,150,800 Joules. This is roughly half the latent heat of vaporization required to vaporize a kilogram of water - 2260 kJ/kg.

Even if I use values from an Isobaric specific heat chart which shows the specific heat increasing substantially at higher temperatures, it still doesn't add up. There's still ~608,000 joules short.

I've read in several studies that higher temperatures decrease the strength of hydrogen bonds. Inversely, low pressures also decrease the strength of hydrogen bonds. Could this decreased strength in Van Der Waals forces be the reason less heat is required to render the heat of vaporization zero at the critical point?

• It's not a homework question. Please re-open. Commented Dec 9, 2021 at 0:42
• Do I have to repost it? Commented Dec 9, 2021 at 0:42

To understand what is going on look at the $$P-H$$ graph of steam. The heat of vaporization is the length of the tie line, that's the line that connects the saturated vapor to the saturated liquid at constant pressure. At 100°C this is the distance $$AC$$ in the graph. Tie lines get shorter as we move up and at the critical point the tie line collapses into a point. The length of the critical tie line is 0, and that's equal to the heat of vaporization at the critical point.
I took the $$PH$$ chart from here.