Seeking help regarding the enthalpy of steam and the heat of vaporization.

Why add the latent heat of vaporization of 130° steam as well? Is it the steam latent heat of 100° or 130°?

Conversely, when using 130° steam to indirectly heat an object, what is the relationship between the phase state of the steam and its temperature? It is 130° steam that first turns into 130° water, and then releases heat. It is still 130° steam that condenses into water at 100°, releasing heat in the process.

Is the pressure 3 kilograms from the start? I just checked – at 130 degrees Celsius, water remains in a liquid state.

It is heated to 3 kilograms, with the water temperature reaching 130°. I want to know the relationship between the phase change of water and temperature during this process. As you said, 130° corresponds to the liquid state; what would be the phase state if it were at a constant pressure of 3 kilograms and an initial pressure of 0? Thank you

Let’s assume that we need to calculate the heat required under normal temperature and pressure conditions: 1. Heat required to raise the temperature from 1.25 degrees to 100 degrees (heat for heating). 2. Heat required for the phase change from liquid to gas at 100 degrees. 3. Heat required to raise the temperature of steam from 100 degrees to 130 degrees. -------------------------------------------------------------- The value specified in your question is 0.3 MPa = 2.96 atm. At this pressure, the substance remains in liquid state throughout the temperature range from 25 degrees to 130 degrees; therefore, the heat required can be calculated directly using the formula: heat = mass * specific heat capacity * temperature difference (130 – 25)

Hello, thank you for getting back to me. If this problem involves water being heated to 130° at 0.3 Mpa to turn into steam, then would adding the vaporization heat of steam at 130° be sufficient? I’m troubling you again – if an object is indirectly heated using steam at 0.3 Mpa and 130°, the phase change of the steam is from steam at 130° to liquid at 130°; does the liquid at 130° then release heat? Thank you so much.

Thank you for your reply. I see. Actually, maybe I described some issues; what I meant was to heat it to 130 degrees under constant pressure to turn it into steam. This process starts to vaporize at 130°, is that correct?

Question 1: With constant pressure at 0.3 MPa, water is heated from 25 degrees to 130 degrees; it remains in liquid state throughout, so it does not reach the point where it turns into steam. Ps. If we assume that it does turn into steam, then the heat required would be equal to the heat needed to raise the temperature of the water from 25 to 130 degrees, plus the latent heat at 130 degrees – this constitutes the total heat required. Question 2: To put it another way, steam is generally used for heating because its latent heat is much greater than the amount of heat that water can absorb during the process of temperature increase; therefore, steam is normally used as a heat source； After being converted from steam to water, this water is then reused (in flash evaporators/tanks, or by using the condensate at 130 degrees Celsius to heat other substances). Question 3 ； To heat water to turn it into steam at a constant pressure depends on the pressure level you want to maintain. From the perspective of phase diagrams, the boiling point is 130 degrees at a pressure of 0.25 MPA; this is why some people mentioned that at 0.3 MPA, the water remains in liquid form. PS: Setting aside the pressure issue for now, if you want water to vaporize at a certain temperature, then the vaporization process occurs at that temperature until all the water has been converted to steam, after which the temperature will start to rise. I hope the reply above can resolve your confusion