I need an answer: the heat required to evaporate 1 kilogram of water
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How much energy is required to heat 1 kilogram of water at 20 degrees Celsius until it evaporates into steam at 150 degrees Celsius, under normal pressure? Please help answer it; a detailed calculation process is needed
This post was last edited by ylb913 on 2018-8-22 06:51. Isn’t atmospheric pressure steam at 100°C? How can it reach 150°C? The energy required to raise temperature from 20°C to 100°C is 80 kcal/kg, while the latent heat of vaporization of water at 100°C is 539 kcal/kg; together these amounts to 619 kcal/kg. It takes about 39 kcal/kg of heat to superheat saturated steam at 100°C to 150°C. What was mentioned above concerns the changes in heat before and after water turns into steam; as for exactly how much heat source is required, there is also the issue of heat utilization efficiency.
Using computational software, the difference in enthalpy values obtained by inputting the two states is the heat that needs to be supplied； Software address: https://bbs.hcbbs.com/thread-2045371-1-1.html
The heat required to heat to 100 degrees is 431.3 KJ, and the enthalpy of vaporization is 2257.6 KJ. The heat needed to heat the steam to 150 degrees is 102 KJ; in total, this amounts to 2864 KJ, which corresponds to 796 W
By consulting the enthalpy table (at standard pressure, i.e., absolute pressure of about 0.1 MPa), the enthalpy of superheated steam at 150 degrees is 2776.32 kJ/kg, while the enthalpy of water at 20 degrees is 83.95 kJ/kg. To heat 1 kg of water from 20 degrees to 150 degrees at standard pressure, an amount of energy of 1 kg x (2776.32 kJ/kg – 83.95 kJ/kg) = 2692.37 kJ is required. Taking into account heat losses during the heating process, with a coefficient of 0.98, the required energy amounts to 2696.37 kJ / 0.98 = 2747.32 kJ