Energy Needed to Evaporate Water at 289.1 torr: Table of Pressures

AI Thread Summary
To evaporate 1 kg of water at a pressure of 289.1 torr, the boiling temperature is 75°C. The energy required for this phase change can be found using steam tables, which provide specific enthalpy values at various pressures. Users can access steam tables through various online resources and academic publications. These tables detail the energy needed for vaporization at different temperatures and pressures. Accurate calculations depend on these references for precise energy requirements.
oldandcurious
Messages
1
Reaction score
0
Knowing the pressure to be 289.1 torr and boiling temperature of water at this pressure to be 75° C, what is the amount of energy needed to evaporate 1kg of water. Where can I find the table for different pressures?
 
Physics news on Phys.org
Hi and welcome.
I think you may be after "Steam tables", as in this link. and elsewhere.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top