Thermodynamics internal energy Problem

AI Thread Summary
The problem involves calculating the increase in internal energy of water transitioning from liquid to vapor at one atmosphere, with a volume change from 0.9 cm³ to 2387 cm³. The latent heat of vaporization for water is given as 2.26 × 10^6 J/kg. The user initially calculated the energy using the formula Q = mL, resulting in 13500 J, but received feedback that this was incorrect. It was suggested that the user may need to adjust their calculations or consider additional factors affecting internal energy. The correct approach requires careful consideration of both heat transfer and work done during the phase change.
liamtcarroll
Messages
8
Reaction score
0
Six grams of water changes from liquid to
vapor at a pressure of one atmosphere. In the
process, the volume changes from 0.9 cm^3
to2387 cm^3. The latent heat of vaporization of
water is 2.26 × 10^6 J/kg

Find the increase in internal energy of the
water.
Answer in units of J

When i did this i did Q=mL. I got 13500J to be my answer and the program I'm using said it was wrong
 
Physics news on Phys.org
I answered it myself i had to subtract my work from my answer
 
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