internal energy

[cowgiljl]

Seem to have ran into a brick wall with this problem
Determine the change in internal energy of 1 kg of water at 100 degrees C when it is fully boiled. Once boiled this volume of water changes to 1671 Liters of steam at 100 degrees C Assume the pressure remains constantat 1 atm

things i know
1 L =1E-3 m3
1atm = 1.013E5 N/m2
1 L = 1000 cm3 = 1E-3 m3
formulsa used
Q=mLv = 1kg*2.26E6 J/k Q=2260000 J
W = -P(Vsteam-Vwater) = (1.013E5)*[(1671000 - 1000)*.001
W = -169171E3 J

change in U = Q+W
2260000-169171000
U = -166911E3 J
If i did make a mistake i think it is where W is .
Thanks Joe

[drag]

Hmm... I can't remember if your calc. of Q is
correct but as far as W goes, if memory serves right,
I believe the conversion should be 1 Atm*Liter = 101 Joule,
and then you get:
W = - P * dV = - 1 (Atm) * (1671-1) (Liter) =
= -1670 * 101 J = -168670 J

Live long and prosper.

[Doc Al]

Originally posted by cowgiljl
W = -P(Vsteam-Vwater) = (1.013E5)*[(1671000 - 1000)*.001
W = -169171E3 J

In your equation, you have the volumes in cubic cm instead of liters. You should have:
W = -P(Vsteam-Vwater) = (1.013E5)*[(1671 - 1)*.001]

(This is consistent with drag's calculation.)

Also, 3 significant figures in your answer is plenty.

[cowgiljl]

Now if my Q is right and I use drags and Doc Als conversion then U is equal to 2.09E6 J
is that correct

these were some of the hw questions i didn't finish

Thanks alot

Joe [:)]