- #1
cowgiljl
- 63
- 1
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
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