Determine the change in internal energy of boiling water

[cowgiljl]

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)*[(1.670 m^3) = - 169272 J

change in U = Q+W
2260000-169272
U = 2090728 J

are these correct?

Now i have not took chenistry yet but it wants to know how many moles of water were converted to steam
I have 2 different answers
1) 54.6 mols using n = PV / RT= 1.013E5*1670 / 8.31*373.15
or
2)18g/mol using the back of the book periodic table (molar mass)
H2O = 1+1(of hydrogen)+16 (oxygen)
which is right if somebody can help me befor 300pm today Please
thanks joe

 

[quantumdude]

Originally posted by cowgiljl
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)*[(1.670 m^3) = - 169272 J

change in U = Q+W
2260000-169272
U = 2090728 J

are these correct?

I haven't checked the numbers, but the procedure is right.

Now i have not took chenistry yet but it wants to know how many moles of water were converted to steam
I have 2 different answers
1) 54.6 mols using n = PV / RT= 1.013E5*1670 / 8.31*373.15
or
2)18g/mol using the back of the book periodic table (molar mass)
H2O = 1+1(of hydrogen)+16 (oxygen)
which is right

The "54.6 mols" and "18g/mol" aren't compatible quantities. The first tells you how many moles you have and the second tells you the mass of a mole.

if somebody can help me befor 300pm today Please
thanks joe

Sorry I was late.

 

[cowgiljl]

Thanks a lot for the reply, at least i am starting to get a grasp on this subject. And doing better in the class

Thanks again Tom


joe