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(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Determine the change in internal energy of boiling water

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