Seem to have ran into a brick wall with this problem(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Internal energy

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