How Is Internal Energy Affected in the Water to Steam Conversion at 100°C?

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SUMMARY

The conversion of 1 kg of water at 100°C to steam at the same temperature and 1 atm results in an increase in internal energy of 2.08 x 106 J. The latent heat of vaporization for water is 2.25 x 106 J/kg, which is the heat added to the system. The work done by the system during this phase change is calculated as 168,765.8 Joules, using the formula W = p(Vv - Vl). The final internal energy increase is determined by the equation ΔU = Q - W.

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  • Understanding of thermodynamics principles, specifically internal energy and phase changes.
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  • Familiarity with the ideal gas law and its implications for work done during volume changes.
  • Basic proficiency in unit conversions, particularly between different states of matter (liquid and gas).
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Homework Statement


Calculate the increase in internal energy of 1 kg of water at 100 degrees Celsius when it is converted into steam at the same temperature and at 1atm. The density of water and steam are 1000kg/m3 and 0.6kg/m3 resp. The latent heat of vaporisation of water is 2.25 x 106 J/kg


The Attempt at a Solution



The heat given to convert 1 kg of water to steam is 2.25 x 106 J.
I don't understand whether work is done by the system or on the system in increasing its volume.
The work done (by or on) the system is given by
W=nRTln(V2/V1) as the temperature remains same.
On adding both the values I got a wrong answer.
The correct answer is 2.08 x 106 J.

Please explain.
 
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Heat added to the system from the environment Q = 2.25x10^6.
Volume if 1 kg of water = 10^-3 m^3
Volume if 1 kg of steam = 1/0.6 = 1.667 m^3
Work done by the system on the environment W = p(Vv - Vl) = 1.013x10^5(1.667 - 0.001)= 168 765.8 Joules.
Hence the increase in internal energy of 1 kg of water at 100 degrees Celsius = Q - W
 

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