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12 May, 00:34

A cylinder with a piston contains 0.8 kg of water at 100 degrees C. What is the change in internal energy of the water when it is converted to steam at 100 degrees C at a constant pressure of 1 atm? The density of water is 1000 kg/m^3 and that of steam is 0.6 kg/m^3. The latent heat of vaporization of water is 2.26E+06 J/kg.

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  1. 12 May, 02:01
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    1672.98 kJ (increase in internal energy)

    Explanation:

    Let's first find the energy received by the water in order for it to vaporize.

    This is: Energy = mass of water * latent heat of vaporization

    Energy = 0.8 * 2.26 * 10^6 = 1808 kJ

    We also need to subtract from this the energy needed for the volume increase. This is calculated the following way:

    Energy needed for volume change = P (Final Volume - Initial Volume)

    Initial volume = 0.8/1000 = 0.0008 m^3

    Final volume = 0.8 / 0.6 = 1.333 m^3

    Pressure = 1 atm = 101325 Pa

    Energy needed for volume change = 101325 (1.333 - 0.0008)

    Energy needed for volume change = 135.02 kJ

    Total Internal Energy Change = 1808 - 135.02 = 1672.98 kJ
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