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I Change in internal energy during water vaporization

  1. Jan 29, 2017 #1
    According to the first law of thermodynamics,

    dQ = dU + dW and you can find dU = nCvdT
    If this is the case then when water at 100°C vaporizes to steam at 100°C shouldn't the change in internal energy be zero because it is dependent on temperature change?


     
  2. jcsd
  3. Jan 29, 2017 #2

    Charles Link

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    It's a change of phase at that temperature, so that the differential heat capacity equation is not applicable. It could be looked at as a discrete jump in the heat capacity as a function of temperature. I believe the 540 cal/gram number is a ## \Delta H ##, but most of this is due to the ## \Delta U ##.
     
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