Change in internal energy during water vaporization

  • #1

Main Question or Discussion Point

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?

 

Answers and Replies

  • #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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