Change in internal energy during water vaporization

[Ian Baughman]

According to the first law of thermodynamics,

dQ = dU + dW and you can find dU = nC_vdT
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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?

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[Charles Link]

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 $\Delta U$.