Hopw many Btu's are required to convert a pound of water at 212% at sea level?

The heat of vaporization of water is 40.65 Kilojoules/Mole or 2257 Kilojoules/Kg. You can convert that to BTU/Lb.

2257\,\frac {KJoules}{Kg} \times \frac {BTU}{KJoule} \times \frac {Kg}{Lb}= ?? \frac {BTU}{Lb}