Final answer:
To vaporize one pound of water, it requires approximately 970 BTUs, which is equivalent to about 1,023,350 J (970 x 1,055 J).
Step-by-step explanation:
The amount of energy required to vaporize one pound of water can be determined using the heat of vaporization of water. The heat of vaporization is the amount of energy needed to convert one pound of water from a liquid to a gas without a temperature change. This value is approximately 970 BTUs for one pound of water. To vaporize one pound of water completely, it would thus require around 970 BTUs.
In terms of Joules, since 1 BTU is equivalent to approximately 1,055 J, converting the energy required to vaporize one pound of water to Joules would involve multiplying the number of BTUs by the conversion factor. Therefore, 970 BTUs would be equivalent to about 1,023,350 J (970 x 1,055 J).