How many BTUs are required to vaporize 1 pound of water at 212 degrees?

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Multiple Choice

How many BTUs are required to vaporize 1 pound of water at 212 degrees?

Explanation:
Latent heat of vaporization is the energy required to turn liquid water into steam at its boiling point without changing its temperature. At 212°F (100°C) and standard atmospheric pressure, water needs about 2257 kJ per kilogram, which converts to roughly 970 BTU for each pound of water. So to vaporize 1 pound of water at 212°F, you must supply about 970 BTU. For context, heating 1 lb of water from room temperature to 212°F (a sensible heat) takes about 180 BTU, so the energy needed to vaporize is much larger. The value around 970 BTU per pound is the standard figure for this process.

Latent heat of vaporization is the energy required to turn liquid water into steam at its boiling point without changing its temperature. At 212°F (100°C) and standard atmospheric pressure, water needs about 2257 kJ per kilogram, which converts to roughly 970 BTU for each pound of water. So to vaporize 1 pound of water at 212°F, you must supply about 970 BTU. For context, heating 1 lb of water from room temperature to 212°F (a sensible heat) takes about 180 BTU, so the energy needed to vaporize is much larger. The value around 970 BTU per pound is the standard figure for this process.

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