The amount of heat removed by evaporating one pound of water in an evaporative cooling tower is approximately:

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Multiple Choice

The amount of heat removed by evaporating one pound of water in an evaporative cooling tower is approximately:

Explanation:
When water changes from liquid to vapor, it absorbs latent heat without a rise in temperature. In evaporative cooling towers you remove heat as this water evaporates, so the heat removed per pound of water is essentially the latent heat of vaporization. For water this value is about 970 BTU per pound at the usual boiling/evaporation conditions, so it’s common to approximate it as 1000 BTU per pound. That’s why evaporating one pound of water removes roughly 1000 BTU of heat. The smaller numbers (a few BTU or a hundred BTU) would imply tiny temperature changes or sensible heat only, which doesn’t reflect the latent heat involved in evaporation. To estimate total cooling, multiply the mass of water evaporated (in pounds) by about 1000 BTU per pound.

When water changes from liquid to vapor, it absorbs latent heat without a rise in temperature. In evaporative cooling towers you remove heat as this water evaporates, so the heat removed per pound of water is essentially the latent heat of vaporization. For water this value is about 970 BTU per pound at the usual boiling/evaporation conditions, so it’s common to approximate it as 1000 BTU per pound.

That’s why evaporating one pound of water removes roughly 1000 BTU of heat. The smaller numbers (a few BTU or a hundred BTU) would imply tiny temperature changes or sensible heat only, which doesn’t reflect the latent heat involved in evaporation. To estimate total cooling, multiply the mass of water evaporated (in pounds) by about 1000 BTU per pound.

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