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As much as 73,489 grams of water can evaporate from the Stingray Bay exhibit on a warm day. How much heat in kilojoules is required to evaporate this much water? Use 2.44 J/g as the heat of vaporization of water. Count your zeros carefully.

Sagot :

Okpalawalter8idn

Heat in kJ is required to evaporate this much water is mathematically given as

He=179313.16kJ

Heat of vaporization of water

Question Parameters:

Heat of vaporization of water = 2.44 kJ per gram

Total amount of water that can be evaporated from the bay on warm day = 75760 grams

Generally the equation for the heat required to evaporate  is mathematically given as

heat required to evaporate this much of water

He= Heat of vaporization x amount of water

Therefore

He= 2.44 kJ/g x 73489 g

He=179313.16kJ

For more information on Heat

https://brainly.com/question/13439286

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