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Sagot :
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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