Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

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 :

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