# Chemistry help needed with Molar Vaporization of water

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# During a strenuous workout, an athlete generates 1860.0 kJ of heat energy. What mass of water would have to evaporate from the athlete's skin to dissipate this much heat? Note: At 100c, the boiling point of water, the molar heat of vaporization of water is 40.67kJ/mol.

Nov 3rd, 2015

Thank you for the opportunity to help you with your question!

Heat generated =1860kJ

Latent heat of vaporization of water=40.67kj/mole

Let x is the no of mole to absorb this heat of vaporization

Heat used=1860.0/40.67moles=45.72 mole

Massif 1 mole of water is 18.015gm

Mass of 45.72 moles of water=18.015x45.72gm=823.65gm

Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 3rd, 2015

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Nov 3rd, 2015
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Nov 3rd, 2015
Nov 23rd, 2017
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