Question- 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 100oC, the boiling point of water, the molar heat of vaporization of water is 40.67 kJ/mol. At 25oC approximately room temperature, the molar heat of vaporization of water is 44.0 kJ/mol.
I just need a general explanation of it please- Thanks