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11 Dec 2019
During a strenuous workout, an athlete generates 2100.0 kJ of heat energy. What mass of water would have to evaporate from the athlete's skin to dissipate this much heat?
Note: At 100°C, the boiling point of water, the molar heat of vaporization of water is 40.67 kJ/mol. At 25°C, approximately room temperature, the molar heat of vaporization of water is 44.0kJ/mol.
During a strenuous workout, an athlete generates 2100.0 kJ of heat energy. What mass of water would have to evaporate from the athlete's skin to dissipate this much heat?
Note: At 100°C, the boiling point of water, the molar heat of vaporization of water is 40.67 kJ/mol. At 25°C, approximately room temperature, the molar heat of vaporization of water is 44.0kJ/mol.
Deanna HettingerLv2
13 Dec 2019
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