How much heat is required to evaporate 100.8 g at 100 degreesCelsius?(The molar heat of vaporization for water is 4.07E4 j/mol)
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During a strenous workout, an athlete generates 2140.0 kj of heat energy. What mass of water would have to evaporate from the athletes 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 25C, approximately room temperature the molar heat ofvaporization of water is 44.0kj/mol
during a strenuous workout an athlete generates 2050.0 kJ of heatenergy. What mass of water would have to evaporate from theathletes skin to dissipate this much heat?note at 100 celsius the molar heat of vaporization of water is40.67 kJ/mol at 25 celsius room temp the molar heat of vaporizationif water is 44.0 kJ/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.