Question

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

Answer 1

the amount of heat generated by athlete = 2140.0 kJ

this amount of heat due to evaporation q = n x delta Hvap

molar heat ofvaporization of water at 25 oC = 44.0kj/mol

moles of water = q / delta Hvap

                        = 2140 / 44

                        = 48.6 mol

mass of water = 48.6 x 18

mass of water = 875.5 g