It has been determined that the body can generate 5500 \(\mathrm{kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise? (The heat of vaporization of water is 40.6 \(\mathrm{kJ} / \mathrm{mol.} )\)

In the given problem, we have: Energy generated per hour = 5500 kJ/hour Duration of exercise = 2 hours Heat of vaporization of water = 40.6 kJ/mol Molar mass of water = 18.015 g/mol The total heat generated during the exercise is: Total heat generated = (5500 kJ/hour) × 2 hours = 11000 kJ Now, we can calculate the moles of water that must evaporate to remove the heat generated: moles of water = Total heat generated / Heat of vaporization of water moles of water = (11000 kJ) / (40.6 kJ/mol) ≈ 270.94 mol Finally, we can determine the mass of water corresponding to these moles: mass of water = moles of water × molar mass of water mass of water = (270.94 mol) × (18.015 g/mol) ≈ 4878.7 g Therefore, approximately 4878.7 g of water must be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise.