An endurance athlete consumes $100 ext{ g}$ of glucose ($ ext{C}_6 ext{H}_{12} ext{O}_6$), which

Question

An endurance athlete consumes $100	ext{ g}$ of glucose ($	ext{C}_6	ext{H}_{12}	ext{O}_6$), which provides a total metabolic energy equivalent of $1560	ext{ kJ}$. During a rigorous athletic event, the athlete utilizes exactly $50\%$ of this gained energy for muscular work and physiological functioning. In order to avoid any storage of excess thermal energy in the body (which would raise core body temperature), the remaining unspent energy must be dissipated entirely through the evaporation of sweat (water). Calculate the mass of water (in $	ext{g}$) that the athlete must perspire and evaporate. The enthalpy of vaporization of water at body temperature is $\Delta H_{	ext{vap}} = 44	ext{ kJ mol}^{-1}$. (Report your answer to one decimal place)

Professor PC Thomas Classes · Answer

The total metabolic energy gained by consuming $100	ext{ g}$ of glucose is $E_{	ext{total}} = 1560	ext{ kJ}$. Since the athlete utilizes $50\%$ of this energy during the athletic event, the excess thermal energy remaining in the body that must be dissipated as heat is: $q_{	ext{excess}} = 0.50 	imes 1560	ext{ kJ} = 780	ext{ kJ}$. To maintain thermal equilibrium without storing energy, this entire heat load must be absorbed by the phase change of water (sweat evaporation): $q_{	ext{excess}} = n_{	ext{water}} 	imes \Delta H_{	ext{vap}}$. Thus, the number of moles of water that must be evaporated is: $n_{	ext{water}} = 780	ext{ kJ} / 44	ext{ kJ mol}^{-1} = 17.727	ext{ mol}$. Using the molar mass of water ($M = 18.015	ext{ g mol}^{-1} \approx 18.0	ext{ g mol}^{-1}$), the mass of perspiration required is: $m_{	ext{water}} = n_{	ext{water}} 	imes M = 17.727	ext{ mol} 	imes 18.0	ext{ g mol}^{-1} = 319.09	ext{ g} \approx 319.1	ext{ g}$.