An athlete is given $100\, g$ of glucose $\left( C _6 H _{12} O _6\right)$ for energy This is equivalent to $1800\, kJ$ of energy The $50 \%$ of this energy gained is utilized by the athlete for sports activities at the event In order to avoid storage of energy, the weight of extra water he would need to perspire is ___ $g$ (Nearest integer) Assume that there is no other way of consuming stored energy Given: The enthalpy of evaporation of water is $45 \,kJ \,m o l^{-1}$ Molar mass of $C H\, \&\, O$ are 12,1 and $16\, g \,mol ^{-1}$
To determine the weight of extra water the athlete needs to perspire, we start by understanding the given data and the task: 1. The athlete consumes \(100\, g\) of glucose yielding \(1800\, kJ\) of energy. 2. \(50\%\) of this energy is used during sports activities, thus \(900\, kJ\) of energy remains. 3. The task is to calculate the weight of water that must be evaporated to dissipate this \(900\, kJ\) of retained energy.
The enthalpy of evaporation of water is given as \(45 \,kJ \,mol^{-1}\). To find out how many moles of water are needed to evaporate the remaining energy, use the formula: Number of moles of water = \(\frac{\text{Remaining energy}}{\text{Enthalpy of evaporation}}\) This becomes: \(=\frac{900\, kJ}{45\, kJ\, mol^{-1}} = 20\, mol\)
The molar mass of water is \(18\, g\, mol^{-1}\). Therefore, the mass of water needed is: Mass of water = \(\text{Number of moles} \times \text{Molar mass of water}\) \(= 20 \,mol \times 18\, g\, mol^{-1} = 360\, g\)
This value of \(360\, g\) lies within the given range \([360, 360]\). Thus, the athlete needs to perspire approximately \(\mathbf{360\, g}\) of extra water to avoid energy storage.
