An athlete is given 180 g of glucose $(C_{6} H_{12} O_{6})$ . He utilizes 50% of the energy due to combustion in the body. In order to avoid storage of energy in the body, calculate the mass of water he would need to perspire.

Given enthalpy of combustion of glucose is -2800 kJ $m o l^{- 1}$ and enthalpy of evaporation of water is 44kJ $m o l^{- 1}$ .

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Solution

Mass of glucose $= 180 g$ .

Molar mass of glucose $= 180 g / m o l$ .

Number of moles of glucose $= \frac{180 g}{180 g / m o l} = 1 m o l$

Enthalpy of combustion of glucose is $- 2800 k J / m o l$

Thus, when 1 mole of glucose is burnt, the energy released is 2800 kJ.

50% of this energy is $2800 \times \frac{50}{100} = 1400 k J$

Enthalpy of evaporation of water is 44 kJ/mol.

44 kJ of energy will evaporate 1 mole of water.

1400 kJ of energy will evaporate $1400 \times \frac{1}{44} = 31.8$ moles of water.

Mass of water needed to perspire $= 31.8 m o l \times 18 g / m o l = 573 g$

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