An athlete is given 100 g of glucose C 6 H 12 O 6 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

Question

An athlete is given 100 g of glucose C 6 H 12 O 6 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 mol - 1 Molar mass of C , H & O are 12.1 and 16 g mol - 1 .

TestHub · Accepted Answer

Overall reaction of metabolism of glucose in the body:
$C_{6} H_{12} O_{6} (s) + 6 O_{2} \to 6 {CO}_{2} (g) + 6 H_{2} O (l)$

Extra energy used to convert $H_{2} O (l)$ into $H_{2} O (g)$ , energy left

$= \frac{1800}{2} = 900 kJ$
Moles of water vaporised from enthalpy of evaporation:

$\Rightarrow 900 = n_{H_{2} O} \times 45$

$n_{H_{2} O} = \frac{900}{45} = 20 mole$

$W_{H_{2} O} = 20 \times 18 = 360 g$

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