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 m o l-1 Molar mass of C . H O are 12,1 and 16 g mol -1.

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 m o l-1 Molar mass of C . H O are 12,1 and 16 g mol -1. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C 6 H 12 O 6( s )+6 O 2 arrow 6 CO 2( g )+6 H 2 O ( l ) Extra energy used to convert H 2 O ( l ) into H 2 O ( l ) into H 2 O ( g ) =(1800/2) =900 kJ ⇒ 900= n H 2 O × 45 n H 2 O =(900/45)=20 text mole W H 2 O =20 × 18=360 g

C6​H12​O6​(s)+6O2​→6CO2​(g)+6H2​O(l) Extra energy used to convert H2​O(l) into H2​O(l) into H2​O(g) =21800​=900kJ ⇒900=nH2​O​×45 nH2​O​=45900​=20 mole WH2​O​=20×18=360g

$C_{6} H_{12} O_{6} (s) + 6 O_{2} \to 6 C O_{2} (g) + 6 H_{2} O (l)$
Extra energy used to convert $H_{2} O (l)$ into $H_{2} O (l)$ into $H_{2} O (g)$
$= \frac{1800}{2} = 900 k J$
$\Rightarrow 900 = n_{H_{2} O} \times 45$
$n_{H_{2} O} = \frac{900}{45} = 20 mole$
$W_{H_{2} O} = 20 \times 18 = 360 g$