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# If $\text{ 150 kJ }$ of energy is needed for a man to walk a distance of $\text{ 1 km }$, then how much of glucose one has to consume to walk a distance of $\text{ 5 km }$, provided that only $\text{ 30 }{\scriptstyle{}^{0}/{}_{0}}\text{ }$ of the energy available for muscular work. (The enthalpy of combustion glucose is $\text{ 3000 kJ mo}{{\text{l}}^{-1}}\text{ }$). A) $\text{ 7}5\text{ g }$ B) $\text{ 30 g }$C) $\text{ 180 g }$D) $\text{ 150 g }$E) $\text{ 4}5\text{ g }$

Last updated date: 20th Jun 2024
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Hint: The glucose provides energy for physical activities. as we know the amount of energy liberated during the combustion of 1 mole of glucose molecule then, The amount of glucose which is required by the man to walk a certain distance is calculated by considering that the 1 mole of glucose contains the $\text{ 180 gm }$ of the glucose.
$\text{ 1 mole of glucose = 180 gm of glucose }$

Glucose provides energy for muscular activities like walking, running, etc.
Here, we have given that a man utilizes a certain amount of energy to walk a distance. the data is given as follows:
The distance travelled by man is $\text{ 1 km }$.
He requires the $\text{ 150 kJ }$ energy to cover a $\text{ 1 km }$ distance.
The enthalpy of combustion glucose is$\text{ 3000 kJ mo}{{\text{l}}^{-1}}\text{ }$.
We are interested to find the energy consumed by a man to walk the $\text{ 5 km }$ distance, provided that only $\text{ 30 }{\scriptstyle{}^{0}/{}_{0}}\text{ }$ of the energy is available for the muscular work.
Let’s first find out the amount of energy required to cover $\text{ 5 km }$ distance.
For walking $\text{ 1 km }$, the man takes the $\text{ 150 kJ }$ of energy then the energy required to covered $\text{ 5 km }$ distance is,
$\text{ }\begin{matrix} \text{Distance travelled by man} & = & \text{Energy required} \\ 1\text{ km} & = & 150\text{ kJ} \\ 5\text{ km} & = & \text{Energy} \\ \text{Energy} & = & \dfrac{\left( 5\text{ km} \right)\left( 150\text{ kJ} \right)}{\left( 1\text{ km} \right)} \\ \text{Energy} & = & 750\text{ kJ } \\ \end{matrix}\text{ }$
Therefore, the energy required to be covered $\text{ 5 km }$ is equal to the$750\text{ kJ}$.
We have given that, only $\text{ 30 }{\scriptstyle{}^{0}/{}_{0}}\text{ }$ of the energy is available for muscular work. Thus the total energy need for $\text{ 5 km }$ will be,
$\text{ }\begin{matrix} \text{Total energy }\!\!\times\!\!\text{ }\dfrac{\text{30}}{\text{100}} & \text{=} & \left( \text{Energy required for muscular work} \right)\text{ } \\ {} & \text{=} & \left( \text{750 kJ} \right)\text{ }\!\!\times\!\!\text{ }\dfrac{\text{100}}{\text{30}} \\ \therefore \text{Total energy needed} & \text{=} & \text{2500 kJ} \\ \end{matrix}\text{ }$

Thus, the total energy required for muscular work is equal to the $\text{2500 kJ}$.
Now, we have given that the 1 mole of the glucose combusts to produce the $\text{ 3000 kJ mo}{{\text{l}}^{-1}}\text{ }$.
$\text{ 1 mole of glucose = 3000 kJ mo}{{\text{l}}^{-1}}\text{ }$

Then, the number of moles of glucose which utilized in the $\text{2500 kJ}$are,
$\text{ }\begin{matrix} \text{1 mole of glucose} & \text{=} & \text{3000 kJ } \\ \text{X mole of glucose} & \text{=} & \text{2500 kJ} \\ \Rightarrow \text{ X mole of glucose} & \text{=} & \dfrac{\left( \text{2500 kJ} \right)\left( \text{1 mole of glucose} \right)}{\left( \text{3000 kJ} \right)} \\ \therefore \text{X mole of glucose} & \text{=} & \dfrac{\text{5}}{\text{6}}\text{ mole} \\ \end{matrix}\text{ }$
Therefore, the number of moles is equal to $\dfrac{\text{5}}{\text{6}}\text{ mole}$.

Now, the 1 mole of the glucose corresponds to the $\text{ 180 gm }$ of the glucose. Then $\text{ }\begin{matrix} 1\text{ mole of glucose} & = & 180\text{ gm of glucose} \\ \dfrac{5}{6}\text{ mole of glucose} & = & \text{ x gm of glucose} \\ \Rightarrow \text{x gm of glucose } & = & \dfrac{5}{6}\text{ }\times \text{ 180 gm} \\ \therefore \text{x gm of glucose} & = & 150\text{ gm} \\ \end{matrix}$Thus, the energy consumed by a man to walk the $\text{ 5 km }$ distance, provided that only $\text{ 30 }{\scriptstyle{}^{0}/{}_{0}}\text{ }$ of the energy is available for the muscular work is equal to the $\text{ }150\text{ gm }$ .
So, the correct answer is “Option D”.

Note: Note that, instead of calculating the number of moles (as done in above) we can solve these problems by only calculating the amount of available energy through the enthalpy of combustion per mole. That is,
$\text{ Actual energy available = }\dfrac{30}{100}\text{ }\times \text{ 3000 = 900 kJ mo}{{\text{l}}^{-1}}$
Here, we consider that the $\text{900 kJ mo}{{\text{l}}^{-1}}$ requires $\text{ 180 gm }$ of glucose. Then for$\text{ 900 kJ }$,
$\text{Amount of glucose = }\dfrac{180}{900}\text{ }\times \text{ 750 = 150 g}$