Question

A person requires $10000{\rm{KJ}}$ energy per day and $\Delta H$ combustion of glucose is $2700KJmo{l^{ - 1}}$. How many grams of glucose is required by the body in a day?
A) $666.67g$
B) $6.67g$
C) $66.67g$
D) $6666.67g$

Answer 1

Hint: We can use the value of $\Delta H$and the given value of energy for glucose to find out the number of moles of glucose required by a human body which can further be multiplied by its molecular mass to calculate the actual mass of the glucose required by the body in a day.

Complete Step by Step Solution:
Heat of combustion: It is defined as the change in energy which occurs when one mole of a substance is burnt completely in the presence of excess oxygen. The organic compounds on complete combustion generally converts into water and carbon dioxide.

In the given case, the heat of combustion is given as $2700KJmo{l^{ - 1}}$ which means on combustion of 1 mole of glucose, $2700KJmo{l^{ - 1}}$ energy is released as heat.
So, number of moles required for $10000{\rm{KJ}}$of energy $ = \frac{1}{{2700}} \times 10000 = 3.7$ moles.

We know that the number of moles is the ratio of the mass of the compound to its molecular mass and the molecular mass of glucose is $180g$.
Therefore, mass of glucose required by a body in a day $ = 3.7 \times 180 \Rightarrow 666.67g$.
Hence, the correct answer is option (C).

Note: It is important to note the energy released also be expressed in terms of calorific value which is defined as the energy contained in a fuel or food which can be determined by calculating the amount of heat produce on complete combustion of the substance in specified quantity and is usually expressed in $JK{g^{ - 1}}$.