Question

An athlete is given 100 g of glucose ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ of energy equivalent to 1560 kJ. He utilizes 50 % of this gained energy in an event. In order to avoid storage of energy in the body what is the weight of water, he would need to perspire? The enthalpy of evaporation of water is 44 kJ/mol.
(a)- 319 g
(b)- 638 g
(c)- 14040 g
(d)- 35.45 g

Answer 1

Hint: Calculate the amount of energy unutilised. It is important to understand that the enthalpy denotes the amount of release after a particular reaction takes place. We will obtain the weight of water to be released by simply dividing the energy unutilised by the enthalpy of evaporation of water.

Complete step by step answer:
- Now, first we are given that 100 g of glucose is equivalent to 1560 kJ.
- As mentioned, 50 % of the energy is utilized. Thus, the energy utilized in the body can be written as:
Energy utilized = $\dfrac{50}{100}\text{ x 1560}$ = 780 k J
- So, we can say that the energy that is not utilized in the body will be calculated by subtracting the energy utilized from the total equivalent energy.
Energy un-utilised = 1560 -780 = 780 kJ
- In other terms, the energy to be released is equal to 780 kJ.
- In the question, it is given that the enthalpy of evaporation of water is 44 kJ/mol.
- We know that the one mole of water is equal to the 18 g of water. Here, 18 g is the molar mass of water.
- Thus, enthalpy of evaporation of water = 44 kJ/mol = 44 kJ for 18 g of water.
- Now, we have to calculate the amount of water required to perspire to avoid the storage of water in the body. It can be written as:
Amount of water = $\dfrac{18}{44}\text{ x 780}$ = 319.09 g
In the last, we can conclude that the amount of water required is 319.09 g approximately.
The correct answer is option “A” .

Note: It is important to understand that when the value of enthalpy is negative, it denotes that energy is released. Thus, for exothermic reactions, the value of enthalpy is high in magnitude and negative in terms of polarity.