Problem 101 - A Complete Resource Book in Chemistry for JEE Main

Question

An athlete is given \(100 \mathrm{~g}\) of glucose of energy equivalent to \(1560 \mathrm{~kJ}\). He utilizes \(50 \%\) of this gained energy in the event. In order to avoid storage of energy in the body, calculate the mass of water he would need to perspire. Enthalpy of \(\mathrm{H}_{2} \mathrm{O}\) for evaporation is \(44 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (a) \(346 \mathrm{~g}\) (b) \(316 \mathrm{~g}\) (c) \(323 \mathrm{~g}\) (d) \(319 \mathrm{~g}\)

Step-by-Step Solution

Answer

(d) 319 g

Step 1: Calculate Energy Utilized

The athlete uses 50% of the energy provided by the glucose. Calculate the energy utilized: \[\text{Energy utilized} = \frac{50}{100} \times 1560 \text{ kJ} = 780 \text{ kJ} \]

Step 2: Calculate Moles of Water Required

Determine how many moles of water need to be evaporated to utilize this energy. Use the formula: \[\text{Energy per mole of } \text{H}_2\text{O} = 44 \text{ kJ/mol}\] \[\text{Moles of } \text{H}_2\text{O} = \frac{780 \text{ kJ}}{44 \text{ kJ/mol}} \approx 17.727 \text{ moles}\]

Step 3: Convert Moles of Water to Mass

Calculate the mass of the water in grams using the molar mass of water (18 g/mol): \[\text{Mass of water} = 17.727 \text{ moles} \times 18 \text{ g/mol} \approx 319.086 \text{ g}\]

Step 4: Select the Closest Answer

Round the calculated mass to the nearest whole number to match the provided options: \[319.086 \text{ g} \approx 319 \text{ g}\] The closest answer choice is (d) 319 g.

Key Concepts

Enthalpy of Evaporation

The enthalpy of evaporation, also known as the heat of vaporization, is the amount of energy required to transform a given quantity of a substance from a liquid into a gas at its boiling point. This concept is crucial in understanding processes like perspiration, where our bodies release heat by converting liquid water to vapor.
For water, the enthalpy of evaporation is 44 kJ/mol. This means it takes 44 kilojoules of energy to evaporate one mole of water, which is approximately 18 grams.
In practical scenarios, like the one provided, this value helps calculate how much energy would be expended or required when water evaporates. Knowing this helps us understand thermodynamic processes involved in cooling mechanisms and energy changes.

Energy Conversion

Energy conversion refers to the process of transforming energy from one form to another. In the context of our body and everyday activities, this often involves converting chemical energy from food into other forms of energy such as kinetic or thermal energy.
In the athlete's scenario, 100 grams of glucose provide 1560 kJ of energy. However, only 50% of this energy is used actively; the rest needs to be dissipated to avoid storage.
This conversion process ensures that the excess energy doesn’t accumulate as extra weight but instead is utilized, for instance, through the evaporation of sweat to cool down. Calculating the utilized portion allows us to find how much energy needs conversion and how this translates to physical processes like perspiration.

Chemical Calculations

Chemical calculations involve using mathematical techniques to solve problems related to chemical substances and their reactions. These calculations can range from determining the mass or moles of substances involved to calculating energy changes.
In the presented exercise, several chemical calculations are required:

• First, determine the energy utilized by the athlete, which is 780 kJ (\[ \frac{50}{100} \times 1560 \text{ kJ} \]).
• Next, calculate the number of moles of water that must evaporate to use this energy: approximately 17.727 moles of water (\[ \frac{780 \text{ kJ}}{44 \text{ kJ/mol}} \]).
• Finally, convert these moles to mass: 319 grams of water is required (\[ 17.727 \text{ moles} \times 18 \text{ g/mol} \]).

Understanding these calculations allows one to discern how thermochemical equations apply to real-life scenarios, such as energy expenditure during physical activity.