Chapter 10, Problem 63A

Interpretation Introduction

Interpretation: The mass in grams and volume inliters of water that would have to be evaporated through perspiration to rid the body of heat generated during two hours needs to be calculated.

Concept Introduction: The change in enthalpy of reaction is always calculated for the balanced chemical equation and can be used to calculate the energy change for the given moles or masses of reactant or product molecules. The enthalpy of vaporization can be defined as the enthalpy change during the conversion of liquid to the vapor state. The relation between the enthalpy of vaporization and heat can be written as:

  $q=n\times \Delta H_v$

Answer to Problem 63A

Mass of water = 4878 g

The volume of water = 4878 mL

Explanation of Solution

Given information:

The heat generated from body = 5500 kJ

Hours = 2

Heat of vaporization = 40.6 kJ/mol

  $q=n\times \Delta H_v$

Here,

q = 5500 kJ

and,

  $\Delta {\text{H}}_{\text{v}}$ = 40.6 kJ/mol

Calculate moles of water in 2 hours:

        $\begin{array}{c}q=n\times \Delta H_v\end{array}$

  $\begin{array}{c}\text{2 hours ×}\frac{\text{5500kJ}}{\text{1hour}}\text{ = }n\text{ × 40}\text{.6kJ/mol}\end{array}$

        $\begin{array}{c}n\text{ = 271 moles of water}\end{array}$

Calculate the mass and volume of water:

Molar mass of water = 18.0 g/mol

Density = 1 g/mL

  $\begin{array}{c}\text{Mass of water = moles × molar mass}\end{array}$

  $\begin{array}{c}\text{Mass of water = 271 mole × 18}\text{.0 g/mol}\end{array}$

  $\begin{array}{c}\text{Mass of water = 4878 g}\end{array}$

 $\begin{array}{c}\text{Consider density = 1 g/mL }\end{array}$

     $\begin{array}{c}\text{Volume = }\frac{\text{mass}}{\text{density}}\text{= }\frac{\text{4878g}}{\text{1g/mL}}\text{=4878mL}\end{array}$

Conclusion

Thus,

mass of water = 4878 g

Volume of water = 4878 mL