Chapter 15, Problem 15.10P

Interpretation Introduction

Interpretation:

Whether heating of $10\text{g}$ of water at $50°\text{C}$ to $100°\text{C}$ or vaporization of $10\text{g}$ of water at $100°\text{C}$ that takes longer time has to be determined.

Concept Introduction:

Molar enthalpy for a process is the quantity of heat energy needed by one mole of a substance to undergo change in the process. The heat energy absorbed by n mole of substance to undergo change is as follows:

  $q=n\left(\Delta H\right)$

Here,

$q$ is the heat energy in $\text{kJ}$

$n$ is the number of moles of substance.

$\Delta H$ is the enthalpy for process in $\text{kJ/mol}$ .

The value of $\Delta H$ is positive always as energy is needed to overcome the interactions between the liquid molecules.

The formula to calculate the time required to transfer heat energy is as follows:

  $\text{time}=\frac{\text{heat}\text{energy}}{\text{heating rate}}$

Answer to Problem 15.10P

The heating of $10\text{g}$ of water at $50°\text{C}$ to $100°\text{C}$ takes only $6.9722\min$ and vaporization of $10\text{g}$ of water at $100°\text{C}$ takes $81.9134\min$. Therefore, vaporization of $10\text{g}$ of water takes much longer time than heating of water from $50°\text{C}$ to $100°\text{C}$.

Explanation of Solution

The formula to calculate the number of moles of water is as follows:

  $\text{number}\text{of}\text{moles}\text{of}\text{water}=\frac{\text{mass}\text{of}\text{water}}{\text{molecular}\text{weight}\text{of}\text{water}}$        (1)

Substitute $18\text{g/mol}$ for molecular weight of water and $10\text{g}$ for mass of water in equation (1) to determine of moles of water.

  $\begin{array}{c}\text{number}\text{of}\text{moles}\text{of}\text{ammonia}=\frac{10\text{g}}{18\text{g/mol}}\\ =0.5555\text{mol}\end{array}$

The value of $T_1$ is $50°\text{C}$.

The value of $T_2$ is $100°\text{C}$.

The temperature difference ($\Delta T$) between $T_2$ and $T_1$ is calculated as follows:

  $\begin{array}{c}\Delta T=T_2-T_1\\ =100°\text{C}-50°\text{C}\\ =50°\text{C}\end{array}$

The size difference on kelvin and Celsius scale is equal so $\Delta T$ is equal to $50\text{K}$.

The formula for heat energy involved in heating $10\text{g}$ of water at $50°\text{C}$ to $100°\text{C}$ is as follows:

  $q=mC\left(\Delta T\right)$        (2)

Substitute $50\text{K}$ for $\Delta T$, $0.5555\text{mol}$ for $n$ and $75.3\text{J}\cdot {\text{mol}}^{-\text{1}}\cdot {\text{K}}^{-\text{1}}$ for $C$ in equation (2) to calculate the heat energy involved in heating $10\text{g}$ of water at $50°\text{C}$ to $100°\text{C}$.

  $\begin{array}{c}q=\left(0.5555\text{mol}\right)\left(75.3\text{J}\cdot {\text{mol}}^{-\text{1}}\cdot {\text{K}}^{-\text{1}}\right)\left(100°\text{C}-50°\text{C}\right)\left(\frac{\text{1}\text{K}}{1°\text{C}}\right)\\ =\left(41.8333\text{J}\cdot {\text{K}}^{-\text{1}}\right)\left(50\text{K}\right)\\ =2091.665\text{J}\end{array}$

The formula to calculate the time required to transfer heat energy is as follows:

  $\text{time}=\frac{\text{heat}\text{energy}}{\text{heating rate}}$        (3)

Substitute $2091.665\text{J}$ for $\text{heat}\text{energy}$ and $5\text{J}\cdot {\text{s}}^{-\text{1}}$ for $\text{heating rate}$ in equation (3)to calculate the time required to transfer heat energy.

  $\begin{array}{c}\text{time}=\frac{2091.665\text{J}}{5\text{J}\cdot {\text{s}}^{-\text{1}}}\\ =418.333\text{s}\left(\frac{0.0166\min }{1\text{s}}\right)\\ =6.9722\min \end{array}$

The formula for heat energy involved in heating $10\text{g}$ of water to vaporization at $100°\text{C}$ is as follows:

  $q_{\text{vap}}=n\left(\Delta H_{\text{vap}}\right)$        (4)

Substitute $40.65\text{kJ/mol}$ for $\Delta H_{\text{vap}}$, $0.5555\text{mol}$ for $n$ in equation (4) to calculate the heat energy involved in heating $10\text{g}$ of water to vaporization at $100°\text{C}$.

  $\begin{array}{c}{q}_{\text{vap}}=\left(0.5555\text{mol}\right)\left(40.65\text{kJ/mol}\right)\left(\frac{1000\text{J}}{\text{1}\text{kJ}}\right)\\ =22581.075\text{J}\end{array}$

The total energy that must be absorbed by $10\text{g}$ of water to vaporizeis calculated as follows:

  $\begin{array}{c}\text{total}\text{energy}=q+q_{\text{vap}}\\ =\left(2091.665\text{J}\right)+\left(22581.075\text{J}\right)\\ =\text{24672}\text{.74}\text{J}\end{array}$

The total energy that must be absorbed by $10\text{g}$ of water to vaporize is $\text{24672}\text{.74}\text{J}$.

Substitute $\text{24672}\text{.74}\text{J}$ for $\text{heat}\text{energy}$ and $5\text{J}\cdot {\text{s}}^{-\text{1}}$ for $\text{heating rate}$ in equation (3) tocalculate the time required to transfer heat energy for $10\text{g}$ of water to vaporize.

  $\begin{array}{c}\text{time}=\frac{\text{24672}\text{.74}\text{J}}{5\text{J}\cdot {\text{s}}^{-\text{1}}}\\ =4934.548\text{s}\left(\frac{0.0166\min }{1\text{s}}\right)\\ =81.9134\min \end{array}$

Therefore vaporization of $10\text{g}$ of water takes much longer time than heating water from $50°\text{C}$ to $100°\text{C}$.