Chapter 6, Problem 6.11E

Calculate the amount of heat necessary to change $10\text{0}\text{.0 g}$ of ice at $-15.{\text{0 }}^{\text{ο}}\text{C}$ to steam at $\text{110}°\text{C}$. You will need the values of the heat capacity for ice, water, and steam, and ${\Delta }_{\text{fus}}H$ and ${\Delta }_{\text{vap}}H$ for ${\text{H}}_{\text{2}}\text{O}$ from Tables 2.1 and 2.3. Is this process exothermic or endothermic?

Interpretation Introduction

Interpretation:

The amount of heat required to change $10\text{0}\text{.0 g}$ of ice at $-15.{\text{0 }}^{\text{ο}}\text{C}$ to steam at ${\text{110 }}^{\text{ο}}\text{C}$ is to be calculated.

Concept introduction:

Sublimation is a process in which the substance in a solid state is directly converted into a vapor state without going through the liquid state.

In a chemical reaction, when an enthalpy is positive the value of $\Delta \text{H}$ is greater than zero which indicates that a system absorbed heat. This reaction is known as endothermic reaction.

The heat capacity defined as the quantity of heat required to raise the temperature of the system by one degree.

Answer to Problem 6.11E

The amount of heat required to change $10\text{0}\text{.0 g}$ of ice at $-15.{\text{0 }}^{\text{ο}}\text{C}$ to steam at ${\text{110 }}^{\text{ο}}\text{C}$ is $306.3\text{2 kJ}$.

The value of heat is positive. Thus, it is an endothermic reaction.

Explanation of Solution

Given

The mass of ice is $10\text{0}\text{.0 g}$.

The temperature of ice is $-15.{\text{0 }}^{\text{ο}}\text{C}$.

The temperature of steam is ${\text{110 }}^{\text{ο}}\text{C}$.

The conversion of ice to steam involves steps, which are the warming of ice at ${\text{0 }}^{\text{ο}}\text{C}$, melting of ice at ${\text{0 }}^{\text{ο}}\text{C}$, warming of water at ${\text{100 }}^{\text{ο}}\text{C}$, evaporation of water at ${\text{100 }}^{\text{ο}}\text{C}$ and warming of steam at ${\text{110 }}^{\text{ο}}\text{C}$.

The required heat capacities and enthalpy of fusion and vaporization are shown below.

$\begin{array}{cccc}& {\text{H}}_{\text{2}}\text{O}\left(s\right)& {\text{H}}_{\text{2}}\text{O}\left(l\right)& {\text{H}}_{\text{2}}\text{O}\left(g\right)\\ \text{Heat capacity,}C\left(\text{J/g}\cdot \text{K}\right)& 2.06& 4.184& 2.04\\ {\Delta }_{\text{fus}}H\left(\text{J/g}\right)& 333.5& & \\ {\Delta }_{\text{vap}}H\left(\text{J/g}\right)& 2260& & \end{array}$

The overall heat required is the sum of each individual step which is shown below.

$q=m{C}_{\text{ice}}\Delta T+m{C}_{\text{fus}}\Delta T+m{C}_{\text{liquid}}\Delta T+m{C}_{\text{vap}}\Delta T+m{C}_{\text{vapour}}\Delta T$ (1)

The temperature of ice is $-15.{\text{0 }}^{\text{ο}}\text{C}$.

The conversion of Celsius to Kelvin is done as,

${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of $-15.{\text{0 }}^{\text{ο}}\text{C}$ to Kelvin is done as,

$\begin{array}{rll}-15.{\text{0 }}^{\text{ο}}{\text{C }}^{\text{ο}}\text{C}& =& -15.\text{0}+2\text{73 K}\\ & =& 2\text{58 K}\end{array}$

Similarly, the conversion of ${\text{110 }}^{\text{ο}}\text{C}$ to Kelvin is done as,

$\begin{array}{rll}{\text{110 }}^{\text{ο}}\text{C}& =& 110+2\text{73 K}\\ & =& 38\text{3 K}\end{array}$

Substitute the values of mass, heat capacities, and temperature in equation (1) to calculate the amount of heat required.

$\begin{array}{rll}q& =& 1\text{00 g}\left(\begin{array}{l}\left(2.0\text{6 J/g}\cdot \text{K}\times \left(2\text{73 K}-25\text{8 K}\right)\right)+\\ \left(\text{333}\text{.5 J/g}+\left(\text{4}\text{.184 J/g}\cdot \text{K}\times \left(\text{373 K}-273\text{ K}\right)\right)\right)\\ +\left(\text{2260 J/g}+\left(\text{2}\text{.04 J/g}\cdot \text{K}\times \left(\text{283 K}-273\text{ K}\right)\right)\right)\end{array}\right)\\ & =& 10\text{0 g}\left(\begin{array}{l}\left(2.0\text{6 J/g}\cdot \text{K}\times 1\text{5 K}\right)+\text{333}\text{.5 J/g}+\left(\text{4}\text{.184 J/g}\cdot \text{K}\times 100\text{ K}\right)\\ +2260\text{ J/g}+\left(\text{2}\text{.04 J/g}\cdot \text{K}\times 10\text{ K}\right)\end{array}\right)\\ & =& 10\text{0g}\left(\text{30}\text{.9 J/g}+\text{333}\text{.5 J/g}+418.4\text{ J/g}+2260\text{ J/g}+20.4\text{ J/g}\right)\\ & =& 306,3\text{20 J}\end{array}$

The conversion of joule into kilojoule is done as,

$\text{1 J}=0.00\text{1 kJ}$

Therefore, the conversion of $306,3\text{20 J}$ into kilojoule is done as,

$\begin{array}{rll}306,3\text{20 J}& =& 306,3\text{20}\times 0.00\text{1 kJ}\\ & =& 306.3\text{2 kJ}\end{array}$

Hence, the amount of heat required to change $10\text{0}\text{.0 g}$ of ice at $-15.{\text{0 }}^{\text{ο}}\text{C}$ to steam at ${\text{110 }}^{\text{ο}}\text{C}$ is $306.3\text{2 kJ}$.

The value of heat is positive. Thus, it is an endothermic reaction.

Conclusion

The amount of heat required to change $10\text{0}\text{.0 g}$ of ice at $-15.{\text{0 }}^{\text{ο}}\text{C}$ to steam at ${\text{110 }}^{\text{ο}}\text{C}$ is $306.3\text{2 kJ}$. The value of heat is positive. Thus, it is an endothermic reaction.