Chapter 6, Problem 131CP

You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data.

$\begin{array}{ll}\text{Specific heat capacity of ice}\hfill & =2.03\text{J/}°\text{C}\cdot \text{g}\hfill \\ \text{Specific heat capacity of water}\hfill & =4.18\text{J/}°\text{C}\cdot \text{g}\hfill \\ \text{Specific heat capacity of steam}\hfill & =2.02\text{J/}°\text{C}\cdot \text{g}\hfill \end{array}$

$\begin{array}{lllll}{\text{H}}_{\text{2}}\text{O}\left(s\right)\hfill & \to {\text{H}}_{\text{2}}\text{O}\left(l\right)\hfill & \hfill & \Delta H_{\text{fusion}}\hfill & =6.02\text{KJ/mol}\left(\text{at 0}°\text{C}\right)\hfill \\ {\text{H}}_{\text{2}}\text{O}\left(l\right)\hfill & \to {\text{H}}_{\text{2}}\text{O}\left(g\right)\hfill & \hfill & \Delta H_{\text{vaporization}}\hfill & =40.7\text{KJ/mol}\left(\text{at 100}\text{.}°\text{C}\right)\hfill \end{array}$

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{-30°C}$ to ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ at $\text{140°C}$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

                                           $\text{C = }\frac{\text{Heat absorbed}}{\text{Temperature change}}\mathrm{......}\text{(1)}$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$\text{Absorbed heat }\left(\text{J}\right)\text{=}\text{Specific heat capacity×Temperature change}\left(\text{c}\right)\text{×mass ofsubstance }\left(\text{g}\right)\mathrm{......}\text{(2)}$

For the above equation heat is:

                                    $\text{S = q×M×T}\mathrm{.....}\text{(1)}$

                                                      q is heat (J)

                                                      M is mass of sample $\left(\text{g}\right)$

                                                      S is specific heat capacity $\left(\text{J/°C·g}\text{.}\right)$

                                                      T is temperature change $\left(\text{C}\right)$

For the process no heat loss to the surroundings means then the heat is

                   $\left(\text{absorbed}\right)\text{-q×M×ΔT=-q×M×ΔT}\text{(released)}\mathrm{......}\text{(3)}$'

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{-30°C}$ to ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ at $\text{140°C}$ is $\text{56}\text{.9}\text{kJ}$.

Explanation of Solution

Explanation

Record the given data:

                 Specific heat capacity of ice $\text{= 2}\text{.03J/°C· g}$

                 Specific heat capacity of water $\text{= 4}\text{.18 J/°C · g}$

                 Specific heat capacity of steam $\text{= 2}\text{.02J/°C· g}$

     Sample water is l.00 mole

                  Initial temperature is $\text{- 30°C}$

                  Final temperature is $\text{- 140°C}$.

                    $\begin{array}{l}{\text{H}}_{\text{2}}{\text{O}}_{\text{(s) }}\to {\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ΔΗ}}_{\text{fusion}}\text{=}\text{6}\text{.02kJ/mol}\text{(at}\text{0°C)}\\ {\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{ΔΗ}}_{\text{vaporization}}\text{=}\text{40}\text{.7kJ/mol}\text{(at}\text{100°C)}\end{array}$

To calculate the required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ from $\text{-30°C}$ to $\text{0°C}$.

                      $\begin{array}{l}{\text{q}}_{\text{(1)}}\text{=}\text{2}\text{.30}\frac{\text{J}}{\text{°C}\text{.g}}\text{×18}\text{.2g×}\left(\text{0-(-30)}\right)\text{°C}\\ \text{=}\text{1}\text{.1}{\text{×10}}^{\text{3}}\text{J}\end{array}$

• The given values are plugging in equation (1) to give the required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ from $\text{-30°C}$ to $\text{0°C}$.
• Required heat for vaporization of 1 mole of water is $\text{1}\text{.1}{\text{×10}}^{\text{3}}\text{J}$.

To calculate the required heat (q) to convert 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{0°C}$  into 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ at $\text{0°C}$.

                          $\begin{array}{l}{\text{q}}_{\text{(1)}}\text{=}\text{1}\text{.00}\text{mol}\text{×6}{\text{.02×10}}^{\text{3}}\text{J/mol}\\ \text{=}\text{6}{\text{.02×10}}^{\text{3}}\text{J}\end{array}$

• The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{0°C}$ into ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ at $\text{0°C}$.
• Required heat (q) to convert 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{0°C}$  into ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ at $\text{0°C}$ is $\text{6}{\text{.02×10}}^{\text{3}}\text{J}$.

To calculate the required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ from $\text{0°C}$ to $\text{100°C}$.

                      $\begin{array}{l}{\text{q}}_{\text{(3)}}\text{=}\text{4}\text{.18}\frac{\text{J}}{\text{°C}\text{.g}}\text{×18}\text{.2g×}\left(\text{100-0}\right)\text{°C}\\ \text{=}\text{7}\text{.53}{\text{×10}}^{\text{3}}\text{J}\end{array}$

• The given values are plugging in equation (1) to give the required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ from $\text{0°C}$ to $\text{100°C}$.
• Required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ from $\text{0°C}$ to $\text{100°C}$ is $\text{7}\text{.53}{\text{×10}}^{\text{3}}\text{J}$.

To calculate the required to convert 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ at $\text{100°C}$ into 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ at $\text{100°C}$.

                   $\begin{array}{l}{\text{q}}_{\text{(4)}}\text{=}\text{1}\text{.00}\text{mol}\text{×40}{\text{.7×10}}^{\text{3}}\text{J/mol}\\ \text{=}\text{4}\text{.07}{\text{×10}}^{\text{4}}\text{J}\end{array}$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ at $\text{100°C}$ into 1 mol ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ at $\text{100°C}$.

To calculate the required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(g\right)}$ from $\text{100°C}$ to $\text{140°C}$.

                          $\begin{array}{l}{\text{q}}_{\text{(5)}}\text{=}\text{2}\text{.30}\frac{\text{J}}{\text{°C}\text{.g}}\text{×18}\text{.2g×}\left(140-100\right)\text{°C}\\ \text{=}\text{1}\text{.5}{\text{×10}}^{\text{3}}\text{J}\end{array}$

• The given values are plugging in equation (1) to give required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(g\right)}$ from $\text{100°C}$ to $\text{140°C}$.
• Required to heat ${\text{H}}_{\text{2}}{\text{O}}_{\left(g\right)}$ from $\text{100°C}$ to $\text{140°C}$ is $\text{1}\text{.5}{\text{×10}}^{\text{3}}\text{J}$.

To calculate the amount of heat required to convert 1.00 mol of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{-30°C}$ to ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ at $\text{140°C}$.

                           $\begin{array}{l}{\text{q}}_{\text{total}}\text{=}{\text{q}}_{\text{1}}{\text{+q}}_{\text{2}}{\text{+q}}_{\text{3}}{\text{+q}}_{\text{4}}{\text{+q}}_{\text{5}}\\ {\text{q}}_{\text{total}}\text{=}\text{1}\text{.1}{\text{×10}}^{\text{3}}\text{J+6}{\text{.02×10}}^{\text{3}}\text{J+}\text{7}\text{.53}{\text{×10}}^{\text{3}}\text{J+4}\text{.07}{\text{×10}}^{\text{4}}\text{J+1}\text{.5}{\text{×10}}^{\text{3}}\text{J}\\ \text{=}\text{5}\text{.69}{\text{×10}}^{\text{4}}\text{J}\\ \text{= 56}\text{.9}\text{kJ}\end{array}$

Conclusion

Conclusion

The heat required to convert 1.00 mol of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{-30°C}$ to ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ at $\text{140°C}$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ at $\text{-30°C}$ to ${\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$ at $\text{140°C}$ was found to be $\text{56}\text{.9}\text{kJ}$.