Chapter 12, Problem 12.72QP

Interpretation Introduction

Interpretation:

The amount of heat needed for the conversion of ice to steam, has to be calculated.

Concept Introduction:

Two types of phase transitions are involved in the conversion of ice to steam. Phase transition is the transfer of one phase to another. To convert ice to steam, first it is needed to convert it into liquid water. Then the liquid water can be converted into steam. By this way ice undergoes two types of phase transitions such as transfer of solid phase of ice to liquid phase. Then transfer of liquid phase to steam which is the gaseous phase.

Answer to Problem 12.72QP

The amount of heat needed to convert 866g of ice to steam is $2317.605\text{kJ}$.

Explanation of Solution

The amount of heat needed to convert ice to steam can be calculated as shown here:

1^st phase transition: Melting of ice to liquid water:

$\text{q}\text{=}\text{m}\text{×}\text{s}\text{×}\text{ΔT}$

$\begin{array}{c}\text{q}\text{=}\text{Heat}\text{transfered}\Rightarrow \text{It can be released or absorbed}\text{. }\\ \text{s}\text{=}\text{specific}\text{heat}\text{of}\text{ice}\text{=}\text{2}\text{.03}\text{J/g}{\text{.}}^{\text{o}}\text{C}\Rightarrow \text{Given}\text{.}\\ \text{m}\text{=}\text{mass}\text{of}\text{ice}\text{=}866\text{g}\Rightarrow \text{Given}\text{.}\end{array}$

$\begin{array}{c}\text{ΔT}\text{=}\text{Temperature}\text{difference}\\ \text{= }\text{Final}\text{Temperature }\text{-}\text{Initial}\text{temperature}\\ \text{=}{\text{T}}_{\text{2}}\text{-}{\text{T}}_{\text{1}}\end{array}$

The initial temperature is given as ${\text{-10}}^{\text{o}}\text{C}$.

In the melting of ice, let the final temperature be its melting point which is ${\text{0}}^{\text{o}}\text{C}$.

$\begin{array}{c}{\text{T}}_{\text{2}}{\text{= 0}}^{\text{o}}\text{C}\\ {\text{T}}_{\text{1}}\text{=}{\text{-10}}^{\text{o}}\text{C}\\ \text{ΔT=}{\text{0}}^{\text{o}}\text{C}\text{-}\left({\text{-10}}^{\text{o}}\text{C}\right)\\ \text{=}{\text{10}}^{\text{o}}\text{C}\end{array}$

Substituting all known values in the formula of heat transferred:

$\begin{array}{c}\text{q}\text{=}\text{m}\text{×}\text{s}\text{×}\text{ΔT}\\ \text{=}866\text{g}\times \text{2}\text{.03}\text{J/g}{\text{.}}^{\text{o}}\text{C}\times {\text{10}}^{\text{o}}\text{C}\\ =17579.8\text{J}\end{array}$

Converting joules into kilojoules:

$\begin{array}{c}\text{1000}\text{J}\text{=}\text{1}\text{kJ}\\ 17579.8\text{J}\text{=}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\text{×}17579.8\text{J}\\ \text{=}17579.8\text{×}{\text{10}}^{\text{-3}}\text{kJ}\\ \text{=}\text{17}\text{.580}\text{kJ}\end{array}$

Therefore, $\text{q}\text{=}\text{17}\text{.580}\text{kJ}$. Let this be $q_1$.

The molar heat of fusion of water is ${\text{ΔH}}_{\text{fus}}=6.01\text{kJ/mol}$.It is the amount of heat energy required to melt one mole of ice to one mole of liquid water.

It is known that $q=\Delta H_{reaction}$

Converting $\Delta H_{reaction}$into $\Delta H_{fus}$:

$\begin{array}{c}\text{866}\text{g}\text{of}\text{ice}\text{=}\text{866}\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}=18\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \therefore \frac{\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}}\times \text{866}\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}\text{=}\text{48}\text{.111}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

Hence,

$\begin{array}{c}\text{866}\text{g}\text{of}\text{ice}=\text{48}\text{.111}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}\text{.}\\ {\text{ΔH}}_{\text{fus}}=6.01\text{kJ/mol}\\ \therefore \frac{6.01\text{kJ}}{1\text{mole}of{\text{H}}_{\text{2}}\text{O}}\times \text{48}\text{.111}\text{mole}of{\text{H}}_{\text{2}}\text{O}\\ =289.147\text{kJ}\end{array}$

Therefore, ${\text{ΔH}}_{\text{fus}}=289.147\text{kJ}$. Let this be $q_2$.

2^nd phase transition: Vaporization of liquid water to steam:

$\text{q}\text{=}\text{m}\text{×}\text{s}\text{×}\text{ΔT}$

$\begin{array}{c}\text{q}\text{=}\text{Heat}\text{transfered}\Rightarrow \text{It can be released or absorbed}\text{. }\\ \text{s}\text{=}\text{specific}\text{heat}\text{of}\text{water}\text{=}\text{4}\text{.184}\text{J/g}{\text{.}}^{\text{o}}\text{C}\\ \text{m}\text{=}\text{mass}\text{of}\text{water}\text{=}866\text{g}\Rightarrow \text{Given}\text{.}\end{array}$

$\begin{array}{c}\text{ΔT}\text{=}\text{Temperature}\text{difference}\\ \text{= }\text{Final}\text{Temperature }\text{-}\text{Initial}\text{temperature}\\ \text{=}{\text{T}}_{\text{2}}\text{-}{\text{T}}_{\text{1}}\end{array}$

Let the initial temperature be ${\text{0}}^{\text{o}}\text{C}$.

In the vaporization of liquid water, let the final temperature be its boiling point which is ${\text{100}}^{\text{o}}\text{C}$.

$\begin{array}{c}{\text{T}}_{\text{2}}\text{= }{\text{100}}^{\text{o}}\text{C}\\ {\text{T}}_{\text{1}}\text{=}{\text{0}}^{\text{o}}\text{C}\\ {\text{ΔT=100}}^{\text{o}}\text{C}\text{-}{\text{0}}^{\text{o}}\text{C}\\ {\text{=100}}^{\text{o}}\text{C}\end{array}$

Substituting all known values in the formula of heat transferred:

$\begin{array}{c}\text{q}\text{=}\text{m}\text{×}\text{s}\text{×}\text{ΔT}\\ \text{=}866\text{g}\times \text{4}\text{.184}\text{J/g}{\text{.}}^{\text{o}}\text{C}\times {\text{100}}^{\text{o}}\text{C}\\ =3623.344\text{J}\end{array}$

Converting joules into kilojoules:

$\begin{array}{c}\text{1000}\text{J}\text{=}\text{1}\text{kJ}\\ 3623.344\text{J=}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\text{×}3623.344\text{J}\\ \text{=}3623.344\text{×}{\text{10}}^{\text{-3}}\text{kJ}\\ \text{=}3.623\text{kJ}\end{array}$

Therefore, $\text{q}\text{=}3.623\text{kJ}$. Let this be $q_3$.

The molar heat of vaporization of water is ${\text{ΔH}}_{\text{vap}}=40.79\text{kJ/mol}$.It is the amount of heat energy required to vaporize one mole of liquid water to one mole of vapor.

It is known that $q=\Delta H_{reaction}$

Converting $\Delta H_{reaction}$into $\Delta H_{fus}$:

$\begin{array}{c}\text{866}\text{g}\text{of}\text{ice}\text{=}\text{866}\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}=18\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \therefore \frac{\text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}}\times \text{866}\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}\text{=}\text{48}\text{.111}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

Hence,

$\begin{array}{c}\text{866}\text{g}\text{of}\text{ice}=\text{48}\text{.111}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}\text{.}\\ {\text{ΔH}}_{\text{vap}}=40.79\text{kJ/mol}\\ \therefore \frac{40.79\text{kJ}}{1\text{mole}of{\text{H}}_{\text{2}}\text{O}}\times \text{48}\text{.111}\text{mole}of{\text{H}}_{\text{2}}\text{O}\\ =1962.448\text{kJ}\end{array}$

Therefore, ${\text{ΔH}}_{\text{fus}}=1962.448\text{kJ}$. Let this be $q_4$.

Heating water from $100^{o}C$ to $126^{o}C$ for converting into steam:

$\text{q}\text{=}\text{m}\text{×}\text{s}\text{×}\text{ΔT}$

$\begin{array}{c}\text{q}\text{=}\text{Heat}\text{transfered}\Rightarrow \text{It can be released or absorbed}\text{. }\\ \text{s}\text{=}\text{specific}\text{heat}\text{of}\text{steam}\text{=}\text{1}\text{.99}\text{J/g}{\text{.}}^{\text{o}}\text{C}\Rightarrow \text{Given}\text{.}\\ \text{m}\text{=}\text{mass}\text{of}\text{water}\text{=}866\text{g}\Rightarrow \text{Given}\text{.}\end{array}$

$\begin{array}{c}\text{ΔT}\text{=}\text{Temperature}\text{difference}\\ \text{= }\text{Final}\text{Temperature }\text{-}\text{Initial}\text{temperature}\\ \text{=}{\text{T}}_{\text{2}}\text{-}{\text{T}}_{\text{1}}\end{array}$

Let the initial temperature be ${\text{100}}^{\text{o}}\text{C}$ which is the boiling point of water. The final temperature is given as ${\text{126}}^{\text{o}}\text{C}$.

$\begin{array}{c}{\text{T}}_{\text{2}}{\text{=126}}^{\text{o}}\text{C}\\ {\text{T}}_{\text{1}}\text{=}{\text{100}}^{\text{o}}\text{C}\\ {\text{ΔT=126}}^{\text{o}}\text{C}\text{-}{\text{100}}^{\text{o}}\text{C}\\ \text{=}{\text{26}}^{\text{o}}\text{C}\end{array}$

Substituting all known values in the formula of heat transferred:

$\begin{array}{c}\text{q}\text{=}\text{m}\text{×}\text{s}\text{×}\text{ΔT}\\ \text{=}866\text{g}\times \text{1}\text{.99}\text{J/g}{\text{.}}^{\text{o}}\text{C}\times {\text{26}}^{\text{o}}\text{C}\\ =44806.84\text{J}\end{array}$

Converting joules into kilojoules:

$\begin{array}{c}\text{1000}\text{J}\text{=}\text{1}\text{kJ}\\ 44806.84\text{J=}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\text{×}44806.84\text{J}\\ \text{=}3623.344\text{×}{\text{10}}^{\text{-3}}\text{kJ}\\ \text{=}44.807\text{kJ}\end{array}$

Therefore, $\text{q}\text{=}44.807\text{kJ}$. Let this be $q_5$.

It is now known that

$\begin{array}{l}{q}_1=\text{17}\text{.580}\text{kJ}\\ q_2=289.147\text{kJ}\end{array}$

$\begin{array}{l}{q}_3=3.623\text{kJ}\\ q_4=1962.448\text{kJ}\end{array}$

$q_5=44.807\text{kJ}$

Therefore,

$\begin{array}{c}{q}_{total}=q_1+q_2+q_3+q_4+q_5\\ =\text{17}\text{.580}\text{kJ}+289.147\text{kJ}+3.623\text{kJ}+1962.448\text{kJ}+44.807\text{kJ}\\ =2317.605\text{kJ}\end{array}$

Hence, the amount of heat needed to conversion of 866g of ice to steam, is $2317.605\text{kJ}$.

Conclusion

Thus the amount of heat required to convert ice to steam was calculated.