Chapter 12, Problem 12.65QP

Interpretation Introduction

Interpretation:

The heat required to convert $\text{866 g}$ of ice at ${\text{-15}}^{\text{°}}\text{C}$ to steam at ${\text{146}}^{\text{°}}\text{C}$ should be calculated.

Concept Introduction:

Specific heat of a substance refers to the quantity of heat per unit mass of substance that rises the temperature of the substance by one degree Celsius.  The quantity of heat added and the specific heat is related as follows –

$\begin{array}{l}\text{q = m}\times \text{s}\times \Delta \text{T}\\ \text{where q = heat added}\\ \text{ m = mass of the substance}\\ \text{ s = specific heat of the substance}\\ \Delta \text{T = change in Temperature}\\ \end{array}$

Molar heat of vaporization is the heat energy required to boil one mole of a substance at its boiling point.

Molar heat of fusion is the heat energy required to melt one mole of the substance at its melting point.

Answer to Problem 12.65QP

The heat required to convert $\text{212}\text{.8 g}$ of ice at ${\text{-15}}^{\text{°}}\text{C}$ to steam at ${\text{138}}^{\text{°}}\text{C}$ is calculated as $\text{758}\text{.32 kJ}$.

Explanation of Solution

Calculate the heat required to warm ice to melting point.

$\begin{array}{l}\text{given data: }\\ \text{mass of ice, m = 866 g}\\ \text{specific heat of ice, s = 2}\text{.03 J/g }{\text{.}}^{\text{°}}\text{C}\\ {\text{T}}_{\text{1}}{\text{ = -15}}^{\text{°}}{\text{C, T}}_{\text{2}}{\text{ = 0}}^{\text{°}}{\text{C ; ΔT = T}}_{\text{2}}{\text{-T}}_{\text{1}}{\text{= 0}}^{\text{°}}{\text{C -(-15}}^{\text{°}}{\text{C) = 15}}^{\text{°}}\text{C}\\ \text{ heat required to warm ice to its melting point , }\\ {\text{ q}}_1\text{ = m s }\Delta \text{T = 866 g}\times \text{ 2}\text{.03 J/g }{\text{.}}^{\text{°}}\text{C}\times {\text{ 15}}^{\text{°}}\text{C}\\ \text{ = 26}\text{.4 kJ}\end{array}$

The ice is warmed from ${\text{-15}}^{\text{°}}\text{C}$ to its melting point ${\text{0}}^{\text{°}}\text{C}$ which are initial and final temperature respectively.  Specific heat of ice is known and the parameters are substituted in the equation $\text{q = m }\times \text{s }\times \Delta \text{T}$, where ‘q’ correlates to the heat utilized in the process of warming ice from ${\text{-15}}^{\text{°}}\text{C}$ to its melting point ${\text{0}}^{\text{°}}\text{C}$.

Calculate the heat required to convert ice to liquid water at its melting point.

$\begin{array}{l}\text{mass of water = 866 g}\\ \text{molar mass of water = 18}\text{.02 g/mol}\\ \text{no}\text{.of moles of water = }\frac{\text{mass of water}}{\text{molar mass of water}}\\ \text{ = }\frac{\text{866 g}}{\text{18}\text{.02 g/mol}}=48.06\text{ mol}\\ \text{we know,}\\ {\text{heat required to convert one mole of ice to liquid water at 0}}^{°}\text{C = 26}\text{.4 kJ/mol}\text{.}\\ \text{therefore,}\\ \text{heat required to convert }48.06{\text{ mol of ice to liquid water at 0}}^{°}\text{C,}\\ {\text{q}}_4\text{ = 6}\text{.0 kJ/mol }\times 48.06\text{ mol = 288}\text{.36 kJ}\end{array}$

Mass of water is converted to number of moles.  The heat required to convert one mole of ice to liquid water at its melting point is known which is termed as molar heat of fusion.  The number of water in given mass of ice (water) is obtained by multiplying with the standard value.

Calculate the heat required to heat water from ${\text{0}}^{\text{°}}\text{C}$ to its boiling point ${\text{100}}^{°}\text{C}$.

$\begin{array}{l}\text{known data: }\\ \text{mass of ice, m = 866 g}\\ \text{specific heat of water, s = 4}\text{.184 J/g }{\text{.}}^{\text{°}}\text{C}\\ {\text{T}}_{\text{1}}{\text{ = 0}}^{°}{\text{C, T}}_{\text{2}}{\text{ = 100}}^{\text{°}}{\text{C ; ΔT = T}}_{\text{2}}{\text{-T}}_{\text{1}}{\text{= 100}}^{\text{°}}{\text{C - 0}}^{\text{°}}{\text{C = 100}}^{\text{°}}\text{C}\\ \text{ heat required to warm ice to its melting point , }\\ {\text{ q}}_3\text{ = m s }\Delta \text{T = 866 g}\times \text{ 4}\text{.184 J/g }{\text{.}}^{\text{°}}\text{C}\times {\text{ 100}}^{\text{°}}\text{C}\\ \text{ = 362}\text{.3 kJ}\end{array}$

Water is heated from its melting point ${\text{0}}^{\text{°}}\text{C}$  to its boiling point ${\text{100}}^{\text{°}}\text{C}$ which are initial and final temperature respectively.  Specific heat of water is known and the parameters are substituted in the equation $\text{q = m }\times \text{s }\times \Delta \text{T}$ where ‘q’ correlates to the heat utilized in the process of heating ice from its melting point ${\text{0}}^{\text{°}}\text{C}$ to its boiling point ${\text{100}}^{\text{°}}\text{C}$.

Calculate the heat required to convert $\text{866 g}$ water to steam at its boiling point.

$\begin{array}{l}\text{mass of water = 866 g}\\ \text{molar mass of water = 18}\text{.02 g/mol}\\ \text{no}\text{.of moles of water = }\frac{\text{mass of water}}{\text{molar mass of water}}\\ \text{ = }\frac{\text{866 g}}{\text{18}\text{.02 g/mol}}=48.06\text{ mol}\end{array}$

$\begin{array}{l}\text{we know,}\\ {\text{heat required to convert one mole of water to steam at 100}}^{°}\text{C = 40}\text{.79 kJ/mol}\text{.}\\ \text{therefore,}\\ \text{heat required to convert 48}{\text{.06 mol of water to steam at 100}}^{°}\text{C,}\\ {\text{q}}_4\text{ = 40}\text{.79 kJ/mol }\times \text{ 48}\text{.06 mol = 1}\text{.96kJ}\end{array}$

Mass of water is converted to number of moles.  The heat required to convert one mole of water to steam at its boiling point is known which is termed as molar heat of vaporization.  The number of water in given mass of ice (water) is obtained by multiplying with the standard value.

Calculate the heat required to raise the temperature of steam from ${\text{100}}^{\text{°}}{\text{C to 146}}^{\text{°}}\text{C}\text{.}$

$\begin{array}{l}\text{known data: }\\ \text{mass of ice, m = 866 g}\\ \text{specific heat of Steam, s = 1}\text{.99 J/g }{\text{.}}^{\text{°}}\text{C}\\ {\text{ T}}_{\text{1}}{\text{ = 100}}^{°}{\text{C, T}}_{\text{2}}{\text{ = 146}}^{\text{°}}{\text{C ; ΔT = T}}_{\text{2}}{\text{-T}}_{\text{1}}{\text{= 146}}^{\text{°}}{\text{C - 100}}^{\text{°}}{\text{C = 46}}^{\text{°}}\text{C}\\ \text{ heat required to warm ice to its melting point , }\\ {\text{ q}}_5\text{ = m s }\Delta \text{T = 866 g}\times \text{ 1}\text{.99 J/g }{\text{.}}^{\text{°}}\text{C}\times {\text{ 46}}^{\text{°}}\text{C}\\ \text{ = 79}\text{.3 kJ}\text{.}\end{array}$

The temperature of steam has to be risen from ${\text{100}}^{\text{°}}\text{C}$  to ${\text{138}}^{\text{°}}\text{C}$ which are initial and final temperature respectively.  Specific heat of steam is known and the parameters are substituted in the equation $\text{q = m }\times \text{s }\times \Delta \text{T}$ where ‘q’ correlates to the heat utilized in the process of rising the temperature of steam.

Calculate the total heat energy utilized to convert $\text{866 g}$ of ice at ${\text{-15}}^{\text{°}}\text{C}$ to steam at ${\text{146}}^{\text{°}}\text{C}$

$\begin{array}{l}{\text{q = q}}_1+{\text{q}}_2+{\text{q}}_3+{\text{q}}_4+{\text{q}}_5\\ \text{q = 26}\text{.4 kJ + 288}\text{.36 kJ + 362}\text{.3 kJ +1}\text{.96 kJ + 79}\text{.3 kJ = 758}\text{.32 kJ}\end{array}$

The total heat energy utilized to convert $\text{866 g}$ of ice at ${\text{-15}}^{\text{°}}\text{C}$ to steam at ${\text{146}}^{\text{°}}\text{C}$ is equivalent to sum of the heat energy required in the processes of required to warm ice to melting point, convert ice to liquid water at its melting point, heating water from ${\text{0}}^{\text{°}}\text{C}$ to its boiling point ${\text{100}}^{°}\text{C}$, converting  water to steam at its boiling point and rising the temperature of steam from ${\text{100}}^{\text{°}}{\text{C to 146}}^{\text{°}}\text{C}\text{.}$

Conclusion

The heat required to convert $\text{866 g}$ of ice at ${\text{-15}}^{\text{°}}\text{C}$ to steam at ${\text{146}}^{\text{°}}\text{C}$ was calculated.