Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $\text{146°C}$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=sm\Delta T$

Here, $q$ is the amount of heat, $s$ is the specific heat, $\Delta T$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$\Delta T=T_{\text{final}}-T_{\text{initial}}$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=n\Delta H_{\text{vap}}$

Here, $n$ is the number of moles and $\Delta H_{\text{vap}}$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=n\Delta H_{\text{fus}}$

Here, $n$ is the number of moles and $\Delta H_{\text{fus}}$ is the molar heat of fusion.

Answer to Problem 85QP

Solution: $\text{2}\text{.72}\times {\text{10}}^{\text{3}}\text{kJ}$

Explanation of Solution

Given information: Specific heat of ice is $\text{2}\text{.03 }\text{J}/\text{g}{.}^{\circ }\text{C}$. Specific heat of steam is $\text{1}\text{.99}\text{J}/\text{g}{.}^{\circ }\text{C}$ and $m=\text{866g}$. The initial temperature is–15°C. Final temperature is $\text{146°C}$

Step $1$:The melting point of ice is $\text{0}°\text{C}$

The energy required to warm ice from –15°Cto $\text{0}°\text{C}$ is calculated using the equation as follows:

$q=sm\Delta T$

Calculate the change in temperatureby using the equation as follows:

$\Delta T=T_{\text{final}}-T_{\text{initial}}$

By substituting values in the above equation as follows:

$\begin{array}{c}\Delta T={\left(0-\left(-15\right)\right)}^{\circ }\text{C}\\ ={15}^{\circ }\text{C}\end{array}$

The specific heat $\left(s\right)$ of ice is $\text{2}\text{.03 }\text{J}/\text{g}{.}^{\circ }\text{C}$.

Substitute the required values in the above equation as follows:

$\begin{array}{c}q=sm\Delta T\\ \text{ =2}\text{.03 }\frac{\text{J}}{\cancel{\text{g}}.°\cancel{\text{C}}}\times 866\cancel{\text{g}}\times \text{15}°\cancel{\text{C}}\\ \text{=2}\text{.637}\times {\text{10}}^{\text{4}}\cancel{\text{J}}\times \frac{\text{1kJ}}{\text{1000 }\cancel{\text{J}}}\\ =26.4\text{ kJ}\\ \end{array}$

Step $2$:The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=n\Delta H_{\text{fus}}$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$\begin{array}{c}\text{Number of moles of water }\left(n\right)\text{=}\frac{\text{866 }\cancel{\text{g}}}{\text{18}\text{.02}\cancel{\text{g}}/\text{mol}}\\ =48.06\text{ mol}\end{array}$

Substitute $48.06\text{ mol}$ for $n$ and $6.01\text{kJ}/\text{mol}$ for $\Delta H_{fus}$ in the above equationto calculate heat energy as follows:

$\begin{array}{c}q=n\Delta H_{\text{fus}}\\ =\left(\text{48}\text{.06 }\cancel{\text{mol}}\right)\left(6.01\text{kJ}/\cancel{\text{mol}}\right)\\ \approx 289\text{ kJ}\end{array}$

Step $3$: The energy required to heat water from $\text{0}°\text{C}$ to $\text{100°C}$ is calculated using the equation as follows:

$q=sm\Delta T$

Calculate the change in temperatureby using the equation as follows:

$\Delta T=T_{\text{final}}-T_{\text{initial}}$

By substituting the values in the above equation as follows:

$\begin{array}{c}\Delta T={\left(100-0\right)}^{\circ }\text{C}\\ ={100}^{\circ }\text{C}\end{array}$

The specific heat of water is $\text{4}\text{.184 }\text{J}/\text{g}{.}^{\circ }\text{C}$.

Substitute the required values in the equation as follows:

$\begin{array}{c}q=sm\Delta T\\ \text{ =4}\text{.184 }\frac{\text{J}}{\cancel{\text{g}}.°\cancel{\text{C}}}\times 866\cancel{\text{g}}\times \text{100}°\cancel{\text{C}}\\ =\text{3}\text{.623}\times {\text{10}}^{\text{5}}\cancel{\text{J}}\times \frac{\text{1kJ}}{\text{1000 }\cancel{\text{J}}}\\ \approx 362.3\text{ kJ}\\ \end{array}$

Step $4$: The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=n\Delta H_{\text{fus}}$

Substitute $48.06\text{ mol}$ for $n$ and $40.79\text{kJ}/\text{mol}$ for $\Delta H_{\text{vap}}$ in the equation to calculate the heat energy as follows:

$\begin{array}{c}q=n\Delta H_{\text{vap}}\\ =\left(\text{48}\text{.06 }\cancel{\text{mol}}\right)\left(40.79\text{kJ}/\cancel{\text{mol}}\right)\\ =1960.4\text{ kJ}\end{array}$

Step $5$: The energy required to heat steam from $\text{100°C}$ to $\text{146°C}$ is calculated using the equation as follows:

$q=sm\Delta T$

Calculate the change in temperatureby using the equation as follows:

$\Delta T=T_{\text{final}}-T_{\text{initial}}$

By substituting the values in the above equation as follows:

$\begin{array}{c}\Delta T={\left(146-100\right)}^{\circ }\text{C}\\ ={46}^{\circ }\text{C}\end{array}$

The specific heat of steam is $\text{1}\text{.99}\text{J}/\text{g}{.}^{\circ }\text{C}$.

Substitute the required values in the equation as follows:

$\begin{array}{c}q=sm\Delta T\\ \text{ =1}\text{.99 }\frac{\text{J}}{\cancel{\text{g}}.°\cancel{\text{C}}}\times 866\cancel{\text{g}}\times \text{46}°\cancel{\text{C}}\\ =\text{7}\text{.927}\times {\text{10}}^{\text{4}}\cancel{\text{J}}\times \frac{\text{1kJ}}{\text{1000 }\cancel{\text{J}}}\\ \approx 79.3\text{ kJ}\\ \end{array}$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$\begin{array}{c}\text{Total energy }=26.4\text{ kJ}+288.5\text{ kJ}+362.3\text{ kJ}+1960.4\text{ kJ}+79.3\text{ kJ}\\ =2716.9\text{ kJ}\\ =\text{2}\text{.72}\times {\text{10}}^{\text{3}}\text{kJ}\end{array}$

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $\text{146°C}$ is $\text{2}\text{.72}\times {\text{10}}^{\text{3}}\text{kJ}$.