Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, $q$ .

The heat absorbed or released by a substance is calculated by the following equation:

$q=m\times C\times \Delta T$ ……(1)

Here, $m$ is the mass, $\Delta T$ is the temperature change, and $C$ is the specific heat capacity.

The expression to calculate the value of $\Delta T$ :

$\Delta T=T_2-T_1$ …..(2)

Here $T_1$ is the initial temperature and $T_2$ is the final temperature.

Substitute (2) in (1)

$q=m\times C\times \left(T_2-T_1\right)$ ……(3)

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is $-\text{98}\text{.27 kJ}$ .

Explanation of Solution

Given Information: Mass of steam is $32.0\text{ g}$ , the temperature of steam is $115.0°\text{C}$ , the

specific heat of ice is $2.03{\text{ J g}}^{-1}°{\text{C}}^{-1}$ , the heat of fusion of ice is $6.01\times {10}^3{\text{ J mol}}^{-1}$ , the

the specific heat of water is $\text{4}{\text{.18 J g}}^{-1}°{\text{C}}^{-1}$ , the heat of vaporization of water is $4.07\times {10}^4{\text{ J mol}}^{-1}$ and the specific heat of steam is $2.02{\text{ J g}}^{-1}°{\text{C}}^{-1}$ .

Calculate the heat lost on cooling steam from ${115.0}^{°}\text{C}$ to ${100.0}^{°}\text{C}$ .

Using equation (3) to calculate the heat required to cool the temperature of ice from $115°\text{C}$ to $100°\text{C}$ .

Substitute $32\text{ g}$ for $m$ , $2.03{\text{ J g}}^{-1}°{\text{C}}^{-1}$ for $C$ as the specific heat of ice, $100°\text{C}$ for $T_2$ and $115°\text{C}$ for $T_1$ in equation (3):

$\begin{array}{c}{q}_1=32\text{ g}\times 2.03{\text{ J g}}^{-1}°{\text{C}}^{-1}\times \left(100°\text{C}-115°\text{C}\right)\\ \text{=}-969.6\text{ J}\end{array}$

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, $n$ .

$\begin{array}{c}{q}_2=n\times \left(-\text{ molar heat of vaporization}\right)\\ \text{=}\left(\frac{\text{32 g}}{\text{18}{\text{.02 g mol}}^{-1}}\right)\times \left(-4.07\times {10}^4{\text{J mol}}^{-1}\right)\\ =-72275.25\text{ J}\end{array}$

Using equation (3) to calculate the heat required to cool the temperature of water from $100°\text{C}$ to $0°\text{C}$ .

Substitute $32\text{ g}$ for $m$ , $\text{4}{\text{.18 J g}}^{-1}°{\text{C}}^{-1}$ for $C$ as the specific heat of ice, $0.0°\text{C}$ for $T_2$ and $100°\text{C}$ for $T_1$ in equation (3):

$\begin{array}{c}{q}_3=32\text{ g}\times 4.18{\text{ J g}}^{-1}°{\text{C}}^{-1}\times \left(0.0°\text{C}-100.0°\text{C }\right)\\ \text{=}-\text{13376 J}\end{array}$

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

$\begin{array}{c}{q}_4=n\times \left(-\text{ molar heat of fusion}\right)\\ \text{=}\left(\frac{\text{32 g}}{\text{18}{\text{.02 g mol}}^{-1}}\right)\times \left(-6.01\times {10}^3{\text{ J mol}}^{-1}\right)\\ =-10672.59\text{ J}\end{array}$

Using equation (3) to calculate the heat required to cool the temperature of ice from $0°\text{C}$ to $-15°\text{C}$ .

Substitute $32\text{ g}$ for $m$ , $\text{2}{\text{.03 J g}}^{-1}°{\text{C}}^{-1}$ for $C$ as the specific heat of ice, $-15°\text{C}$ for $T_2$ and $0°\text{C}$ for $T_1$ in equation (3):

$\begin{array}{c}{q}_5=32\text{ g}\times 2.03{\text{ J g}}^{-1}°{\text{C}}^{-1}\times \left(\left(-15.0°\text{C}\right)-0.0°\text{C}\right)\\ \text{=}-\text{ 974}\text{.4 J}\end{array}$

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

$\begin{array}{c}{q}_{total}=q_1+q_2+q_3+q_4+q_5\\ \text{= }-969.6\text{ J }-72275.25\text{ J }-13376\text{ J }-10672.59\text{ J }-974.4\text{ J}\\ \text{= }-98267.84\text{ J}\\ \text{= }-\text{9}8.27\text{ kJ}\end{array}$

The total heat released in the process is $-\text{9}8.27\text{ kJ}$ .

Conclusion

The total heat released in the process is $-\text{9}8.27\text{ kJ}$ .