Chapter 12, Problem 12.29P

Interpretation Introduction

Interpretation:

The heat that is required to convert $2.500\text{ kg}$ of solid ${\text{SO}}_2$ at the melting point to gas at $60°\text{C}$ is to be calculated.

Concept introduction:

Specific heat capacity $\left(c\right)$ of a substance is the amount of heat needed to raise the temperature of $1\text{g}$ of a substance by $1\text{K}$. The formula to calculate heat required is as follows:

$q=\left(\text{mass}\right)\left(c\right)\left(T_2-T_1\right)$ (1)

Here,

$T_2$ is the final temperature.

$T_1$ is the initial temperature.

$q$ is the heat released or absorbed.

$c$ is the specific heat capacity of the substance.

The formula to calculate heat at phase change is as follows:

$q=\left(\text{mol}\right)\left(\Delta H\right)$ (2)

Here,

$\Delta H$ is the enthalpy change.

Answer to Problem 12.29P

The heat that is required to convert $2.500\text{ kg}$ of solid ${\text{SO}}_2$ at the melting point to gas at $60°\text{C}$ is $1.606\times {10}^6\text{J}$.

Explanation of Solution

The formula to calculate the number of moles of ${\text{SO}}_2$ is as follows:

$\text{Number}\text{of moles}=\left(\frac{\text{given}\text{mass}}{\text{molecular}\text{mass}\text{of}{\text{SO}}_2}\right)$ (3)

Substitute $2.500\text{ kg}$ for a given mass and $\mathrm{64.06.02}\text{g/mol}$ for molecular mass of ${\text{SO}}_2$ in the equation (3).

$\begin{array}{c}\text{Number}\text{of moles}=\left(\frac{2.500\text{ kg}}{64.06\text{g/mol}}\right)\left(\frac{1000\text{g}}{1\text{kg}}\right)\\ =39.0259132\text{mol}\end{array}$

Substitute $39.0259132\text{mol}$ for $\text{mol}$ and $8.619\text{kJ/mol}$ for $\Delta H_{\text{fus}}^{°}$ in the equation (2) to change the phase of solid ${\text{SO}}_2$ at $-73°\text{C}$ to liquid ${\text{SO}}_2$ at $-73°\text{C}$.

$\begin{array}{c}{q}_1=\left(39.0259132\text{mol}\right)\left(8.619\text{kJ/mol}\right)\\ =336.364346\text{kJ}\left(\frac{1000\text{J}}{1\text{kJ}}\right)\\ =336364.3\text{J}\end{array}$

Substitute $2.500\text{ kg}$ for mass, $0.995\text{J/g}\cdot °\text{C}$ for $c_{\text{liquid}}$, $-10°\text{C}$ for $T_2$ and $-73°\text{C}$ for $T_1$ in the equation (1) to calculate the heat required to increase the temperature of liquid ${\text{SO}}_2$ from $-73°\text{C}$ to $-10°\text{C}$.

$\begin{array}{c}{q}_2=\left(2.500\text{ kg}\right)\left(\frac{1000\text{g}}{\text{1}\text{kg}}\right)\left(0.995\text{J/g}\cdot °\text{C}\right)\left(-10°\text{C}-\left(-73°\text{C}\right)\right)\\ =156712.5\text{J}\end{array}$

Substitute $39.0259132\text{mol}$ for $\text{mol}$ and $25.73\text{kJ/mol}$ for $\Delta H_{\text{vap}}$ in the equation (2) to change the phase of liquid ${\text{SO}}_2$ at $-10°\text{C}$ to gas ${\text{SO}}_2$ at $-10°\text{C}$.

$\begin{array}{c}{q}_3=\left(39.0259132\text{mol}\right)\left(25.73\text{kJ/mol}\right)\\ =1004.13675\text{kJ}\left(\frac{1000\text{J}}{1\text{kJ}}\right)\\ =1004136.75\text{J}\end{array}$

Substitute $2.500\text{ kg}$ for mass, $0.622\text{J/g}\cdot °\text{C}$ for $c_{\text{gas}}$, $60°\text{C}$ for $T_2$ and $-10°\text{C}$ for $T_1$ in the equation (1) to calculate the heat required to increase the temperature of water from $-10°\text{C}$ to $60°\text{C}$.

$\begin{array}{c}{q}_4=\left(2.500\text{ kg}\right)\left(\frac{1000\text{g}}{\text{1}\text{kg}}\right)\left(0.622\text{J/g}\cdot °\text{C}\right)\left(60°\text{C}-\left(-10°\text{C}\right)\right)\\ =108850\text{J}\end{array}$

The formula to calculate the total heat required is as follows:

$q_{\text{total}}=q_1+q_2+q_3+q_4$ (4)

Substitute $336364.3\text{J}$ for $q_1$, $156712.5\text{J}$ for $q_2$, $004136.75\text{J}$ for $q_3$ and $108850\text{J}$ for $q_4$ in the equation (4).

$\begin{array}{c}{q}_{\text{total}}=336364.3\text{J}+156712.5\text{J}+004136.75\text{J}+108850\text{J}\\ =1,606,063.5\text{J}\\ \approx 1.606\times {10}^6\text{J}\end{array}$

Conclusion

The heat that is required to convert $2.500\text{ kg}$ of solid ${\text{SO}}_2$ at the melting point to gas at $60°\text{C}$ is $1.606\times {10}^6\text{J}$.