Chapter 23, Problem 4PEB

A parcel of air with a volume of 9.1 $\times$ 10⁴ km3 that contains 5.7 $\times$ 10⁷ kg of water vapor rises to an altitude where all the water in the parcel condenses and then freezes. What is the change in temperature of the parcel of air due to freezing? (Assume the density of the air at the condensation altitude is 7.2 $\times$ 10² g/m³.)

To determine

The change in temperature of the parcel of air due to freezing.

Answer to Problem 4PEB

Solution:

$\underset{\_}{3.17\times {10}^{-3}°\text{C}}$.

Explanation of Solution

Given data:

The volume of parcel of air containing $5.7\times {10}^7\text{ kg}$ water vapour is $9.1\times {10}^4{\text{ km}}^3$. The density of the air at the condensation altitude is $7.2\times {10}^2{\text{ g/m}}^3$.

Formula used:

Write the expression for thermal energy absorbed or released during the phase change of the substance.

$Q=mL$ …...…... (1)

Here, $Q$ denotes thermal energy or heat, $m$ and $L_v$ represents mass and latent heat for the substance.

Write the expression for density.

$\rho =\frac{m}{V}$

Here, $V$ is the volume.

Rearrange the expression for density in terms of mass.

$m=\rho V$ …...…... (2)

Write the expression for sensible heat.

$Q_{\text{sensible heat}}=mc\Delta T$ …...…... (3)

Here, $c$ represents specific heat capacity and $\Delta T$ represents the change in temperature.

Compare equation (2) and (3) and write the expression for sensible heat in terms of volume and density.

$Q_{\text{sensible heat}}=\rho Vc\Delta T$ …...…... (4)

Explanation:

The parcel of air contains water vapour. At the condensation altitude, two stages of phase change occur. First, the water vapour condenses and change to liquid, and then it undergoes another phase change and freezes. The heat released from the two-phase change is absorbed by the surrounding air. The solution can be obtained in three steps.

Step 1: Consider the condition for the phase change when the water vapour undergoes condensation. The latent heat for vapourization of water vapour is $540\text{ cal/g}$. Calculate the value of released thermal energy during condensation.

$Q_{\text{condensation}}=m{L}_v$

Here, $L_v$ is the latent heat of vapourization of water vapour.

Substitute $5.7\times {10}^7\text{ kg}$ for $m$ and $540\text{ cal/g}$ for $L_v$.

$\begin{array}{ccc}\hfill Q_{\text{condensation}}& =& \left(5.7\times {10}^7\text{ kg}\right)\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\left(540\text{ cal/g}\right)\hfill \\ \hfill & =& 30.78\times {10}^{12}\text{ cal}\hfill \end{array}$

Step 2: Consider the condition for the phase change when the condensed water vapour freezes further. The latent heat of fusion for water is $80\text{ cal/g}$. Calculate the value of released thermal energy during freezing.

$Q_{\text{freezing}}=m{L}_f$

Substitute $5.7\times {10}^7\text{ kg}$ for $m$ and $80\text{ cal/g}$ for $L_f$.

$\begin{array}{ccc}\hfill Q_{\text{freezing}}& =& \left(5.7\times {10}^7\text{ kg}\right)\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\left(80\text{ cal/g}\right)\hfill \\ \hfill & =& 4.56\times {10}^{12}\text{ cal}\hfill \end{array}$

Step 3: The heat released from the two-phase change is absorbed by the surrounding air. Therefore, the total heat released during the two-phase change is equal to the sensible heat. Calculate the total heat released during the phase change from water vapour to ice.

$Q=Q_{\text{condensation}}+Q_{\text{freezing}}$

Substitute $30.78\times {10}^{12}\text{ cal}$ for $Q_{\text{condensation}}$, $4.56\times {10}^{12}\text{ cal}$ for $Q_{\text{freezing}}$.

$\begin{array}{ccc}\hfill Q& =& 30.78\times {10}^{12}\text{ cal}+4.56\times {10}^{12}\text{ cal}\hfill \\ \hfill & =& 35.34\times {10}^{12}\text{ cal}\hfill \end{array}$

This heat is equal to the sensible heat.

Therefore, calculate the change in temperature of the air parcel due to freezing.

Substitute $7.2\times {10}^2{\text{ g/m}}^{\text{3}}$ for $\rho$, $9.1\times {10}^4{\text{ km}}^3$ for $V$, $0.17\text{ cal/g}\cdot °\text{C}$ for $c$, and $35.34\times {10}^{12}\text{ cal}$ for $Q_{\text{sensible heat}}$ in the equation (4).

$\begin{array}{ccc}\hfill 35.34\times {10}^{12}\text{ cal}& =& \left(7.2\times {10}^2{\text{ g/m}}^{\text{3}}\right)\left(9.1\times {10}^4{\text{ km}}^3\right){\left(\frac{1000\text{ m}}{1\text{ km}}\right)}^3\left(0.17\text{ cal/g}\cdot °\text{C}\right)\Delta T\hfill \\ \hfill 35.34\times {10}^{12}\text{ cal}& =& \left(1.11384\times {10}^{16}\text{ cal/}°\text{C}\right)\Delta T\hfill \\ \hfill \Delta T& =& \frac{35.34\times {10}^{12}\text{ cal}}{1.11384\times {10}^{16}\text{ cal/}°\text{C}}\hfill \\ \hfill & =& 3.17\times {10}^{-3}°\text{C}\hfill \end{array}$

Conclusion:

The change in temperature due to freezing is $\underset{\_}{3.17\times {10}^{-3}°\text{C}}$.