Question

One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 µm. Forthat range, give the lower value and the higher value, respectively, for the following. (a) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.0 km and radius 1.0 km? (b) How many 1-liter pop bottles would that water fill? (c) Water has a density of$\mathbf{1000}\mathbf{}\mathbf{kg}\mathbf{}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. How much mass does the water in the cloud have?

Short answer

(a) The total volume of water in the cloud is from $2.0\times {10}^3{\mathrm{m}}^3\mathrm{to}2.0\times {10}^4\mathrm{m}3$.

(b) The number of bottles can be fill is $2.0\times {10}^6\mathrm{to}2.0\times {10}^7$.

(c) The mass of water in the cloud is from$2.0\times {10}^6\mathrm{kg}\mathrm{to}2.0\times {10}^7\mathrm{kg}$

Step-by-step solution

Given data

Radius of a drop of water,$r=10\mu \mathrm{m}$

Height of the cloud,$h=3.0\mathrm{km}$

Radius of the cloud, $R=1.0\mathrm{km}$

Water drops in one cubic centimeter of cumulus cloud are $50\mathrm{to}500$.

Density of water, $p=1000\mathrm{kg}/{\mathrm{m}}^3$.

Understanding the density of a material

The density of a material (in this case water) is defined as the mass per unit volume.

The expression for density is given as:

$p=\frac{m}{V}$ … (i)

Here, $p$is the density, $m$is the mass and $V$is the volume.

(a) Determination of the volume of water in the cloud

The volume of the cloud is calculated as:

$V={\mathrm{\pi R}}^{2}h$

$$\begin{gathered} =3.14\times {\left(1.0\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\right)}^2\times \left(3.0\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\right) \\ =9.42\times {10}^9\times {\mathrm{m}}^3 \end{gathered}$$

Since $1{\mathrm{cm}}^3={10}^{‐6}{\mathrm{m}}^3$and the number of water droplets in one cubic cm are 50 to 500 therefore, the number of water droplets in one cubit meter are $50\times {10}^6$to $500\times {10}^6$.

So, the number of total water droplets in entire cloud are,

$\left(9.42\times {10}^9{\mathrm{m}}^3\right)\times \left(50\times {10}^6\right)=4.71\times {10}^{17}$to $\left(9.42\times {10}^9{\mathrm{m}}^3\right)\times \left(500\times {10}^6\right)=4.71\times {10}^{18}$

drops.

The volume of each drop is calculated as:

role="math" localid="1654517265789" $$\begin{gathered} V_{\mathrm{drop}}=\frac{4}{3}{\mathrm{\pi r}}^3 \\ =\frac{4}{\mathrm{\beta m}}\times \mathrm{\beta m}14\times \left(10\times \frac{{10}^{‐6}\mathrm{m}}{1}\right) \\ =4.20\times {10}^{‐15}{\mathrm{m}}^3 \end{gathered}$$

Now, multiply the volume of the each drop with the number of drops to find the total volume of the water in the cloud.

role="math" localid="1654517499825" $$\begin{gathered} V_{\mathrm{water}}=4.20\times {10}^{‐15}{\mathrm{m}}^3\times =4.71\times {10}^{17} \\ \approx 2.0\times {10}^3{\mathrm{m}}^3 \end{gathered}$$

Or,

$$\begin{gathered} V_{\mathrm{water}}=4.20\times {10}^{‐15}{\mathrm{m}}^3\times =4.71\times {10}^{18} \\ \approx 2.0\times {10}^4{\mathrm{m}}^3 \end{gathered}$$

Thus, the total volume of water in the cloud is from$2.0\times {10}^3{\mathrm{m}}^3$to$2.0\times {10}^4{\mathrm{m}}^3$.

(b) Determination of the number of bottles filled

Since, $1\mathrm{L}={10}^3{\mathrm{cm}}^3\mathrm{or}{10}^{‐3}{\mathrm{m}}^3$therefore, the number of bottles filled will be,

$$\begin{gathered} N=\frac{2.0\times {10}^3{\mathrm{m}}^3}{{10}^{‐3}{\mathrm{m}}^3} \\ =2.0\times {10}^6 \end{gathered}$$

Or,

$$\begin{gathered} N=\frac{2.0\times {10}^4{\mathrm{m}}^3}{{10}^{‐3}{\mathrm{m}}^3} \\ =2.0\times {10}^7 \end{gathered}$$

Thus, the number of bottles filled are $2.0\times {10}^6$ to$2.0\times {10}^7$.

(c) Determination of the mass of water

Using equation (i), the mass of water is given as:

$m=V\times p$

Substitute the values in the above expression.

role="math" localid="1654518531625" $$\begin{gathered} m=2.0\times {10}^3{\mathrm{m}}^3\times {10}^3\mathrm{kg}/{\mathrm{m}}^3 \\ =2.0\times {10}^6\mathrm{kg} \end{gathered}$$

Or,

$$\begin{gathered} m=2.0\times {10}^4{\mathrm{m}}^3\times {10}^3\mathrm{kg}/{\mathrm{m}}^3 \\ =2.0\times {10}^7\mathrm{kg} \end{gathered}$$

Thus, the mass of the water in cloud is from $2.0\times {10}^6\mathrm{kg}$to $2.0\times {10}^7\mathrm{kg}$.