Q26P - Fundamentals Of Physics

Question

One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 µm. For that 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?

Step-by-Step Solution

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 1: 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, $p=1000 \mathrm{kg} / {\mathrm{m}}^3$.

Step 2: 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.

Step 3: (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: 

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

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

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

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 $2.0\times {10}^4 {\mathrm{m}}^3$.

Step 4: (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  $2.0\times {10}^7$.

Step 5: (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.

$$\begin{gathered} m = 2.0\times {10}^3 {\mathrm{m}}^3 \times {10}^{3 }\mathrm{kg}/{\mathrm{m}}^3 \\ = 2.0\times {10}^6 \mathrm{kg} \end{gathered}$$ 

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 $2.0\times {10}^7 \mathrm{kg}$.